builder places a 3kg hammer on the top of a ladder, which is 4m above the ground. Calculate the gravitational potential energy of the hammer while on the ladder.
Answer:
[tex]E=mgh[/tex]
[tex]m=3kg[/tex]
[tex]h=4m[/tex]
[tex]g=9.8m/s^{2}[/tex]
[tex]E= 3*4*9.8=117.6J[/tex]
Explanation:
Only substitute amounts to formula.
Hope this helps ;)
Cheers :D
2. If a vehicle with four wheels is traveling in a linear path through soft dirt, how many and which tires would you expect to leave tire track impressions
Given our understanding of the situation, for this vehicle with four wheels is traveling in a linear path through soft dirt, we would expect for only two wheels to leave tire track impressions.
The vehicle has four wheels, and one might think initially that it makes sense for all four wheels to create tire track impressions on the dirt. This assessment is correct, all four tires will create impressions, However, not all four tires will leave a said impression.
This is due to the fact that as the vehicle advances, the tires in the back will erase the impressions left by the front tires. Therefore, only two tire track impressions will be left.
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TRUE OR FALSE! PLZ HELP
Magnitude is a measure of how strong a force is.
A. True
B. False
Answer:
True
Explanation:
Magnitude is the "value" the greater the value the greater the force is and vice versa
Answer:
TRUE
Explanation: I did the test
Hope this helps
Two ends of a steel wire of length 8m and 2mm radius are fixed to two rigid supports. Calculate the increase in tension when the temperature falls by 10°C. Given linear expansivity of steels = 12x10^_6 per kelvin and Young's modules for steel =2x10^11 n/m^2
The increase in tension on the steel wire is 8,484.75 N.
The given parameters;
original length of the wire, l = 8 mradius of the wire, r = 2 mmThe area of the steel wire is calculated as follows;
[tex]A = \pi r^2\\\\A = \pi \times (2\times 10^{-3})^2\\\\A = 1.257 \times 10^{-5} \ m^2[/tex]
The extension of the steel wire is calculated as follows;
[tex]\Delta l = \alpha \times l\times \Delta T\\\\\Delta l = (12\times 10^{-6}) \times (8) \times (10 + 273)\\\\\Delta l = 0.027 \ m[/tex]
The increase in tension on the steel wire is calculated as follows;
[tex]E = \frac{stress}{strain } = \frac{\ F/A}{\Delta l/l} \\\\E = \frac{F\times l}{A \times \Delta l} \\\\F = \frac{E\times A \times \Delta l }{l} \\\\F = \frac{(2\times 10^{11}) \times (1.257\times 10^{-5})\times 0.027}{8} \\\\F = 8,484.75 \ N[/tex]
Thus, the increase in tension on the steel wire is 8,484.75 N.
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How does climate change?
Answer:
Con el cielo del agua
Explanation:
espero que te ayude
The same type of engine is placed in iwo different airplanes The first airplane is twice as heavy as the second suchane Wrich
statement is correct about the mass of the airplane and its acceleration, assuming a constant not forest (1 point)
Increasing the acceleration of the airplane causes the enging to provide more force
Decreasing the mass of the airplane will result in the airplane accelerating loss
O Decreasing the mass of the airplane will result in the airplane accelerating more
increasing the acceleration of the airplane causes the engine to provide les force
Answer:
Explanation: If the engines in both of the airplanes are at the same Weight then The engines in both of the airplanes would stil have the same mass because mass stays the same everywhere
Since mass is constant everywhere, if the engines in both airplanes are at the same weight, the engines in both airplanes would still have the same mass.
What is Weight?Weight is the force an object exerts, mass is the amount of matter that makes up an object. Newton is the SI unit of weight while kilogram is the SI unit of mass.
Given that in question the same type of engine is placed in two different air-planes The first air-plane is twice as heavy as the second Suchane Wrich statement is correct about the mass of the air-plane and its acceleration, assuming a constant net force.
Increasing the acceleration of the air-plane causes the enging to provide more force.
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Calculate the altitude above the surface of Earth, in meters, at which the acceleration due to gravity is g
Answer:
By definition the acceleration due to gravity at the surface is g:
The altitude above the surface is zero for an acceleration of g.
according to Newton's __ law, an object with no net force acting on it remains at rest or in motion with a constant veloctiy
Answer:
i think its his law of inertia
Explanation:
this law is about motion
Question 2 (4 points)
Listen
A 14,000kg freight train, coasting at 8.1m/s, strikes another car, and the two move
forward together at 3.6m/s.
What is the mass of the second car?
Paragraph
B
1
U
A
+
Answer:
Explanation:
I will ASSUME that the struck car was initially at rest. Would not have to be, but a different mass will result if it was
Conservation of momentum
14000(8.1) + m(0) = (14000 + m)(3.6)
14000(8.1) = 14000(3.6) + m(3.6)
m = 14000(8.1 - 3.6) / 3.6
m = 17500 kg
The mass of the second car is 17500 kg.
What is law of conservation of momentum?The law of conservation of momentum asserts that, unless an external force is applied, the total momentum of two or more bodies operating upon one another in an isolated system remains constant. As a result, momentum cannot be gained or lost.
Newton's third law of motion has a direct impact on the idea of momentum conservation.
mass of the first car = 14000 kg.
Initial speed of first car = 8.1 m/s.
Final speed of the two cars = 3.6 m/s.
From law of conservation of momentum; we can write:
Total initial momentum = total final momentum
14000 kg ×8.1 m/s + m×0 = (14000 + m) kg × (3.6 m/s)
14000(8.1) = 14000(3.6) + m(3.6)
m = 14000(8.1 - 3.6) / 3.6
m = 17500 kg
Hence, the mass of the second car is 17500 kg.
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who the football games yesterday 2021
Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a horizontal surface of negligible friction, releasing the block, and measuring the velocity v of the block as it leaves the spring, as shown in Figure 1. The experiments indicate that as x increases, so does v in a linear relationship. The surface is now lifted so that the surface is at an angle θ above the horizontal. Which of the following indicates how the relationship between v and x changes?
Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.
The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.Reasons:
The energy given to the block by the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]
According to the principle of conservation of energy, we have;
On a flat plane, energy given to the block = [tex]0.5 \cdot k \cdot x^2[/tex] = kinetic energy of
block = [tex]0.5 \cdot m \cdot v^2[/tex]
Therefore;
0.5·k·x² = 0.5·m·v²
Which gives;
x² ∝ v²
x ∝ v
On a plane inclined at an angle θ, we have;
The energy of the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]
The force of the weight of the block on the string, [tex]F = m \cdot g \cdot sin(\theta)[/tex]
The energy given to the block = [tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)[/tex] = The kinetic energy of block as it leaves the spring = [tex]\mathbf{0.5 \cdot m \cdot v^2}[/tex]
Which gives;
[tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2[/tex]
Which is of the form;
a·x² - b = c·v²
a·x² + c·v² = b
Where;
a, b, and c are constants
The graph of the equation a·x² + c·v² = b is an ellipse
Therefore;
As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.Please find attached a drawing related to the question obtained from a similar question online
The possible question options are;
As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of xThe relationship is no longer linear and v will be more for the same value of xThe relationship is still linear, with lesser value of vThe relationship is still linear, with higher value of vThe relationship is still linear, but vary inversely, such that as x increases, v decreasesLearn more here:
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To become a healthy individual, a person must have a properly executed routine to achieve his/her fitness goals. On the contrary, in what instance does a person become unhealthy
please.
I mark u brainliest answer
Answer:
destroy
Explanation:
cuz if anyhow throw rubbish, it will affect/destroy the world environment
Answer:
bathing regularly 2pollute 3 throwing waste in land water water pollution
This drawing of the field lines between these two charged plates is correct.
True or false
Answer:
True
Explanation:
Since positive and negative forces are both opposites of each other they attract. But if the forces are the same where both are positive or negative thye repel do to being the same charges in that case the picture would show lines moving in opposite directions.
A plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again. The apparent weight of the gas is the difference between these two masses. The gas is squeezed out of the bag to determine its volume by the displacement of water. What is the actual weight of the gas
The actual weight of the gas = apparent weight + weight.
The actual weight = [tex]W_{A}[/tex] + W
Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.
If the apparent weight of the gas is the difference between these two masses, then let the apparent weight = [tex]W_{A}[/tex]
The gas is squeezed out of the bag to determine its volume by the displacement of water. Since
density = mass / volume
The density of water is 1000 kg/[tex]m^{2}[/tex]
we can get the mass of the gas by making m the subject of the formula.
W = mg
The actual weight of the gas = apparent weight + weight
That is,
The actual weight = [tex]W_{A}[/tex] + W
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Throwing all over the place will___________the environment.
Answer:
pollute the environment
a. Block on a smooth incline.
A block of mass m= 3.8 kg on a smooth inclined plane of angle 36° is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 3 kg hanging vertically. Take the positive direction up the incline and use 9.81 m/s2
for g.
What is the acceleration of each block to 1 decimal place?
Answer:
Explanation:
F = ma
3(9.81) - 3.8(9.81sin36) = (3 + 3.8)a
a = 1.10566...
a = 1.1 m/s²
the 3.8 kg mass will move up slope and the 3 kg mass will fall at that acceleration.
Using the expression for the total energy of this system, it is possible to show that after the switch is closed, d2qdt2=−kq, where k is a constant. Find the value of the constant k.
The value of the constant K is [tex]\mathbf{K = \dfrac{1}{LC}}[/tex]
According to Kirchhoff's loop rule, the total algebraic sum of potential differences in any loop, combining voltage provided by voltage sources as well as resistive components, must equal zero.
Thus, the relation for Kirchhoff's loop rule can be expressed as:
[tex]\mathbf{\dfrac{q}{c}- L\dfrac{dI}{dt} = 0}[/tex]
We all know that the current in the nonconstant charge flow can be written as:
[tex]\mathbf{I = \dfrac{dq}{dt}}[/tex]
∴
Replacing the current (I) into Kirchhoff's loop rule, we have:
[tex]\mathbf{ L\dfrac{d}{dt} ( \dfrac{dq}{dt})= -\dfrac{q}{c}}[/tex]
[tex]\mathbf{ \dfrac{d^2q}{dt^2}= -\dfrac{q}{Lc} \ \ ---(1)}[/tex]
From the given question, when the switch is closed
[tex]\mathbf{ \dfrac{d^2q}{dt^2}= -kq\ \ ---(2)}[/tex]
Then, the charges on the capacitor start to b, resulting in the rise of the current in the circuit.
∴
By equating both equations (1) and (2);
[tex]\mathbf{K = \dfrac{1}{LC}}[/tex]
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help meee please please please
A racecar can be slowed with a constant acceleration of -14 m/s2. If the car is going 75 m/s, how many meters will take to stop?
Answer:
Explanation:
v² = u² + 2as
s = (v² - u²)/2a
s = (0² - 75²) / (2(-14))
s = 200.8928
s = 200 m
Answer:
Identification
Explanation:
:-;..........
II) A 0.40-kg ball, attached to the end of a horizontal ord, is rotated in a circle of radius 1.3 m on a friction- less horizontal surface. If the cord will break when the tension in it exceeds 60 N, what i~ the maximum speed the ball can have
Hi there!
In this instance, the object spinning in a horizontal circle will experience a net force in the horizontal direction due to tension.
The net force is equivalent to the centripetal force, so:
∑F = T
mv²/r = T
Solve for v:
v = √rT/m
v = 13.96 m/s
What’s 1 + 1 many half a window
Answer:
1
+
1
is
2
Explanation:
because if you add one to a one it would end up being 2 <3
Answer: 2
Explanation: 1+1=2
This was THE hardest question I've ever decided to answer
the wheel of bicycle has a radius of 25cm. what will be the magnitudes of the angular displacement in radian and revolution respectively, when the wheel has rolled a distance of 350cm on straight level road?
Answer:
Explanation:
350 cm / 25 cm = 14 radians
14 rad / 2π rad/rev = 2.23 revolutions
If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0
meters, calculate the power required to lift it?
A car driving on level ground at 20.0 m/s slams on its brakes and skids to a halt. If the coefficient of kinetic friction between the car’s tires and the road is 0.750, how far does the car skid before stopping? How far would the car have skidded if it had been moving at 40.0 m/s?
The distance the car has skidded if it had been moving at 40.0 m/s is 27.2m
The linear force acting on the car is opposed by the frictional force. Hence;
[tex]F=F_f[/tex]
[tex]ma = -\mu R\\ma =-\mu mg[/tex]
m is the mass of the car
a is the acceleration
[tex]\mu[/tex] is the coefficient of friction
R is the normal force
Given the following parameters
[tex]a=-\mu g[/tex]
[tex]a =-0.75(9.8)\\a=-7.35m/s^2[/tex]
Get the distance the car has skidded if it had been moving at 40.0 m/s
[tex]v^2=u^2+2as[/tex]
[tex]0^2=20^2+2(-7.35)s\\0=400-14.7s\\s=\frac{400}{14.7}\\s= 27.2m[/tex]
Hence the distance the car has skidded if it had been moving at 40.0 m/s is 27.2m
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For a certain transverse standing wave on a long string, an antinode is at x = 0 and an adjacent node is at x = 0.20 m. The displacement y(t) of the string particle at x = 0 is shown in the figure, where the scale of the y axis is set by ys = 4.3 cm. When t = 0.90 s, what is the displacement of the string particle at (a) x = 0.30 m and (b) x = 0.40 m ? What is the transverse velocity of the string particle at x = 0.30 m at (c) t = 0.90 s and (d) t = 1.3 s?
The expressions for the traveling and standing wave to find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
The traveling wave is a disturbance in the medium that moves at constant speed, in the case of a transverse wave the expression for the perpendicular oscillation is:
y = A sin (kx - wt)
Where y is the oscillation perpendicular to the direction of the displacement, A the amplitude, k in wave number and w the angular velocity.
Standing waves are formed when a traveling wave collides with an obstacle and is reflected, in this case the sum of the two waves gives a wave that does not shift in time and fulfills the relationship
[tex]\frac{\lambda}{2} = \frac{L}{n}[/tex]
Where λ is the wavelength, L the distance between the reflection points and n the number of nodes.
Indicates that for the standing wave the distance between an antinode and the node is x = 0.20 m, therefore
[tex]\frac{\lambda}{4} = \frac{L}{1}[/tex]
λ = 4L
λ = 4 0.20
λ = 0.80 m
The wave number.
k = [tex]\frac{2\pi }{\lambda }[/tex]
k = [tex]\frac{2 \pi }{0.80 }[/tex]
k = 2.5π i m⁻¹
In the associated traveling wave, from the graph we can see that the period of the wave is:
T = 2.8 s
the angular velocity is related to the period.
[tex]w=\frac{2\pi}{T} \\w = \frac{2\pi }{2.8}[/tex]
w = 0.714π rad/s
indicate the maximum displacement that is the amplitude of the wave.
A = [tex]y_s[/tex]
A = 4.3 cm
Let's write the equation of the traveling wave.
y = 4.3 sin [π (2.5 x - 0.714 t)]
with this expression we can answer the questions.
a) the displacement of the particle for x = 0.30 m
y = 4.3 sin (π (2.5 0.30 - 0.714 t))
y = 4.3 sin π( 0.75 - 0.714 t(
Remember that the angles must be in radians. For time t = 0 the displacement is
y = 4.3 0.707
y = 3.04 cm
b) The displacement for x = 0.4m
y = 4.3 sin (π 2.5 0.4)
y = 0 cm
c) the transverse velocity of the wave at x = 0.30 m for the time of t = 0.90s
the speed of the wave is
[tex]v= \frac{dy}{dt} \\v= A w cos ( kx - wt)[/tex]
v = 4.3 0.714π cos π(2.5 0.3 - 0.714 t)
v = 9.65 cos π(0.75 - 0.714 t)
For time t = 0.90 s the velocity is:
v = 9.65 cos π(0.75 - 0.714 0.9)
v = 9.65 0.9436
v = 9.11 cm / s
d) The velocity for time t = 1.3 s
v = 9.65 cos π(0.75 - 0.714 1.3)
v = 9.65 0.9999
v = 9.65 cm / s
In conclusion, using the expressions for the traveling and standing wave, we can find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
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If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k
Recall the definition of the cross product with respect to the unit vectors:
i × i = j × j = k × k = 0
i × j = k
j × k = i
k × i = j
and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)
Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have
a × b = (8i + j - 2k) × (5i - 3j + k)
a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)
… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)
… … … … + 8 (i × k) + (j × k) - 2 (k × k)
a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)
a × b = - 5k - 10j - 24k - 6i - 8j + i
a × b = -5i - 18j - 29k
Answer:
Explanation:
If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k
A 260-kg object and a 560-kg object are separated by 3.10 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 49.0-kg object placed midway between them.
C is the correct answer:))))))))
PLEASEEEEEEEE HELPPPPPPP
Answer: 62.25
Explanation: F = ma
F = 7.5 * 8.3
F = 62.25
Thus, the answer is 62.25 Newtons.
Would appreciate brainliest <3
A single paragraph can be selected by triple in the paragraph. ture or false
Answer:
The answer for the question should be True