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Llana [10]
3 years ago
15

A car is being towed at a constant velocity on a horizontal road using a horizontal chain. The tension in the chain must be equa

l to the weight of the car in order to maintain a constant velocity.
True or False?
Physics
1 answer:
lukranit [14]3 years ago
3 0

Answer:

<em>The statement is false</em>

Explanation:

<u>Dynamics</u>

To analyze the situation stated in the question, we must apply some basic knowledge about dynamics, specifically Newton's laws that explain how acceleration is produced or not, depending on the forces acting on a system of particles.

The second Newton's law states the acceleration of an object of mass m can be found by knowing the net force Fn applied to it with the formula

F_n=m.a

Considering the kinetics equations, we know that

\displaystyle a=\frac{v_f-v_o}{t}

Where vf-vo is the change of velocity in time t. If an object has zero acceleration, it means its velocity doesn't change, it has constant velocity.

Under such conditions, then the net force is also zero.

We know the car is being towed horizontally at a constant velocity, meaning it has a net force equal to zero in the horizontal direction. In other words, all the horizontal forces are balanced.

We also know the chain is applying a horizontal force to tow the car, so there must be another force opposing tho that force making the car moving at a constant velocity. Although it's not mentioned, the friction force must be acting to stop the car and compensating the tension T exerted by the chain.

The friction force is given by

F_r=\mu N

Where \mu is the friction coefficient and N is the normal force, which is equal to the weight of the object W at the conditions of the problem.

Knowing the net force is zero, then

\mu W=T

If the tension is equal to the weight of the car, then

\mu T=T

simplifying

\mu=1

Thus, the only way the tension and the weight are equal is when the friction coefficient is 1, so the initial assumption is false

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fiasKO [112]
Let's use the mirror equation to solve the problem:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}
where f is the focal length of the mirror, d_o the distance of the object from the mirror, and d_i the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for d_i by using the numbers given in the text of the problem:
\frac{1}{12 cm}= \frac{1}{5 cm}+ \frac{1}{d_i}
\frac{1}{d_i}= -\frac{7}{60 cm}
d_i = -8.6 cm
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.
6 0
3 years ago
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A bird takes off from the tree and after 5 s is flying at a speed of 10 m/s. Calculate
Volgvan

Answer:

10miles/second

Explanation:

Change in velocity of the bird is expressed as the difference between the final velocity and initial velocity of the body.

Change in velocity = Final velocity - initial velocity

Since the bird takes off from the tree, the initial velocity of the bird = 0miles/sec

Final velocity = 10miles/secs

Change in velocity = 10-0

Change in velocity = 10miles/second

6 0
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Taigas, tundra and tropical rainforests share what common abiotic component?
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water

Explanation:

hope this helps

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2 years ago
Complete this sentence. Average speed is worked out from dividing _________ by ________
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Average speed is worked out from dividing distance by time.
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2 years ago
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1.   A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second.
Alla [95]

Answer:

a.Distance = 150 m

b. Displacement = 50 m

Time lapsed = 5 seconds

Explanation:

a. Distance is the change in the position of an object.

The distance covered by the car = 100 + 50

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b. Since displacement is a vector quantity,

Displacement of the car = 100 - 50

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c. Time elapsed is the time taken for the motion of the car starting from when its starts to when it stops.

Thus, the time elapsed = 4 + 1

                        = 5 seconds

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