The coefficient of kinetic friction that would allow the block to return to equilibrium and stop is given by (kx)/mg.
What coefficient of kinetic friction?
Let's analyze the motion of the block in order to determine the coefficient of kinetic friction that would allow the block to return to equilibrium and stop.
First, let's consider the forces acting on the block when it is pulled from equilibrium and released. Initially, when the block is pulled a distance x from equilibrium, the only force acting on it is the force exerted by the spring, which is given by Hooke's Law:
F_spring = -kx
where;
k is the spring constant and x is the displacement of the block from its equilibrium position.When the block is released, it will experience an acceleration due to the force exerted by the spring. As it moves towards equilibrium, the spring force will decrease until it becomes zero when the block reaches the equilibrium position. At this point, the only force acting on the block will be the kinetic friction force, which opposes the motion of the block.
The equation of motion for the block can be given by Newton's second law:
F_net = ma
where;
F_net is the net force acting on the block, m is the mass of the block, and a is its acceleration.Since the only force acting on the block is the kinetic friction force when it reaches equilibrium, we can equate the net force with the kinetic friction force:
F_friction = ma
where;
F_friction is the kinetic friction force.Now, let's express the kinetic friction force in terms of the coefficient of kinetic friction μ_k and the normal force N. The normal force is the force exerted by the surface on the block and is equal in magnitude to the weight of the block, which is given by:
N = mg
where;
g is the acceleration due to gravity.The kinetic friction force can be expressed as:
F_friction = μ_kN
Substituting the expression for N, we get:
F_friction = μ_kmg
Now, equating this with ma, we have:
μ_kmg = ma
Canceling the mass from both sides, we get:
μ_kg = a
Now, we can substitute the expression for acceleration a with the acceleration due to the spring force using Hooke's Law:
μ_kg = (kx)/m
Finally, solving for the coefficient of kinetic friction μ_k, we get:
μ_k = (kx)/mg
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10J of energy is transformed when a A) Mass of 10kg moves a distance of 1m B) Mass of 10kg is given a velocity of 10m/s C) Force of 10N acts for 10se D) Force of 10N displaced a body through a distance of 1m.
The correct option is D) Force of 10N displaced a body through a distance of 1m.
This is because work is defined as the product of force and displacement in the direction of force. Mathematically, work (W) = force (F) x displacement (d) x cos(theta), where theta is the angle between the force vector and the displacement vector.WHY ABC INCORRECT ? In option A, the work done would be W = force x displacement x cos(theta) = 0, since there is no force acting on the object.In option B, the work done would be W = (1/2) x mass x velocity^2 = 500 J, which is not equal to 10 J.In option C, the work done would be W = force x displacement x cos(theta) = 100 J, which is also not equal to 10 J.In option D, the work done would be W = force x displacement x cos(theta) = 10 x 1 x cos(0) = 10 J, which is the given amount of energy. Therefore, option D is the correct answer.
What is the significance of a intersection of the red and green curves
The significance of a intersection of the red and green curves (B). Consumption becomes greater than supply is the correct option.
The context in which two curves are being studied determines the importance of their junction, regardless of whether they are red and green or any other colour. Here are a few potential meanings:
The intersection point represents the solution or point of similarity between the two functions, if the red and green curves represent mathematical functions or equations. Or, to put it another way, the variables or parameters that make the two functions equal. Problems or equations involving the two functions can be solved by locating the intersection point .
Graphical Analysis: The intersection point on a graph of the red and green curves represents the point at which the two curves cross each other. These ideas could be used to analyse the relationship between the two curves, such as locating crucial spots, pinpointing sites of convergence or divergence, calculating distances or relative locations, etc.
Red and green curves may occasionally have symbolic meanings based on their colours. Red, for instance, may be a symbol of peril, ardor, or heat, whereas green may be a symbol of growth, serenity, or nature. Depending on the situation, the meeting or conflict of these attributes could be represented by the intersection of these curves.
Therefore, the correct option is (B).
What is the significance of the intersection of the red and green curves?
A. Supply becomes greater than consumption
B. Consumption becomes greater than supply
C. Supply and consumption are both zero.
D. Consumption increases but supply remains stable,
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Josh was blowing bubbles through a straw into a glass of tap water. As Josh exhaled into the straw, he blew carbon dioxide gas into the water. The harder he blew, the more bubbles entered the water. The more bubbles, the ______________ the solution gas/water solution.
Responses
Josh was blowing bubbles through a straw into a glass of tap water. As Josh exhaled into the straw, he blew carbon dioxide gas into the water. The harder he blew, the more bubbles entered the water. The more bubbles, the more saturated the solution gas/water solution.
This is because the bubbles increase surface area of gas-water Interface, allowing for more gas molecules to come into contact with water and dissolve. As more gas dissolves, concentration of the gas in the water increases, and the solution becomes more saturated . This effect is known as Henry's law, which states that amount of gas that dissolves in a liquid is directly proportional to the partial pressure of the gas above liquid.
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Four point masses 2kg, 4kg, 6kg and are placed at the corners of Square ABCD of 2cm long respectively. Find the Position of centre of mass of the system from the corner
To find the position of the center of mass of the system, we need to first calculate the coordinates of the center of mass along both the x-axis and y-axis.
Let's begin by finding the coordinates of the center of mass along the x-axis:
We can use the formula:
x_cm = (m1x1 + m2x2 + m3x3 + m4x4) / (m1 + m2 + m3 + m4)
where m1, m2, m3, and m4 are the masses of the point masses at each corner, and x1, x2, x3, and x4 are the x-coordinates of each point mass.
In this case, we have:
m1 = 2kg, x1 = 0cm
m2 = 4kg, x2 = 2cm
m3 = 6kg, x3 = 2cm
m4 = 4kg, x4 = 0cm
Substituting these values into the formula, we get:
x_cm = (2kg x 0cm + 4kg x 2cm + 6kg x 2cm + 4kg x 0cm) / (2kg + 4kg + 6kg + 4kg)
x_cm = 24 / 16
x_cm = 1.5cm
Therefore, the x-coordinate of the center of mass is 1.5cm.
Now let's find the coordinates of the center of mass along the y-axis:
We can use the formula:
y_cm = (m1y1 + m2y2 + m3y3 + m4y4) / (m1 + m2 + m3 + m4)
where m1, m2, m3, and m4 are the masses of the point masses at each corner, and y1, y2, y3, and y4 are the y-coordinates of each point mass.
In this case, we have:
m1 = 2kg, y1 = 0cm
m2 = 4kg, y2 = 0cm
m3 = 6kg, y3 = 2cm
m4 = 4kg, y4 = 2cm
Substituting these values into the formula, we get:
y_cm = (2kg x 0cm + 4kg x 0cm + 6kg x 2cm + 4kg x 2cm) / (2kg + 4kg + 6kg + 4kg)
y_cm = 20 / 16
y_cm = 1.25cm
Therefore, the y-coordinate of the center of mass is 1.25cm.
So the center of mass of the system is located at the point (1.5cm, 1.25cm) from the corner.
If each galaxy is 150 kpc across, how long does the event last?
In a galaxy collision, two similar-sized galaxies pass through each other with a combined relative velocity of 1200 km/s
If each galaxy is 150 kpc across, the event lasts for approximately 7.7 billion years.
How long does the event last?To calculate the time for the event, we need to know the distance traveled by the galaxies during the collision. Since each galaxy is 150 kpc across and they are passing through each other, the total distance traveled is twice the diameter of one galaxy, or 300 kpc.
To find the time, we can use the formula:
time = distance / speed
where distance is 300 kpc and speed is 1200 km/s. However, we need to convert the distance to the same units as the speed, so let's convert kpc to km:
1 kpc = 3.086 × 10^16 meters
1 meter = 3.281 × 10^-6 miles
1 mile = 1.609 × 10^3 meters
1 kpc = 3.086 × 10^19 miles
So, 300 kpc is equal to:
300 kpc × 3.086 × 10^19 miles/kpc = 9.258 × 10^21 miles
Now we can plug in the values:
time = distance / speed
time = 9.258 × 10^21 miles / 1200 km/s
time ≈ 2.432 × 10^14 seconds
So the event lasts for approximately 2.432 × 10^14 seconds. To convert to a more meaningful unit, we can divide by the number of seconds in a year:
time ≈ 7.7 billion years
Therefore, the event lasts for approximately 7.7 billion years.
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A 150kg of rectangular tank of dimension 20m X 10m X 15m is placed on a level floor. What maximum and minimum pressure dose the tank exerts on the floor? Take g = 10 m/s^2
Explanation:
Weight of tank = 150 kg * 10 m/s^2 = 1500 N
since we are not given which of the dimensions are the bottom of the tank
smallest area would be 10 x 15 = 150 m^2
largest area would be 20 x 15 = 300 m^2
Max pressure would then be 1500 N / 150m^2 = 10 pascal
Min pressure would be 1500 N / 300 m^2 = 5 pascal
If |Qh| is the total energy that enters the system by heat in one cycle, find the ratio |Qh|/(PiVi)
If |Qh| is the total energy that enters the system by heat in one cycle.
The ratio of |Qh|/(PiVi) is a measure of the efficiency of a heat engine, where |Qh| is the total energy that enters the system by heat in one cycle, Pi is the initial pressure, and Vi is the initial volume of the system.
In a heat engine, energy is transferred from a high-temperature reservoir (|Qh|) to a low-temperature reservoir (|Qc|) to perform work. The efficiency of the heat engine is given by the ratio of the work output (W) to the heat input from the high-temperature reservoir (|Qh|), which can be written as
Efficiency = W/|Qh|
Using the first law of thermodynamics, we can relate the work output to the difference between the heat input and the heat output.
W = |Qh| - |Qc|
Substituting this into the efficiency equation, we get
Efficiency = (|Qh| - |Qc|)/|Qh|
Rearranging this expression, we get
Efficiency = 1 - |Qc|/|Qh|
The quantity |Qc|/|Qh| is known as the heat rejection ratio, which is the ratio of the heat output to the heat input. Since energy cannot be created or destroyed, the total energy entering the system by heat (|Qh|) is equal to the sum of the work done by the system (W) and the energy rejected as heat (|Qc|) we get
|Qh| = W + |Qc|
Substituting this into the efficiency equation, we get
Efficiency = W/(W + |Qc|)
We can also write the work output as the product of the pressure and volume change.
W = PiVi - PfVf
Where Pi and Vi are the initial pressure and volume, and Pf and Vf are the final pressure and volume. Substituting this into the efficiency equation and simplifying, we get
Efficiency = (PiVi - PfVf)/(PiVi)
Rearranging this expression, we get
(PiVi - PfVf)/(PiVi) = |Qh|/(PiVi + |Qc|)
Since the heat engine operates in a cycle, the final volume and pressure are the same as the initial volume and pressure, and |Qc| is equal to zero. Therefore, we can simplify the expression to
(PiVi - PiVi)/(PiVi) = |Qh|/(PiVi)
Which simplifies further to
Efficiency = 1 - (|Qc|/|Qh|) = 1 - 0 = 1
Hence, the maximum efficiency of a heat engine is 1, which is achieved when all the energy transferred by heat is converted into work. In other words, the ratio of |Qh|/(PiVi) represents the maximum theoretical efficiency of a heat engine, which is known as the Carnot efficiency.
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a box of mass 8.0 kg rests on a horizontal rough surface. a string attached to the box passes over a smooth pulley and supports a 2.0 kg mass at its other end, when the box is released, a frictional force of 6.0 n acts on it. (a) draw free-body diagrams of the 8.0 kg box and the 2.0 kg mass. (b) determine the (i) acceleration of the system, (in) tension in the string.
a. The free-body diagrams of the 8.0 kg box and the 2.0 kg mass is attached below.
bi) the acceleration of the system is 0.4 m/s^2
bii) the tension in the string is 0.8 N.
How do we calculate?Net force = ma
Net force = T - F_friction
where T is the tension force and F_friction is the frictional force.
Substituting the given values, we have:
T - F_friction = ma
T - 6.0 N = (8.0 kg + 2.0 kg) a (since the tension force acts on both masses)
T - 6.0 N = 10.0 kg a
(ii) To find the tension in the string, we can apply Newton's second law of motion to the 2.0 kg mass:
Net force = ma
where m is the mass of the mass and a is the acceleration of the system. The net force on the mass is the tension force:
Net force = T
Substituting the given values, we get:
T = ma
T = 2.0 kg x a
To solve for T, we need to find the acceleration of the system.
To find both a and T, we can use the equations we derived above and solve them simultaneously:
T - 6.0 N = 10.0 kg a (equation 1)
T = 2.0 kg x a (equation 2)
Substituting equation 2 into equation 1, we get:
2.0 kg x a - 6.0 N = 10.0 kg a
Solving for a, we get:
a = 0.4 m/s^2
Substituting a into equation 2, we get:
T = 2.0 kg x 0.4 m/s^2 = 0.8 N
Therefore, the acceleration of the system is 0.4 m/s^2 and the tension in the string is 0.8 N. of the system is 0.4 m/s^2 and the tension in the string is 0.8 N.
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Find the size of the image formed in the situation shown in figure.
A) 0.5 cm
B) 0.6 cm
C) 1.2 cm
D) 1 cm
Given
distance of the object u = -40 cm.
radius of curvature of the refracting surface R = -20 cm.
The height of the object h = 1 cm.
To Find
The size of the image
Solution
We will now write the following data using the sign convention:
The object's distance is u = -40 cm.
The refracting surface has a radius of curvature of R = -20 cm.
The object's height is h = 1 cm.
We know that the lens maker formula is: n2vn1u=n2n1R.
v denotes the image formation distance.
The refractive index of the second medium is n2=1.33, and the refractive index of the first medium is n1=1.
In the preceding equation, we will now substitute the known values.
1.33v−1−40cm=1.33−1−20cmv=32.05cm
We know that the magnification ratio can be expressed as vu=Hh.
H denotes the height of the created image.
We'll now swap the given and obtained values.
⇒32.05cm divide by 40cm=H divide by 1cm
⇒H=0.6cm
As a result, the best choice is (B).
Further information: The rules of reflection apply to reflection from a concave mirror. The normal to the point of incidence is drawn along the radius of the mirror, that is, by connecting the point of incidence to the centre of curvature.
The development of an image in a concave mirror is mostly determined by the distance between the object and the mirror. The concave mirror produces both real and virtual images. A virtual and magnified picture is generated when the object is placed very close to the mirror.
The complete question is -
Find the size of the image formed in the situation shown in figure.
photo
A) 0.5 cm
B) 0.6 cm
C) 1.2 cm
D) 1 cm
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What was Swimmer C’s average speed?
The average speed of swimmer C in the graph below is 4.8 km/h.
How do you calculate the average speed of swimmer C in the graph?When we look at the information provided on the graph, for swimmer C
distance traveled = 2400 m
travel time is 30 minutes.
If we should convert Distance into kilometers, it would be:
2400 m = 2400 / 1000 = 2.4 km
If we are to convert Time to the hour:
30 minutes equals 30/60 = 0.5 hour
To calculate the average speed, we use the formula;
Distance traveled / time taken = average speed
2.4 km/0.5 hour = 4.8 km/hr average speed
The above answer is in response to the full question below;
In the chart, Swimmer C' speed is between 1,400 and 1,600 what is his average speed?
4.8 km/h
2.4 km/h
2400 km/h
4800 km/h
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3) A ball is thrown from a cliff with a speed of 28.00m/s at an angle 48.0 degrees above
the horizontal direction. The cliff is 25.00m high.
a) what is the highest height attained by the ball?
ad profesor
b) At what time it hits the ground?
c) What are the magnitude and direction of velocity when it hits the ground?
If a ball is thrown from a cliff with a speed of 28.00m/s at an angle 48.0 degrees above the horizontal direction. the highest height attained by the ball is 45.06 meters.
What is the highest height?a) We can use the following equation to find the time it takes for the ball to reach this point:
vf = vi + at
where vf is the final velocity (which is zero at the highest point), vi is the initial velocity (which is the vertical component of the initial velocity), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time it takes to reach the highest point.
First, let's find the vertical component of the initial velocity:
vi = 28.00 m/s * sin(48.0°) = 19.84 m/s
Now, we can use the equation above to find the time it takes for the ball to reach the highest point:
0 = 19.84 m/s + (-9.8 m/s^2)t
t = 2.02 s
The time it takes for the ball to reach the highest point is 2.02 seconds. To find the highest height attained by the ball, we can use the following equation:
y = yi + vit + (1/2)at^2
where y is the height at any time t, yi is the initial height (25.00 m in this case), vi is the initial vertical velocity (19.84 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
At the highest point, the vertical displacement (y - yi) is equal to the highest height attained by the ball. We can use the time we just found to calculate this displacement:
y - yi = vi*t + (1/2)at^2
y - 25.00 m = 19.84 m/s * 2.02 s + (1/2)(-9.8 m/s^2)(2.02 s)^2
y - 25.00 m = 20.06 m
y = 45.06 m
Therefore, the highest height attained by the ball is 45.06 meters.
b) To find the time it takes for the ball to hit the ground, we can use the same equation we used to find the time to reach the highest point, but this time we'll solve for the time when y = 0 (i.e., when the ball hits the ground):
0 = 19.84 m/s + (-9.8 m/s^2)t
t = 2.03 s
Therefore, the time it takes for the ball to hit the ground is 2.03 seconds.
c) At the instant the ball hits the ground, its vertical velocity is -vi = -19.84 m/s (i.e., the negative of its initial vertical velocity), and its horizontal velocity is v = 28.00 m/s * cos(48.0°) = 18.08 m/s (i.e., the horizontal component of its initial velocity). The magnitude of its velocity can be found using the Pythagorean theorem:
v = sqrt[(18.08 m/s)^2 + (-19.84 m/s)^2] = 26.03 m/s
The direction of its velocity can be found using the inverse tangent function:
theta = tan^-1((-19.84 m/s) / (18.08 m/s)) = -49.57°
The negative sign indicates that the velocity vector makes an angle of 49.57° below the horizontal direction.
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How fast can the 145 A current through a 0.170 H inductor be shut off if the induced emf cannot exceed 78.0 V?
The maximum rate at which the current can be shut off in the inductor is 459.0 A/s.
What is an inductor?An inductor is a passive electronic component that stores energy in a magnetic field when an electric current flows through it.
What are some applications of inductors?Inductors are used in a variety of electronic applications such as power supplies, filters, and signal processing circuits. They are also used in transformers to change the voltage of alternating current (AC) power.
To calculate the maximum rate at which the current can be shut off in the inductor, we can use the formula:
emf = -L(di/dt)
where emf is the induced electromotive force, L is the inductance, and di/dt is the rate of change of current.
Rearranging the formula, we get:
di/dt = -emf/L
Substituting the given values, we get:
di/dt = -78.0 V / 0.170 H = -459.0 A/s
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describe the energy changes in a mass spring system that is oscillating horizontally explain how this changes of the system is vibrating vertically ?
The energy changes in a mass spring system that is oscillating horizontally is between kinetic energy and potential energy.
What is the energy change in the spring?In a mass spring system oscillating horizontally, the energy changes between kinetic energy and potential energy as the spring expands and contracts and the mass moves back and forth.
At the point of maximum displacement to one side, the mass has zero velocity and maximum potential energy. As the mass moves back to its equilibrium position, the potential energy decreases and is converted into kinetic energy. At the equilibrium position, the potential energy is at its minimum while the kinetic energy is at its maximum.
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(Figure 1) shows a 11 kg cylinder held at rest on a 20∘ slope.
What is the tension in the cable?
Express your answer with the appropriate units.
What is the magnitude of the static friction force?
Express your answer with the appropriate units.
The tension force T = μs×mg×cos(20∘)sin(20∘) and the cylinder is at rest, the force of friction must be equal in magnitude and opposite in direction to the component of the weight of the cylinder that is parallel to the slope.
What is tension force?
Based on the given information, we can use the concepts of static equilibrium and free-body diagrams to solve for the tension in the cable and the magnitude of the static friction force.
The cylinder is at rest, the sum of the forces acting on it must be zero in both the x and y directions.
In the x direction, there are two forces: the tension force T in the cable and the force of friction Ff acting up the slope. The component of the weight of the cylinder that is parallel to the slope also acts in the x direction, but it cancels out with the force of friction since the cylinder is at rest. Therefore, we have:
T - Ff = 0
In the y direction, there are two forces: the weight of the cylinder mg acting straight down and the component of the weight that is perpendicular to the slope. This component is given by mg×cos(20∘), where cos(20∘) is the cosine of the angle between the weight vector and the normal vector to the slope. This component is balanced by the normal force N from the slope. Therefore, we have:
N - mg×cos(20∘) = 0
Now we can solve for the tension force T and the force of friction Ff. From the first equation, we have:
T = Ff
Substituting this into the second equation, we have:
N - mg×cos(20∘) = 0
N = mg×cos(20∘)
Since the cylinder is at rest, the force of friction must be equal in magnitude and opposite in direction to the component of the weight of the cylinder that is parallel to the slope. This component is given by mg×sin(20∘), where sin(20∘) is the sine of the angle between the weight vector and the normal vector to the slope. Therefore, we have:
Ff = (μs)×N
where μs is the coefficient of static friction between the cylinder and the slope. The magnitude of the static friction force is therefore:
Ff = μs×mg×cos(20∘)sin(20∘)
To solve for the tension force T, we can use the fact that T = Ff:
T = μs×mg×cos(20∘)sin(20∘)
Without knowing the value of the coefficient of static friction or the angle of the slope, we cannot provide a numerical answer for the tension force or the magnitude of the static friction force.
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Perceived threat is the combination of actual vulnerability and perceived severity, making it a dynamic concept.
True
False
Answer:
True.
Explanation:
Perceived threat is the perceived risk of harm or loss resulting from a particular situation or event, and it is influenced by both the perceived severity (the magnitude of the harm or loss that is expected to result if the event occurs) and the perceived vulnerability (the perceived likelihood of the event occurring and affecting the individual). Since both perceived severity and perceived vulnerability can change over time, perceived threat is indeed a dynamic concept that can vary in different circumstances.
2. A soil was found to have the following water retention characteristics: Water content at -10 kPa (field capacity) = 0.15 kg kg-1; water content at -1500 kPa (wilting point) = 0.06 kg kg-1. The bulk density was found to be 1250 kg m-3. Assume the density of water to be 1000 kg m-3. 2.1 Calculate the plant available water capacity (PAWC), expressing your results in depth units. [2] 2.2 The crop growing in this field has a rooting depth of 40 cm. Calculate the plant available water (PAW) for this crop? [1] 2.3 A farmer irrigated his soil to field capacity then allowed it to dry out so that the PAW was 65% of its original value before he reapplied irrigation. Calculate the volume of water (per hectare of land) that the farmer must apply to raise the soil water content from the depletion level back to field capacity for the rooting depth of the crop.[3
Answer:
Given:
Water content at field capacity = 0.15 kg kg-1
Water content at wilting point = 0.06 kg kg-1
PAWC = Water content at field capacity - Water content at wilting point
PAWC = 0.15 kg kg-1 - 0.06 kg kg-1
PAWC = 0.09 kg kg-1
To convert kg kg-1 to depth units, we need to multiply by the bulk density of the soil, and divide by the density of water.
Bulk density of soil = 1250 kg m-3
Density of water = 1000 kg m-3
PAWC in depth units = (PAWC * Bulk density of soil) / Density of water
PAWC in depth units = (0.09 kg kg-1 * 1250 kg m-3) / 1000 kg m-3
PAWC in depth units = 0.1125 m
So, the Plant Available Water Capacity (PAWC) is 0.1125 m or 11.25 cm.
2.2 Plant Available Water (PAW) for the crop with rooting depth of 40 cm can be calculated as:
PAW = PAWC * Rooting depth
PAW = 0.1125 m * 0.40 m
PAW = 0.045 m or 4.5 cm
So, the Plant Available Water (PAW) for the crop is 0.045 m or 4.5 cm.
2.3 To raise the soil water content from the depletion level back to field capacity for the rooting depth of the crop, the farmer needs to apply water equivalent to the difference between PAW at depletion level and PAWC, per hectare of land.
Given:
PAW at depletion level = 65% of PAWC = 0.65 * 0.1125 m = 0.073125 m
PAWC = 0.1125 m
Volume of water to be applied = (PAWC - PAW at depletion level) * Area of land
Let's assume the area of land is 1 hectare, which is equivalent to 10,000 m^2.
Volume of water to be applied = (0.1125 m - 0.073125 m) * 10,000 m^2
Volume of water to be applied = 0.039375 m * 10,000 m^2
Volume of water to be applied = 393.75 m^3
So, the farmer must apply 393.75 m^3 of water per hectare of land to raise the soil water content from the depletion level back to field capacity for the rooting depth of the crop.
Which term best describes the light generated by a laser?
A) diffused
B) coherent
C) dispersive
D) longitudinal
Answer:
coherent
Explanation:
...the laser light sticks together in a tight beam....does not disperse
Answer:
B. Coherent
Explanation:
The term that best describes the light generated by a laser is B) coherent. Lasers produce light that is coherent, meaning that the waves of light are in phase with one another, and travel in a narrow, concentrated beam. This is in contrast to light produced by other sources, which is typically diffused and travels in many different directions.
A water rocket is launched with an initial velocity of 10m/s at an angle of 30. How far will it travel?
The water rocket is launched with an initial velocity of 10 m/s at an angle of 30° and reaches a height of 3.75m.
From the given,
the velocity of rocket = 10 m/s
angle = 30°
The distance/ height traveled by the rocket=?
H = u²sinθ² / 2g
u is the initial velocity of the rocket
g is the acceleration due to gravity = 10 m/s²
H = (10)²sin²(60°)/(2×10)
= (100 ×3/4)×(1/20)
=3.75 m
Thus the rocket covers a distance of 3.75 m.
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e =1.60×10−19C
me=9.11×10−31kg
k=8.99×109N⋅m2/C2
A point charge q = -0.55 nC is fixed at the origin. Where must an electron be placed in order for the electric force acting on it to be exactly opposite to its weight? (Let the y axis be vertical with the positive direction pointing up.)
Answer:
3.33×10^-5 meters above the origin along the positive y-axis
Explanation:
The weight of an object is given by its mass multiplied by the acceleration due to gravity, which is 9.81 m/s² near the surface of the earth. For an electron with mass me, its weight is:
W = me * g
= 9.11×10^-31 kg * 9.81 m/s²
= 8.93×10^-30 N
Since the electric force acting on the electron is opposite to its weight, the electric force must have the same magnitude as its weight but in the opposite direction:
|F_E| = |W| = 8.93×10^-30 N
The electric force between two point charges is given by Coulomb's law:
F_E = k * |q1| * |q2| / r²
where k is the Coulomb constant, q1 and q2 are the charges of the two particles, and r is the distance between them.
In this problem, q1 is the fixed charge of -0.55 nC at the origin, and q2 is the charge of the electron, which we want to find. Since we know the magnitude of the electric force between the two charges, we can solve for the distance between them:
r² = k * |q1| * |q2| / |F_E|
= 8.99×10^9 N⋅m²/C² * 0.55×10^-9 C * |q2| / (8.93×10^-30 N)
= 6.95×10^6 * |q2|
Taking the square root of both sides, we get:
r = 2.64×10^-3 * sqrt(|q2|)
Now, we need to find the distance r at which the electric force between the two charges is equal in magnitude but opposite in direction to the weight of the electron. Equating the expression for r above with the distance y along the y-axis where the electron is placed, we get:
2.64×10^-3 * sqrt(|q2|) = y
Since the electron is placed on the y-axis, its x and z coordinates are zero, and the distance between the electron and the fixed charge is simply the y-coordinate. The electric force between the charges will be attractive (i.e., in the negative y direction), so the direction of the force vector will be opposite to the positive y direction. Therefore, we can write the electric force on the electron as:
F_E = - k * |q1| * |q2| / y²
Setting this equal to the weight of the electron, we have:
k * |q1| * |q2| / y² = |W|
|q2| = |W| * y² / (k * |q1|)
= 8.93×10^-30 N * y² / (8.99×10^9 N⋅m²/C² * 0.55×10^-9 C)
= 1.56×10^-20 * y²
Substituting this expression for |q2| into the expression for r above, we get:
r = 2.64×10^-3 * sqrt(1.56×10^-20 * y²)
= 1.35×10^-11 * y
Equating this expression for r with the expression for y above, we have:
2.64×10^-3 * sqrt(1.56×10^-20 * y²) = y
Squaring both sides and simplifying, we get:
y = 3.33×10^-5 m
Therefore, the electron must be placed at a distance of 3.33×10^-5 meters above the origin, along the positive y-axis, in order for the electric force acting on it to be exactly opposite to its weight.
To summarize, we used Coulomb's law to relate the electric force between the electron and the fixed charge at the origin to the distance between them, and equated this force with the weight of the electron. We then solved for the distance at which the two forces are equal in magnitude but opposite in direction, and found that the electron must be placed 3.33×10^-5 meters above the origin along the positive y-axis.
Hope this helps!
A particle was moving in a straight line with a constant acceleration. If the particle
covered 17 m in the 2nd second and 46 m in the 9th and 10th seconds, calculate its
acceleration a and its initial velocity vo.
a. a=4.44m/s², vo-12.56m/s
b. a=0.8m/s², vo=15.8m/s
C. a=5.33m/s², vo=9m/s
d. a=0.67m/s², vo=16.33m/s
Answer:
Displacement S= 17mHere the velocity will increase Displacement S= 17mHere the velocity will increase so it will be C or B
Explanation:
I don't know whether I am correct or not. Please help me. Thank you.
The kinetic energy of the truck would be transferred to the spring such that the spring would make some movement rapidly.
What is the principle of the conservation of energy?As we look at the principle of the conservation of mechanical energy, we have to know that the total energy of the system is a constant. The implication of this is that the kinetic energy of the truck and the potential energy of spring must remain the same.
The second law of motion can be applied here to obtain the average force as follows;
F.t = mv - mu
mv = final momentum
mu = initial momentum
F.t = impulse
Then we have that;
F = mv - mu/t
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Problem 2.5. (6 pts) A wave is modeled by the wave function y(z,t) = (0.30 m)sin [(-18.00=t)]. What are the (a) amplitude, (b) wavelength, (c) wave speed, (d) period, (e) initial phase shift and (f) frequency of the wave?
A wave is modeled by the wave function y(z,t) = (0.30 m)sin [(-18.00t)]. then
amplitude is 0.30 m, (b) wavelength is undefined , (c) wave speed is undefined , (d) period is 0.34s , (e) initial phase shift is 0° and (f) frequency of this wave is 2.94 Hz.
Wave is is a disturbance in a medium that carries energy as well as momentum . wave is characterized by amplitude, wavelength and phase. Amplitude is the greatest distance that the particles are vibrating. especially a sound or radio wave, moves up and down. Amplitude is a measure of loudness of a sound wave. More amplitude means more loud is the sound wave.
Wavelength is the distance between two points on the wave which are in same phase. Phase is the position of a wave at a point at time t on a waveform. There are two types of the wave longitudinal wave and transverse wave.
The displacement of the wave is given by,
y(z,t) = A sin(kz - ωt +φ)
Comparing with our wave,
y(z,t) = (0.30 m)sin [(-18.00t)]
a) Amplitude = 0.30 m
Phase φ = 0°
b) for wavelength,
k = 2π/λ
k = 0 when λ = infinite
λ is not defined
c) speed of the wave,
c = λf = infinite = undefined
d) for for period
ω = 18
ω = 2π/T
T = 2π/ω
T = 2π/18 = 0.34s
f) for frequency
f = 1/T
f = 1/0.34 = 2.94 Hz
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According to the Doppler effect what appears to happen when the light source moved further away from the observers? The electromagnetic appear more red in color the waves lengthennand the frequency appears nTo increase the electromagnetic wave appears more blue in color the waves compress and the frequency appears to increase
According to the Doppler effect, when the light source moves further away from the observers, the electromagnetic wave appears more red in color, the wavelength lengthens, and the frequency appears to decrease.
What is the Doppler effect?The Doppler effect is the change in frequency and wavelength of waves that occurs when the source of the waves and/or the observer moves relative to each other.
How is the Doppler effect used in real-world applications?The Doppler effect is used in real-world applications such as radar and sonar systems to measure the speed and direction of moving objects, as well as in medical imaging technologies such as ultrasound. It is also used to determine the motion of stars and other celestial objects.
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The momentum of object 1 is (8.37,13.9) kg m/s. The momentum of object 2 is (32.7,12.39) kg m/s. Object 1 collides with object 2 and they stick together. What is the x-component of the final momentum in kg m/s after the collision if the net external force acting on the system is zero?
Answer:
x = 41.07 kg m/s
Explanation:
Momentum is conserved so,
(8.37, 13.9) + (32.7,12.39) = (41.07, 26.29)
What is the wavelength of a radar signal that has a frequency of 33 GHz?
Explanation:
For electromagnetic waves c = wavelength * frequency
c = speed of light = 3 x 10^8 m/ s
3 x 10^8 m/s = wl * 33 x 10^9 Hz
wl = .009 m ( or 9 mm)
If a radar signal that has a frequency of 33 GHz, Then the wavelength of the radar signal is 9 mm.
What is wavelength?Wavelength is a fundamental concept in the study of waves, which are disturbances that propagate through space or a medium. It is defined as the distance between two consecutive points in a wave that are in phase, meaning that they have the same position in their respective cycles.
In other words, the wavelength is the spatial period of a wave, which is the distance over which the wave repeats itself. It is commonly represented by the Greek letter lambda (λ) and is measured in meters (m), although it can also be expressed in other units such as nanometers (nm) or micrometers (μm).
Wavelength is a key property of waves, as it determines many of their characteristics and behavior. For example, the wavelength of an electromagnetic wave (such as light or radio waves) determines its color or frequency, and thus its energy and ability to interact with matter. Similarly, the wavelength of a sound wave determines its pitch, and thus its perceived tone and musical quality.
The relationship between wavelength, frequency, and velocity is described by the wave equation, which states that the velocity of a wave is equal to the product of its wavelength and frequency. This relationship is important for understanding how waves behave and interact with their environment, such as when they are reflected, refracted, or diffracted.
So, the wavelength is a crucial concept in the study of waves, as it defines their properties and behavior. It is the distance between two consecutive points in a wave that are in phase, and is measured in meters or other units. The relationship between wavelength, frequency, and velocity is described by the wave equation, which is fundamental to the study of waves in various fields such as physics, engineering, and communication.
Here in the Question,
The wavelength of a radar signal can be calculated using the formula:
wavelength = speed of light/frequency
where the speed of light is approximately 3 x 10^8 meters per second.
Plugging in the given frequency of 33 GHz (33 x 10^9 Hz), we get:
wavelength = 3 x 10^8 / (33 x 10^9)
wavelength = 0.009090909... meters
Therefore, By rounding to the nearest millimeter, the wavelength of the radar signal is approximately 9 mm.
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You have a piece of cork with a volume of 2 cm3 and a density of 210 kg/m3. You hold it under water and release it.
a. What is the mass of the piece of cork? (1 point)
b. What is the buoyant force on the cork after you release it? (1 point)
c. What is the net force on the cork after you release it? (1 point)
d. What is the acceleration of the cork after you release it? (1 point)
Answer:
a. m = 4.2•10⁻⁴ kg
b. B = 0.01962 N
c. ∑F = 0.0155 N (upward)
d. a = 6.51•10⁻⁶ m/s² (upward)
Explanation: I'm unsure if you have the correct values, but either way, the steps are below:
a. Mass:
Volume • Density = Mass
(2 cm³)(210 kg/m³)(10⁻⁶ m³/cm³)= 0.00042 kg or 4.2•10⁻⁴ kg
b. Buoyant Force:
Fluid Density • Displaced Volume • Gravity = Buoyant Force
**The Fluid is water, so the fluid density = 1 g/cm³, and the displaced volume is the same as the volume of the cork.**
ρ • V • g = B
[(1 g/cm³)(10⁻³ kg/g)](2 cm³)(9.81 m/s²) =
(10⁻³)(2)(9.81) = 0.01962 N
c. Net Force:
∑F: B - W = ma
mass • gravity = W
(4.2•10⁻⁴ kg)(9.81 m/s²) = 0.0041202 N
(0.01962) - (0.0041202) = 0.0154998 N
d. Acceleration:
F = ma
F/m = a
(0.0155•10⁻² kg•m/s²)/(4.2•10⁻⁴ kg) = 0.000006509916 m/s²
or 6.51•10⁻⁶ m/s²
––––––––––––––––––––––––––––––––––––––
The reason the values are so small is because N = kg•m/s², so all mass units need to stay in kg, even though it would be simpler to have them in grams.
A ball of mass m = 0.275 kg swings in a vertical circular path tied to a string of L = 0.850 m long. (a) What are the forces acting on the ball at any point on the path? (b) Draw force diagrams for the ball when it is at the bottom of the circle and when it is at the top. (c) If its speed is 5.20 m/s at the top of the circle, what is the tension in the string there? (d) If the string breaks when its tension exceeds 22.5 N, what is the maximum speed the ball can have at the bottom before that happens?
Given that a mass, [tex]\bold{m=0.275 \ kg}[/tex], swings in a circular path due to an attached string with a length, [tex]\bold{l=0.850 \ m}[/tex]. Refer to IMAGE #1 for a picture of the situation.
We are asked to answer the following...
(a) What forces always act on the mass throughout its swing?
(b) Draw force diagrams or free-body diagrams showing what forces are acting on the mass at the bottom of its swing vs the top.
(c) Given a velocity, [tex]\bold{\vec v = 5.20 \ m/s}[/tex], when the mass is at the top of its swing, find the tension in the string, [tex]\vec T[/tex].
(d) Given the maximum tension,[tex]\bold{\vec T_{max}=22.5 N}[/tex], what is the fastest the mass can travel at the bottom of its swing, [tex]\vec v_{max}[/tex]?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For part (a):
To answer part a lets list of all forces that can act on an object and then narrow them down.
List of forces:
1. Weight or the force of gravity which is the mass of an object multiplied by the magnitude of the acceleration from gravity
([tex]\vec w= m||\vec g|| \ where \ \vec g=-9.8 \ m/s^2[/tex]).
2. Tension force which can be descried as a pulling force.
3. Normal force which is the force an object applies to prevent an object from traveling through its surface.
4. Friction force is a force that resists an objects motion.
5. Spring force is a force provided by a spring.
6. Air resistance force which is caused by air, can be interpreted as a friction force.
7. Electrical force is a force provided by charges.
8. Magnetic force which is a force provided by a magnetic field.
9. Buoyant force which is an upward force a fluid exherts on an object, can be interpreted as a normal force.
Right away we can eliminate 5,7,8, and 9. As we are not dealing with any springs, electricity, magnets, or fluids. We are also going to assuming there is no resistive forces acting on the ball so we can eliminate 4 and 6. The mass is not sitting on something so we can eliminate 3.
That leaves us with the gravitational force (weight) and the tension force. Which logically would make sense, gravity would act on the mass at all points in its path and the string will always provide a pulling force keeping the mass from unhooking and flying away in a straight line.
For part (b):
Refer to IMAGE #2
For part (c):
Refer to IMAGE #3
Add up forces acting in the centripetal axis. Note that the centripetal axis acts like the y axis and the tangential axis acts like the x axis.
[tex]\vec v = 5.20 \ m/s[/tex]
[tex]\Sigma \vec F_c: \vec Tsin90 \textdegree +\vec wsin90\textdegree=m\vec a_c[/tex] where [tex]\vec a_c[/tex] is centripetal acceleration. [tex]\vec a _c = \frac{\vec v^2}{r}[/tex]
[tex]\Longrightarrow \vec T(1) +\vec w(1)=m(\frac{\vec v^2}{r} ) \Longrightarrow \vec T+\vec w=m(\frac{\vec v^2}{r} ) \Longrightarrow \vec T=m(\frac{\vec v^2}{r} ) -\vec w[/tex]
[tex]\Longrightarrow \vec T=m(\frac{\vec v^2}{r} ) -m ||\vec g|| \Longrightarrow \vec T=m((\frac{\vec v^2}{r} ) - ||\vec g||) \Longrightarrow \vec T=(0.275)((\frac{ (5.20)^2}{0.850} ) - 9.8) \Longrightarrow \boxed{\boxed{\vec T = 6.05 N}} \therefore Sol.[/tex]
Thus, the tension in the string is found.
For part (d):
Refer to IMAGE #4
Add up forces acting in the centripetal axis. Note that the centripetal axis acts like the y axis and the tangential axis acts like the x axis.
[tex]\vec T_{max}=22.5 \ N[/tex]
[tex]\Sigma \vec F_c: \vec T_{max}sin90 \textdegree +\vec wsin90\textdegree=m\vec a_c[/tex]
[tex]\Longrightarrow \vec T_{max}sin90 \textdegree +\vec wsin90\textdegree=m(\frac{v^2_{max}}{r} ) \Longrightarrow \vec T_{max}(1) +\vec w(1)=m(\frac{v^2_{max}}{r} )[/tex]
[tex]\Longrightarrow \vec T_{max} +\vec m||\vec g||=m(\frac{v^2_{max}}{r} ) \Longrightarrow \vec v_{max}= \sqrt{r(\frac{\vec T_{max}+m||\vec g||}{m}) }[/tex]
[tex]\Longrightarrow \vec v_{max}= \sqrt{(0.850)(\frac{22.5+(0.275)(9.8)}{0.275}}) \Longrightarrow \boxed{\boxed{\vec v_{max}=8.82 \ m/s} } \therefore Sol.[/tex]
Thus, the max speed is found.
1. A rich merchant finds himself stranded in the middle of a frozen lake. Don't ask how he got there. Think Twilight Zone... The surface
is perfectly frictionless. All he has is his clothing and a large bag of gold coins. How can he save himself? Write down your answer.
2. You are hammering a nail into a hard piece of wood. You are using one of your little brothers light, toy hammers and getting nowhere fast. Finally, you grab a hammer with a heavier head and your task goes much easier. Which one of Newtons laws did you finally remember? explain.
A pool of water has a rectangular base of 5 m by 10 m and the water is 6 m deep. What is the pressure on the base of the pool taken g = 10m/ss
How has the expansion of the universe affected the cosmic microwave background radiation?
A. It has caused it to heat up.
B. It has caused it to cool.
C. It has caused it to blue shift.
D. It has caused it to red-shift.
Answer:
The expansion of the universe has caused cosmic microwave background radiation to cool. As the universe expands, the radiation is stretched out, causing its wavelength to increase and its temperature to decrease. This phenomenon is known as cosmic microwave background radiation redshift.
the answer is D. It has caused it to red-shift.