All categories

3 years ago

A car accelerates uniformly from rest to a speed of 7.6 m/s in 4.6 s. How far does the car travel in that time?

Physics

1 answer:

uysha [10]3 years ago

4 0

Answer:

The car will travel a distance of 17.45 meters.

Explanation:

Given:

Initial velocity $(V_i)$ = 0

Final velocity $(V_f)$ = 7.6 m/s

Time taken = 4.6 s

Acceleration = (Final velocity - Initial Velocity )/time

$a=\frac{(V_f-Vi)}{t}= \frac{7.6-0}{4.6}=1.65\ ms^-2$

We have to calculate total distance traveled by the car.

Let the distance traveled be 'd'

Equation of motion:

$d=V_i(t)+\frac{at^2}{2}$

Plugging the values.

⇒ $d=V_i(t)+\frac{at^2}{2}$

⇒ $d=0+\frac{1.65*(4.6)^2}{2}$

⇒ $d=17.45\ m$

The car will travel a distance of 17.45 meters for the above case.

You might be interested in

A 40.0 N crate is pulled up a 5.0 m inclined plane. The angle of inclination of the plane to horizontal is 37o. If there is a co

Digiron [165]

The net gain in potential energy by the crate is 120 J.

<h3>What is potential energy?</h3>

Potential energy is the energy that an object has stored in it as a result of its position or condition. Potential energy exists in a stretched spring, a book held above your head, and a bicycle on top of a hill. The joule, which is represented by the letter "J," is the standard unit for calculating potential energy.

<h3 /><h3>Calculation:</h3>

We use the idea of potential energy, which is given by the equation,

PE = mgh

where,

m = mass of the crate

g = gravitational acceleration

h = height

Given,

mg = 40 N

L = 5 m

angle of inclination, ∅ = 37°

h = L sin∅

h = 5 (sin 37°)

h = 5 (3/5)

h = 3 m

Put the values in the above formula,

PE = mgh

PE = 40 (3)

PE = 120 J

Therefore, the net gain in potential energy by the crate is 120 J.

<h3 />

Learn more about the calculation of work and force here:

brainly.com/question/28043403

#SPJ4

4 0

2 years ago

An ___ wave carries energy through matter

garik1379 [7]

A sound wave is the answer

4 0

3 years ago

Calculate The pressure produced by a force of 392 N acting on an area of 8.0 m^2

Nonamiya [84]

Answer:

Explanation:

500-0.05=499.95

500-100+5=405

100+100=10000

5 0

2 years ago

A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

Olegator [25]

Answer:

366.90149 m/s

923.821735 J

324.734 J

Initial Kinetic energy > Final kinetic energy

Explanation:

$m_1$ = Mass of block = 0.072 kg

$m_2$ = Mass of bullet = 4.67 g

$u_1$ = Initial Velocity of block = 0

$u_2$ = Initial Velocity of bullet = 629 m/s

$v_1$ = Final Velocity of block = 17 m/s

$v_2$ = Final Velocity of bullet

In this system the linear momentum is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.072\times 0+4.67\times 10^{-3}\times 629-0.072\times 17}{4.67\times 10^{-3}}\\\Rightarrow v_2=366.90149\ m/s$

Final Velocity of bullet is 366.90149 m/s

The initial kinetic energy

$K_i=\frac{1}{2}m_2u_2^2\\\Rightarrow K_i=\frac{1}{2}4.67\times 10^{-3}\times 629^2\\\Rightarrow K_i=923.821735\ J$

The final kinetic energy

$K_f=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2\\\Rightarrow K_f=\frac{1}{2}4.67\times 10^{-3}\times 366.90149^2+\frac{1}{2}0.072\times 17^2\\\Rightarrow K_f=324.734\ J$

Initial Kinetic energy > Final kinetic energy

3 0

3 years ago

Where is the switch located on this diagram?

Fed [463]

For this case, the switch is located at point B of the diagram.
Remember that point D is the universal symbol for resistance.
In A what you have is a source of power and in C what you have is a cable.
Therefore, the answer for this case is B.

6 0

3 years ago