I need help on this question !

A constant 10.0-N horizontal force is applied to a 20.0-kg cart at rest on a level floor. If friction is negligible, what is the

Otrada [13]

Answer:

2.83 m/s

Explanation:

Given:

Mass of the cart (m) = 20.0 kg

Initial velocity of the cart (u) = 0 m/s

Final velocity (v) = ? m/s

Displacement of the cart (S) = 8.0 m

Horizontal force acting on the cart (F) = 10.0 N

Surface is frictionless. So, only horizontal force is the force acting on the cart.

Now, as per work-energy theorem, the work done by the net force acting on a body is equal to the change in the kinetic energy of the body.

Here, the work done by the horizontal force is given as:

$W_{net}=FS=(10\ N)(8.0\ m)=80\ Nm$

The change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}(20.0\ kg)(v^2-0)\\\\\Delta K=10v^2$

Now, from work-energy theorem:

$\Delta K=W_{net}\\\\10v^2=80\\\\v^2=\frac{80}{10}\\\\v=\sqrt{8}=2.83\ m/s$

Therefore, the speed of the cart when it has been pushed 8.0 m is 2.83 m/s.

5 0

3 years ago

What is an isotope?

NikAS [45]

Answer:

A

Explanation:

they are atoms of the same element that have the same number of protons but different number of neutrons.

7 0

3 years ago

The positions of four objects as a function of time are shown

KengaRu [80]

There is no movement in line C and the greatest velocity occurs at line D. The answers are:

1. 0.5 m/s

2. 0.25 m/s

3. 14m and -2m

4. -1 m/s

<h3>What is Position - time Graph ?</h3>

Position time graph is the graph of distance or displacement against time. The slope of the graph is velocity.

The given positions of four objects as a function of time are shown

on the graph to the right.

1.) The velocity of object A will be the slope m of the line A.

Slope m = Δx / Δt

m = (4 - 0) / (8 - 0)

m = 4 / 8

m = 0.5 m/s

Velocity at A = 0.5 m/s

2.) The average velocity of object B will be the slope m of the line B.

Slope m = Δx / Δt

m = (6 - 4) / (8 - 0)

m = 2 / 8

m = 0.25 m/s

The average velocity of object B is 0.25s

3.) The object moved a total distance during the first eight seconds will be 4m for A, 2m for B, and 8m for D

Total distance = 4 + 2 + 8 = 14m

It’s net displacement during the same time will be 2. That is,

Displacement = 8 - 6 = -2m

4.) The greatest speed occurred at line D. The velocity of the object moving at the greatest speed will be the slope of the line D

V = -Δx / Δt

V = -8/8

V = -1 m/s

Therefore, there is no movement in line C and the greatest velocity occurs at line D.

Learn more about velocity time graph here :brainly.com/question/769606

#SPJ1

6 0

2 years ago

A boy lifted a 50 newton rock 1 meter. How much work was done?

dem82 [27]

Answer:

The answer is 50 Nm

Explanation:

<h3><u>Given</u>;</h3>

Applied Force = 50 Newton
Total Displacement = 1 meter

Work done = ?

Here,

W = F • d

W = 50 • 1

W = 50 Nm

Thus, Work done is 50 Nm

<u>-TheUnknownScientist 72</u>

5 0

3 years ago

Read 2 more answers

Your friends sit in a sled in the snow. If you apply a force pf 75 N to them, they have an acceleration of 0.9 m/s ^ 2. What is

Elza [17]

Answer:

Mass of the sled in the snow 83.33 kg.

<u>Explanation</u>:

Given that,

Force applied to move the sled in the snow (F) = 75N

$\text { Acceleration }(a)=0.9 \mathrm{m} / \mathrm{s}^{2}$

We know that

Newton's second law of motion is

$\text { Force }=\text { mass } \times \text { acceleration }$

F = ma (Or "force" is equal to "mass" times "acceleration".)

So if we move this around we can isolate mass and get mass

$\text { Mass }=\frac{\text { force }}{\text { accelearation }}$

$\mathrm{M}=\frac{75}{0.9}$

M = 83.33 kg

Mass of the sled in the snow <u>83.33 kg.</u>

3 0

3 years ago