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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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The correct answer for this problem is c
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A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time
saul85 [17]

Answer:

  t = 0.714 s and  x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

         vₓ = 7.0 m / s

Let's find the time it takes to get to the river

         y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

        0 = y₀ + 0 - ½ g t²

        t = \sqrt{\frac{2y_o}{g}  }

        t = ra 2 2.5 / 9.8

        t = 0.714 s

the distance traveled is

       x = vₓ t

       x = 7.0 0.714

       x = 5.0 m

3 0
3 years ago
When do we say a body has goined heat​
svetoff [14.1K]

Answer:

your mom and mark me brainlyist if I was right

3 0
2 years ago
Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

8 0
3 years ago
What is the mass of a bird if it has a potential energy of 147J when flying at a height of 30 m above the ground?
Artist 52 [7]

Answer:

0.5kg

Explanation:

Given parameters:

Potential energy = 147J

Height  = 30m

Unknown:

Mass of the bird  = ?

Solution:

Potential energy is the energy due to the position of a body. Now, the expression for finding the potential energy is given as;

      P.E  = mgH

m is the mass

g is the acceleration due to gravity = 9.8m/s²

H is the height

      147  = m x 9.8 x 30

         m = 0.5kg

5 0
3 years ago
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