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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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7 0
3 years ago
What should be changed to make the following sentence true? ""acrophobia is characterized by intense fear
Bess [88]

Answer:Agoraphobia

Explanation:

Acrophobia is a fear associated with heights i.e. fear of height

while Phobia characterized by intense fear is called Agoraphobia which is a anxiety disorder that can make you feel trapped so "Acrophobia" must be replace with "Agoraphobia" to make the sentence right.          

5 0
4 years ago
what relationship must exist between an applied force and the velocity of a moving obkect if uniform circular motion is to resul
Leona [35]

Answer:

See explanation

Explanation:

The centripetal force keeps an object moving in a circular orbit at constant velocity. The velocity of an object undergoing uniform motion is always tangential to the circle while the centripetal force is directed towards the center of the circle.

This now implies that the direction of the force acting on a body undergoing circular motion at constant velocity is perpendicular to the direction in which the object is being displaced.

5 0
3 years ago
PLEASE HELP ME
Ipatiy [6.2K]

Answer:

the time period is 22.63 years

Explanation:

The computation of the number of years taken is as follows:

According to the Keplers law

T^2 \propto R^3

Here T denotes the time period

And, R denotes the average distance among the sun and planet

Now let us assume for earth

R_1 & T_1 be 1 year

So, for Planet

R_2 = 8R_1 , T_2 = ?

Now applied Keplers law

(\frac{T_2}{T_1})^2 \propto (\frac{R_2}{R_1})^3\\\\(T_2)^2 = (\frac{8R_1}{R_1})^3 \times 1^2\\\\(T_2)^2 = 512\\\\T_2 = 22.674

So, the time period is 22.63 years

8 0
3 years ago
A river flows south with a speed of 2 m/s. A motorboat crosses the river due east with a velocity of 4.2 m/s relative to the wat
Rina8888 [55]

Answer:

a) Vb = 4.65m/s at 25.46° due south of east

b) t = 119s

c) d = 238m

d) 28.43° due north of east.

e) Vb = 3.69m/s due east

f) t = 135.5s

Explanation:

Velocity relative to the earth is:

V_b = V_{b/w} + V_w = [4.2,0]+[0,-2]=[4.2,-2]m/s

Vb = 4.65m/s < -25.46°

Since the distance to travel is 500m:

t = X / V_{b-x} = 500/4.2 =119s

The distance on the y-axis is given by:

Y = V_{b-y}*t=2*119=238m

If the final position is directly east from the starting position:

V_b = V_{b/w}+V_w

V_b= [V_{b-x}, 0] = [4.2*cos\beta ,4.2*sin\beta ]+[0,-2]

From the y-components:

0=4.2*sin\beta -2 Solving for β:

β=28.43°

With this angle, the velocity would be:

Vb = 4.2*cos(28.43°) = 3.69m/s

And the time it would take it to cross:

t = 500/3.69 = 135.5s

5 0
3 years ago
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