The force needed is 29.2 N.
To calculate the force needed to accelerate a 9.35 kg bowling ball at a rate of 3.12 m/s²,
you can use the equation F = ma, where F is the force, m is the mass of the object (in this case, the bowling ball), and a is the acceleration. Putting in the values, you get:
F = (9.35 kg) x (3.12 m/s²)
= 29.2 N
This works out to be approximately 29.2 Newtons of force.
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Mechanical energy starts the Ferris wheel, charging the lights along the seats, showing Choose energy. The full Ferris wheel begins to move going from Choose to Choose energy.
Mechanical energy is used to start the Ferris wheel, providing the initial rotation and motion to the wheel.
The motion of the wheel then causes the lights along the seats to be powered, allowing them to light up as the wheel turns. The wheel is then able to generate kinetic energy as it moves, which it then uses to continue turning.
As the wheel turns, it transfers its energy from mechanical to kinetic, and this energy is used to keep the wheel moving from start to finish. Kinetic energy is also used to power the centrifugal force which keeps the riders in their seats as the wheel turns, allowing them to enjoy the ride.
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Expression for kinetic energy is __________ while that of potential energy is __________
Particles q1 = -8.99 uC, q2 = +5.16 uC, and q3 = -89.9 uC are in a line. Particles q1 and q2 are separated by 0.220m and particles q2 and q3 are separated by 0.330m. What is the net force on particle q1?
Answer: -39200.45 Nm^2/C^2 This is rough work
Explanation:
So, the net force will be felt on particle [tex]\sf{q_1}[/tex] approximately 15.39 N to the right.
IntroductionHi! Here I will explain the net force of three charges that interacts. Remember that this question can be solved using Coulomb's Law. Coulomb's Law states that the interaction force between two interacted charges will be proportional to the magnitude of the charges that have specific interaction but inversely proportional to the square of the value of the distance between each other. So, with this concept, we can calculate the magnitude of coulomb force with this equation:
[tex] \boxed{\sf{\bold{F = k \cdot \frac{q_1 \cdot q_2}{(r_{12})^2}}}} [/tex]
Formula UsedOn this condition, the net force will be felt only by the first charge, so we must define the position of the first charge as a fixed one. So, we can calculate the magnitude of coulomb force concerning the first charge with this equation:
[tex] \boxed{\sf{\bold{\sum F_{q1} = k \cdot (\frac{q_1 \cdot q_2}{(r_{12})^2} + \frac{q_1 \cdot q_3}{(r_{13})^2})}}} [/tex]
With the following condition:
[tex] \sf{\sum F_{q1}}[/tex] = the net force will be felt by the first charge (N)k = constant of coulomb ≈ [tex] \sf{9 \times 10^{9}}[/tex] N.m²/C².[tex] \sf{q_1}[/tex] = first charge (C)[tex] \sf{q_2}[/tex] = second charge (C)[tex] \sf{q_3}[/tex] = third charge (C)[tex] \sf{r_{12}}[/tex] = the distance between first and second charge (m)[tex] \sf{r_{13}}[/tex] = the distance between first and third charge (m)Problem SolvingWe know that:[tex] \sf{q_1}[/tex]= first charge = -8.99 [tex] \sf{\mu C}[/tex] = [tex] \sf{-8.99 \times 10^{-6}}[/tex] C.[tex] \sf{q_2}[/tex]= second charge = +5.16 [tex] \sf{\mu C}[/tex] = [tex] \sf{+5.16 \times 10^{-6}}[/tex] C.[tex] \sf{q_3}[/tex]= third charge = -89.9 [tex] \sf{\mu C}[/tex] = [tex] \sf{-8.99 \times 10^{-5}}[/tex] C.[tex] \sf{r_{12}}[/tex] = the distance between first and second charge = 0.220 m = [tex] \sf{2.2 \times 10^{-1}}[/tex] m.[tex] \sf{r_{13}}[/tex] = the distance between first and third charge = 0.220 + 0.550 m = [tex] \sf{5.5 \times 10^{-1}}[/tex] m.What was asked?[tex] \sf{\sum F_{q1}}[/tex] = the net force will be felt by the first charge = ... NStep by Step:[See the attached image to know how to solve this problem]
Conclusion:So, the net force will be felt on particle [tex]\sf{q_1}[/tex] approximately 15.39 N to the right.
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I honestly have absolutely no idea where to begin solving this problem. I know the equation for the electric field (kQ/r2), but I have no idea how to work with the information given to get a net electric force. Can anyone show me where to start?
Two charged particles are located on the x axis.
What are charged particles?
Charged particles are particles that have either a positive or negative electrical charge. These particles can be found in many forms in nature, such as electrons, protons, and ions. These particles are a key component of electricity, and can be used as a source of energy. Charged particles can be found in atoms and molecules, where they interact with other particles and affect their behavior. They can also be found in plasma, a state of matter that is made up of ions and electrons.
1. Start by writing out the equation for the electric field at the origin:
E = (kQ)/r^2
2. Substitute in the values for Q, k, and r, and solve for the unknown charge:
E = (2keQ)/a^2
E = (2k(+Q))/(3a)^2
+Q = (2kE)/(3a)^2
Since the electric field has a magnitude of 2keQ / a2, the unknown charge can be either positive or negative:
+Q = ±(2kE)/(3a)^2
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What is the mass of a softball if it hit the catcher's glove traveling at
48.7 m/s2 and struck with a force of 51.3-Newton?
Answer:
To solve this problem, you can use the formula F = ma, where F is the force applied to an object, m is the mass of the object, and a is the acceleration of the object. In this case, the force applied to the object is 51.3 newtons and the acceleration of the object is 48.7 m/s2. So, you can set up the equation like this:
51.3 newtons = m x 48.7 m/s2
To solve for the mass of the object, you can divide both sides of the equation by 48.7 m/s2:
51.3 newtons / 48.7 m/s2 = m x 48.7 m/s2 / 48.7 m/s2
This simplifies to:
1.05 kg = m
So, the mass of the softball is approximately 1.05 kilograms.
Runner A is initially 2.0 mi west of a flagpole and is running with a constant velocity of 8.0 mi/h due east. Runner B is initially 9.0 mi east of the flagpole and is running with a constant velocity of 4.0 mi/h due west. How far are the runners from the flagpole when they meet?
The definition of constant velocity is the speed of a moving object that doesn't change over time. The system travelling at a constant speed is subject to zero net acceleration.
What distinguishes instantaneous velocity from average velocity?In contrast to instantaneous velocity, which is determined by slope of tangent line, average speed is defined as the shift in location (or displacement) during the course of transit.
Which of the four main claims about how a body can accelerate even when it has no velocity is untrue?Speed is the measure of velocity in magnitude. As a result, it is impossible to have two objects moving at the same velocity. The right response is therefore option A.
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calculate the kinetic energy of a 50 kg object that is moving at a speed of 12 m/s
Hello there!
Answer:
3600 J
Explanation:
[tex]E = \frac{mv^2}{2}[/tex]
So m = 50 kg and v = 12 m/s then
[tex]E = \frac{50 * 12^2}{2} J = 3600 J[/tex]
Mass = 50 kg
Velocity = 12 m/s
FindThe kinetic energy of an object.
SolutionThe formula that is used to find the kinetic energy of an object is given by :
[tex]K = \frac{1}{2}mv^{2} \\K = \frac{1}{2} X 50 X 12^{2} \\K = \frac{1}{2} X 50 X 144[/tex]
Answer= 3, 600 JIf A and B are non-zero vectors, is it possible for A×B and A·B both to be zero?
It is important to note that the dot product and cross product of any two non-zero vectors cannot both be zero therefore A×B and A·B cannot both be zero.
What are non-zero vectors?In vector space, a non-zero vector. V is a vector that differs from the zero vector in V. If u is a non-zero vector in V and an is a scalar, i.e. an element of the field over which V is defined, then au=0 is only possible if a=0.
The product of any two non-zero vectors is non-zero in this vector context. It is referred to as Hamilton's Quaternions. Over the real numbers, quaternions form a four-dimensional vector space, and their multiplication is a combination of dot product and cross product.
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A series circuit has 1200 ohms of total resistance with 12 v as the power supply. What is the total of the circuit
According to Ohm's Law, the total current flowing through a circuit with a resistance of 1200 ohms and a 12 V power supply is equal to 0.01 ampere.
Ohm's law is what?Electric current is inversely correlated with resistance according to Ohm's Law and proportionate to voltage. 'The law can be expressed mathematically as V = IR, in which 'V' is the voltage variation 'I' is the flowing current in amps, & the resistance (R) in ohm.
Given data :
Resistance = 1200 ohm
power supply = 12 V
By using Ohm's law
V = I × R
I = V / R
I = 12 / 1200
I = 0.01 ampere.
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The seatback must be adjusted far enough for your __________ to rest on the top of the steering wheel when your arm is comfortably extended.
The seatback must be adjusted far enough for your wrist to rest on the top of the steering wheel when your arm is comfortably extended.
What is the motion of steering wheel?The majority of modern automobiles, compact trucks, and SUVs have a rack and pinion steering. This transforms the steering wheel's rotational action into the linear motion which rotates the wheels and directs your course. A steering pinion, a circular gear, is used in the system to lock teeth on a rod (the rack).
What kind of basic device is a steering wheel?The basic device at work in steering, doorknobs, turbines, and bicycle wheels is a wheel and axle. An inclined plane wound around a cylinder is known as a screw.
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Bird bones have air pockets in them to reduce their weight. This also gives them an average density significantly less than that of the bones of other animals. Suppose an ornithologist weighs a bird bone in air and in water and finds its mass is 47.0 g47.0 g and its apparent mass when submerged is 3.60 g3.60 g (the bone is watertight). What mass of water is displaced
The mass of water displaced when the mass of bird bone in water and air is given, is calculated to be 43.4 g.
Given that,
Weight of bird bone in air = True weight = 47 g
Apparent weight of the bird bone when submerged = 3.6 g
The equation for apparent weight is known to be,
Apparent weight = True weight - Buoyant force
Making Buoyant force as subject, we have,
Buoyant force = 47 g - 3.6 g = 43.4 g
F b = mw × g = (m wat - m air)× g
mw = m wat - m air = 47 g - 3.6 g = 43.4 g
Thus, the mass of the water displaced is 43.4 g.
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if a bullet has a mass of 0.013 (13 gram) and moves at 1800 m/s what is it's momentum
Answer:
23.4 kg m/s
Explanation:
p = mv = 0.013kg × 1800m/s = 23.4 Kg.m/s
You bought a can of apricot jam. The contents marked as: net weight: 500g and volume: 400ml is the statement correct?why? and what is the mass of the jam you bought? calculate the density of jam in g/cm
The statement is not correct the mass of the jam is 500g and the density of the jam is 0.125gm/cm³
How to calculate the density?The statement is not correct because mass and volume cannot be used together to describe the contents of a can. Mass is a measure of how much matter is in an object and volume is a measure of the amount of space taken up by an object. So, in this case, the can of apricot jam should just be marked with the net weight in grams.The mass of the jam you bought is 500g.To calculate the density of the jam, we need to divide the mass by the volume. The density of the jam is 500g/400ml = 1.25g/ml. To put this in g/cm3, we need to divide the density by 1000, which gives us 1.25g/ml = 0.125g/cm3.To learn more about density refer to:
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How much current flows through the bottom wire in the figure(Figure 1) ?
Express your answer to two significant figures and include the appropriate units.
The current through the bottom of the wire is 0.415 A and the negative sign of the current indicates that current flows from left to right from the bottom wire.
The current through various parts of the circuit is as shown in figure.
Here we have to calculate the current I_5.
Applying Kirchoff's loop rule on each of the loops in the circuit, Consider the left triangle, for it, we have
[tex]$$\begin{aligned}& \left(I_1\right) 6 \Omega+\left(I_3\right) 12 \Omega-9 \mathrm{~V}=0 \\& \Rightarrow 6 I_1+12 I_3=9 \\& \Rightarrow I_1=\frac{9}{6}-\frac{12}{6} I_3 \\& \Rightarrow I_1=1.5-2 I_3-----(1)\end{aligned}$$[/tex]
For the centre of the triangle, we have
[tex]$$\begin{aligned}& \left(I_4\right) 24 \Omega-\left(I_3\right) 12 \Omega=0 \\& \Rightarrow I_4=\frac{12}{24} I_3 \\& \Rightarrow I_4=\frac{I_3}{2}-----(2)\end{aligned}$$[/tex]
And for the right triangle, we have
[tex]$$\begin{aligned}& \left(I_2\right) 10 \Omega+\left(I_4\right) 24 \Omega-15 \mathrm{~V}=0 \\& \Rightarrow I_2=\frac{15}{10}-\frac{24}{10} I_4 \\& \Rightarrow I_2=1.5-2.4 I_4------(3)\end{aligned}$$[/tex]
The junction rule applied at the left corner gives
I_1 =I_3+I_5
=I_5 =I_1-I_3
=1.5-2 I_3-(I_3)=1.5-3 I_3--------(4)
And applying the junction rule at the right corner,
I_4 =I_2+I_5
I_5 =I_4-I_2
=I_4-(1.5-2.4 I_4)=3.4 I_4-1.5-------(5)
Using equation (2) [tex]$I_4=\frac{I_3}{2}$[/tex], equation (5) can be written as
[tex]$$\begin{aligned}I_5 & =3.4\left(\frac{I_3}{2}\right)-1.5 \\\end{aligned}$$[/tex]==1.7 I_3-1.5
Solving equations (4) and (6), we have
1.5-3 I_3=1.7 I_3-1.5
(1.5+1.5)=(1.7+3.0) I_3
3.0=4.7 I_3
[tex]\Rightarrow I_3=\frac{3.0}{4.7} \mathrm{~A}[/tex]
Hence the current in the bottom wire is given by
[tex]I_5 & =1.7\left(\frac{3.0}{4.7}\right)-1.5 \\[/tex] =1.085-1.5 =-0.415 A
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The table shows the values Jane obtained when she measured the thickness of a steel pipe. The known thickness of the pipe is 1.32 centimeters. Which statement about her results is true?
As the values are closed to each other, Jane's measurements are precise.
What is precision?The precision of a substance is defined as the degree to which two or more measurements agree with one another.
It is incredibly precise but not always accurate to measure something if you measured it five times and get around 1.50 cm each time. Accuracy is not necessary for precision. There are various categories of precision:
Persistence: The variation that results when the conditions are maintained constant and repeated measurements are done over a brief period of time.Reproducibility: When employing the same measurement method with various instruments and operators over longer time periods, variation occurs.Learn more about precision here:
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a person who believes that nothing is known or can be known of the existence or nature of God or of anything beyond material phenomena; a person who claims neither faith nor disbelief in God.
Why are dynamic flexibility tests not used as often as?
Dynamic flexibility tests are not used as often as static flexibility tests because dynamic flexibility tests involve more subjective measurements.
Static flexibility tests are more common in comparison to dynamic tests because static flexibility tests are easier to replicate. It is difficult to standardize motion-based tests such as dynamic flexibility tests between various subjects, while static flexibility tests are simple and easy to standardize it is because the metrics are fixed and not open to interpretation.
Since, dynamic flexibility tests are hard to perform because they measure the amount of stress on a muscle that is in motion. And it is difficult to assess this among multiple people and to get an accurate picture. That is why these tests are not used as often as static flexibility tests.
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Complete question
Why are dynamic flexibility tests not used as often as static flexibility tests?
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A passenger elevator operates at an average speed of 8 m/s. if the 60th floor is 219 m above the first floor,how long does it take the elevator to go from the first floor to the 60th floor?
Answer:
27s
Explanation:
Time = distance/speed
Time = 219/8
Time = 27.375 s
A block is resting on an slope. (Figure 3) Which of the following forces act on the block? Check all that apply. O weight O static friction O normal force O kinetic friction O force of push
The correct option is A, B, and C. A block resting on a slope is Weight, static friction, and Normal force.
Static friction is a force that continues an item at relaxation. Static friction definition can be written as: The friction skilled while individuals try to move a desk-bound object on a surface, without surely triggering any relative movement between the frame and the surface on which it's miles.
Static friction is a form of friction force that acts on a body when there's no relative motion between the object and the surface. So, it can act even if the frame is in motion however there should be no relative motion.
A pressure acting on an item is said to be a static force if it no longer exchanges the scale, function, or route of that precise object. The pressure implemented to a structure acts as a load to that precise structure, that is why static force is likewise called a static load.
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Complete Question:
A block is resting on a slope. (Figure 3) Which of the following forces act on the block?
Check all that apply.
A). Weight
B). Static friction
C). Normal force
D). Kinetic friction
E). Force of push
How much work is done by gravity to cause a 12.0 kg stone to drop from a height of 22.0 m to a height of 3.00 m?
Answer:
Work done is 2280 Joules.
Explanation:
Work done can be found using the formula
[tex]W=Fs\\[/tex]
where:
W is work done ( in Joules)
F is force ( in Newtons)
s is displacement ( in Meters)
Assuming we are on Earth, we can find the force acting on the object using the mass with the following:
Mass × 10 = Force
12.0 × 10 = 120N
Moreover, we can find displacement by simply subtracting the final distance from the starting distance:
22.0m - 3.00m = 19.0m
Therefore,
[tex]W = Fs[/tex]
[tex]W = 120N[/tex] × [tex]19m[/tex]
[tex]W= 2280J[/tex]
To tighten a loose screw, a torque of 90 N.m must be applied. A wrench that is 30cm long is used to tighten it. Find the minimum force needed
The minimum force needed to apply a torque of 90 N.m to the screw using a wrench that is 30 cm long is 300 N.
What is force?
Force is a quantity that describes the interaction between two objects or systems. It is a vector quantity, meaning it has both magnitude (size or strength) and direction. Forces can cause changes in an object's motion, shape, or internal energy.
To find the minimum force needed to tighten the loose screw, we can use the equation:
F = T / r
where F is the force, T is the torque (90 N.m in this case), and r is the distance from the center of the screw to the point where the force is applied (the length of the wrench, 30 cm in this case).
We need to convert the length of the wrench from cm to meters before calculating the force:
r = 30 cm / 100 cm/m = 0.3 m
Now we can plug in the values into the equation and find the force:
F = 90 N.m / 0.3 m = 300 N
So the minimum force needed to apply a torque of 90 N.m to the screw using a wrench that is 30 cm long is 300 N.
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The speed of light is approximately $3\times10^5$ kilometers per second. How long does it take sunlight to reach Jupiter
it takes about 26 seconds for speed of light with approximate speed of 3\times10^5 kilometers per second to reach Jupiter.
To calculate the time it takes light to reach Jupiter, we need to know the distance between the Sun and Jupiter. The average distance between the Sun and Jupiter is approximately 778 million kilometers. We can use the formula distance = speed x time to calculate the time it takes for sunlight to reach Jupiter. We know that the speed of light is approximately 3x10^5 kilometers per second and the distance between the Sun and Jupiter is 778 million kilometers.time = distance / speed
time = (77810^6 km) / (310^5 km/s)
time = 26 sec (approximately)
Therefore, it takes about 26 seconds for light to reach Jupiter.
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A car moving with an initial speed v collides with a stationary car that is one- half as massive. After the collision, the first car moves moves in the same direction as before with the speed v/3. a) Find the final speed of the second car b) What type of collision is it (elastic or inelastic)
The final speed of the second car is '2v' and the collision is elastic collision.
What exactly are elastic collisions?The elastic collision is one in which the system does not experience a net loss of kinetic energy as the result of a collision. In elastic collisions, momentum & kinetic energy are both conserved.
v′ = m1v1 + m2v2 /m1+m2
v ′ = m 1 v 1 + m 2 v 2/ m 1 + m 2
in which m1 is mass of body 1, v1 is its initial speed, m2 is mass of body 2, & v2 is the final speed at which the two objects will be moving after colliding.
m1v +0 = mv/3 +m/3v2
v2 = 2v
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A cannonball is launched horizontally from the top of a 166 m
cliff with an original speed of 35.9 m/s.
Determine the horizontal displacement (in m) of the cannonball.
The horizontal displacement of the cannonball is 65.75 meter.
What is speed?Speed is characterized as the rate at which an object's position changes in any direction. The distance travelled in relation to the time it took to travel that distance is how speed is defined. Due to having no direction and only having magnitude, speed is a scalar quantity With SI unit meter/second.
original speed: u= 35.9 m/s.
acceleration due to gravity: g = 9.8 m/s².
Initial height of the cannonball = 166 m.
Hence, the horizontal displacement of the cannonball = u²/2g
= 35.9²/(2×9.8) meter
= 65.75 meter.
so, the displacement of the cannonball is 65.75 meter.
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In an electrical circuit, two resistors of 2Ω & 4Ω have been connected in series to a 6V battery. The heat dissipated by the 4Ω resistor in 5s will be
R= 2+4=6Ω
in series connection, same current and different voltage
total voltage= total current × total resistance
6=I×6
I=6/6= 1A
voltage across 4Ω resistor= 1×4=4v
Therefore, heat dissipated by the 4Ω resistor in 5s= (v²/R)×t= (4²/4)× 5 = 16/4 × 5 = 20J
given the masses of p1, p2, p3, and p4 are respectively 1kg, 2kg, 3kg, and 4kg, determine the value of the greatest horizontal range of the four projectiles, assuming they land at the same height that they fire at.
As Projectile range is independent of mass, so They all will have same range.
When an object is close to Earth's surface, its gravitational attraction is calculated using the formula F = mg, where m is the object's mass and g is a constant whose value varies depending on the location. On the surface of the earth, it averages 9.81 meters per second. And when one moves away from the surface in either an inner or an outward direction, this value drops.
Although the strength of the earth's gravitational pull changes depending on the mass of the object, the acceleration caused by it is constant (of course close to the surface). Consequently, when two items of different masses are dropped from the same height, they hit the ground simultaneously (neglect air resistance).
Any motion of a projectile simply consists of vertical and horizontal motion. Regardless of the direction the projectile is going in, gravity controls vertical motion since it only acts in that direction. Additionally, the acceleration caused by gravity is not reliant on the mass of the item. Therefore, the mass has no bearing on the vertical motion.
A projectile moves horizontally in a very straightforward manner. A projectile is not affected by horizontal forces. As a result, its horizontal velocity does not change. Additionally, the range is influenced by horizontal velocity and flying time. Flight time is influenced by vertical motion.
We now know that mass has no bearing on flying duration or range. And if you run into a projectile motion problem, simply divide the problem into horizontal and vertical motions. And handle it the same way you would a single-direction motion. greatly simplifies matters.
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Two expressions were used for the drag force experienced by a moving object in a liquid. One depended upon the speed, while the other was proportional to the square of the speed. In which types of motion would each of these expressions be more applicable than the other one
The expression for drag force that is proportional to the speed of the object is known as linear drag, while the expression that is proportional to the square of the speed is known as quadratic drag.
What is linear drag?Linear drag is generally more applicable to objects moving at low speeds, such as a swimmer moving through water. In this case, the drag force is directly proportional to the speed of the object, and the drag coefficient is constant.Quadratic drag is generally more applicable to objects moving at high speeds, such as a bullet moving through air. In this case, the drag force is proportional to the square of the speed of the object, and the drag coefficient increases as the speed of the object increases. This type of drag is caused by the formation of a turbulent boundary layer around the object, which results in a significant increase in drag force at high speeds.So linear drag is more applicable to low speed and quadratic drag is more applicable to high speed.
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How are static and passive stretching similar?
Static stretching and passive stretching are very similar in that they both involve moving the limbs into and holding the stretch posture.
Static stretching involves the individual holding their own limbs in place, while passive stretching involves a partner assisting with the movement and holding of limbs. When you stretch to your limit and hold that position, you are engaging in a form of stretching known as static stretching. Passive stretching, on the other hand, is also a form of static stretching; however, during this technique, you remain relaxed and do not actively participate in increasing your range of motion.
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A car is traveling for 1638 km for a time period of 6.5 hours. What is the average speed of the car?
Answer:
252 km/hr
70 m/s
Explanation:
s = d/t
answer in km/hr:
s = 1638 km / 6.5 hr = 252 km/hr
if you need the answer in m/s:
s = ((1638 km)(1000 m/km)) / ( (6.5 hr)(60 min/hr)(60 s/min))
= 70 m/s
A car is travelling along a horizontal road in an easterly direction. The car has a mass of 2300 kg and the drag force (friction) acting on the car is a constant 5000 N to the west.
The car is initially travelling at a speed of 10 m s‒1 and 20 s later is travelling at a speed of 14 m s‒1.
What is the gravitational force on the car?
What is the vertical normal (support/reaction) force acting on the car?
What is the horizontal force on the cars tyres from the road?
The gravitational force on the car is 23000 N, the vertical normal (support/reaction) force acting on the car is 0 N, and the horizontal force on the car's tyres from the road is 5460 N.
Mass of the car, m = 2300 kg
Force acting on the car, F = 5000 N
The initial speed of the car, u = 10 m/s
The final speed of the car, v = 14 m/s
Time, t = 20 s
The gravitational force on the car is calculated by the formula given as
F = mg
F = 2300 × 10
F = 23000 N
Now, as there is no vertical movement in the car therefore the vertical force acting on the car is 0 N.
The horizontal force on the car's tyres from the road
= force acting on the car + net force on the car
= 5000 + ma
= 5000 + 2300 × (v-u)/t
= 5000 + 2300 × (14 - 10)/20
= 5000 + 2300 × 4/20
= 5000 + 460
= 5460 N
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