All categories

2 years ago

PLEASE HELP ME,, I WOULD BE SO HAPPY

Physics

2 answers:

Juliette [100K]2 years ago

6 0

Answer:

Energy is force times distance. For your problem, no matter how long you push, the wall still goes nowhere, so there is no obvious energy transfer. so in conclusion, you actually didn't do anything :(

Explanation:

Goshia [24]2 years ago

6 0

Answer is
Work = (Distance / Time) Example 35 Miles per Gallon.

Since the wall did NOT move you did did not actually do any work.

I hope this helps you.

You might be interested in

To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

7 0

3 years ago

What variable is represented on the y-axis?

gladu [14]

Distance/ Time which means Distance is on horizontal and time is on vertical

8 0

3 years ago

What are three disadvantages of making and using plastic?

Gemiola [76]

Harmful to the environment
non bio degradable
harmful chemicals <span />

4 0

2 years ago

Read 2 more answers

A worker pushes a 50 kg crate a distance of 7.5 m across a level floor. He

Taya2010 [7]

a) 73.5 N

b) 551.3 J

c) -551.3 J

d) 0 J

e) 0 J

f) 0 J

g) 0 J

Explanation:

There are two forces acting on the crate:

- The push of the worker, F, in the forward direction

- The frictional force, $F_f=\mu mg$ , in the backward direction, where

$\mu=0.15$ is the coefficient of friction

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration due to gravity

According to Newton's second law of motion, the net force on the crate must be equal to the product of mass and acceleration, so:

$F-F_f=ma$

However, the crate here is moving with constant velocity, so its acceleration is zero:

$a=0$

So the previous equation becomes:

$F-F_f=0$

And we can find the magnitude of the applied force:

$F=F_f=\mu mg=(0.15)(50)(9.8)=73.5 N$

The work done by the applied force on the crate is

$W_F=Fd cos \theta$

where:

F is the magnitude of the force

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here we have:

F = 73.5 N

d = 7.5 m

$\theta=0^{\circ}$ (the force is applied in the same direction as the displacement)

Therefore,

$W_F=(73.5)(7.5)(cos 0^{\circ})=551.3 J$

The work done by friction on the crate is:

$W_{F_f}=F_f d cos \theta$

where in this case:

$F_f=73.5 N$ is the magnitude of the force of friction

d = 7.5 m is the displacement of the crate

$\theta=180^{\circ}$ , because the displacement is forward and the force of friction is backward, so they are in opposite direction

Therefore, the work done by the force of friction is:

$W_{F_f}=(73.5)(7.5)(cos 180^{\circ})=-551.3 J$

To find the normal force, we analyze the situation of the force along the vertical direction.

We have two forces on the vertical direction:

- The normal force, N, upward

- The force of gravity, $mg$ , downward, where

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration due to gravity

Since the crate is in equilibrium in this direction, the vertical acceleration is zero, so the two forces balance each other:

$N-mg=0\\N=mg=(50)(9.8)=490 N$

The work done by the normal force is:

$W_N=Nd cos \theta$

In this case, $\theta=90^{\circ}$ , since the normal force is perpendicular to the displacement of the crate; therefore, the work done is

$W_N=(490)(7.5)(cos 90^{\circ})=0$

The work done by the gravitational force is:

$W_g=F_g d cos \theta$

where:

$F_g=mg=(50)(9.8)=490 N$ is the gravitational force

d = 7.5 m is the displacement of the crate

$\theta=90^{\circ}$ is the angle between the direction of the gravitational force (downward) and the displacement (forward)

Therefore, the work done by gravity is

$W_g=(490)(7.5)(cos 90^{\circ})=0 J$

The total work done on the crate can be calculated by adding the work done by each force:

$W=W_F+W_{F_f}+W_N+W_g$

where we have:

$W_F=+551.3 J$ is the work done by the applied force

$W_{F_f}=-551.3 J$ is the work done by the frictional force

$W_N=0$ is the work done by the normal force

$W_g=0$ is the work done by the force of gravity

Substituting,

$W=+551.3+(-551.3)+0+0=0 J$

So, the total work is 0 J.

According to the work-energy theorem, the change in kinetic energy of the crate is equal to the work done on it, therefore:

$W=\Delta E_K$

where

W is the work done on the crate

$\Delta E_K$ is the change in kinetic energy of the crate

In this problem, we have:

$W=0$ (total work done on the crate is zero)

Therefore, the change in kinetic energy of the crate is:

$\Delta E_K = W = 0$

5 0

3 years ago

What is Kinetic Energy and Potential energy❓<br><br>Ty!!

diamong [38]

The energy which is possessed by an object because of its motion is called Kinetic energy.

Example:

Wind mills. Energy is formed by the motion of wind mills, so it is kinetic energy.

-----------------------------------------------------------------------

The energy which is possessed by an object because of its position is called Potential energy.

Example:

A raised weight

-----------------------------------------------------------------------

6 0

1 year ago