1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in
valkas [14]

Answer:

Volume of the sample: approximately \rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}.

Average density of the sample: approximately \rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}.

Assumption:

  • \rm g = 9.81\; N \cdot kg^{-1}.
  • \rho(\text{water}) = \rm  1.000\times 10^{3}\; kg \cdot m^{-3}.
  • Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to \rm 17.50 - 11.20 = 6.30\; N.

That's also equal to the weight (weight, m \cdot g) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with g.

\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg.

Assume that the density of water is \rho(\text{water}) = \rm  1.000\times 10^{3}\; kg \cdot m^{-3}. To the volume of water displaced from its mass, divide mass with density \rho(\text{water}).

\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}.

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}.

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with g.

\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg.

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}.

3 0
3 years ago
What is the main difference between protons and neutrons?
Studentka2010 [4]
Protons have a positive energy charge, while neutrons have a neutral charge
4 0
3 years ago
Read 2 more answers
According to one set of measurements, the tensile strength of hair is 196 MPa , which produces a maximum strain of 0.380 in the
slega [8]

Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

A=\dfrac{\pi}{4}\times d^2

Put the value into the formula

A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2

A=1.96\times10^{-9}\ m^2

We need to calculate the magnitude of the force

Using formula of force

F=\sigma A

Put the value into the formula

F=196\times10^{6}\times1.96\times10^{-9}

F=0.38416\ N

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

strain=\dfrac{\Delta l}{l_{0}}

\Delta l=strain\times l_{0}

Put the value into the formula

\Delta l=0.380\times l_{0}

Length after expansion is 12 cm

We need to calculate the original length

Using formula of length

l=l_{0}+\Delta l

Put the value into the formula

I=l_{0}+0.380\times l_{0}

l=1.38l_{0}

l_{0}=\dfrac{l}{1.38}

l_{0}=\dfrac{12\times10^{-2}}{1.38}

l_{0}=0.0869\ m

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

4 0
3 years ago
Assume your mass is 60 kg. The acceleration due to gravity is 9.8 m/s 2 . How much work against gravity do you do when you climb
Andre45 [30]

Answer:

W=1705.2 J

Explanation:

Given that

mass ,m= 60 kg

Acceleration due to gravity ,g= 9.8 m/s²

Height ,h= 2.9 m

As we know that work done by a force given as

W = F . d

F=force

d=Displacement

W=work done by force

Now by putting the values

F= m g (Acting downward  )

d= h  (Upward)

W= m g h    ( work done against the force)

W= 60 x 9.8 x 2.9 J

W=1705.2 J

Therefore the answer will be 1705.2 J.

8 0
3 years ago
A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant dece
seraphim [82]

Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s

final velocity of the car when it stops, v = 0

Apply the following kinematic equation to solve for the deceleration of the car.

v² = u² + 2as

0 = 88.02² + (2 x 450)a

-900a = 7747.5204

a = -7747.5204 / 900

a = -8.61 ft/s²

|a| = 8.61 ft/s²

Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .

4 0
2 years ago
Other questions:
  • A ball which is dropped from the top of a building strikes the ground with a speed of 30m/s. Assume air resistance can be ignore
    9·1 answer
  • An object is thrown vertically upward at 27.1 m/s. The velocity of the object 3.4 seconds later is ____ m/s. Round your answer t
    10·1 answer
  • An Earth satellite needs to have its orbit changed so the new orbit will be twice as far from the center of Earth as the origina
    7·1 answer
  • Billy pushes 15.0 kg block at a rate of 2.5 m/s/s eastward. What is the net force Billy applied to the block?
    7·2 answers
  • A toboggan approaches a snowy hill moving at 12.4 m/s. The coefficients of static and kinetic friction between the snow and the
    9·1 answer
  • This player along with the point guard, are considered the leaders of the team.
    5·2 answers
  • Ik this is a lot but I am super confused, could you pls helpppp
    14·2 answers
  • A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 25.0 N/m. The
    10·1 answer
  • You push a cart with mass 30 kg forward, giving it an acceleration of 5 m/s2How much force did you apply? O A. 0.17 N O B. 35 N
    10·1 answer
  • Answer all of these questions and you will get the brainlist
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!