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Anvisha [2.4K]
3 years ago
6

29.5 inches is _______ meters. 74.9 74.9 11.6 11.6 0.749 0.749 0.116

Physics
1 answer:
Valentin [98]3 years ago
5 0

Answer:

0.749 m

Explanation:

1 m = 39.37 inches

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How can you use graphs to calculate the displacement of an object?
tekilochka [14]

Answer:

c. find the slope of the velocity time graph

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3 years ago
The time constant of an RC circuit is 2.7 s. How much time t is required for the capacitor (uncharged initially) to gain 0.63 of
gayaneshka [121]

Answer:

2.7s

Explanation:

The solution of time required is shown below:-

In the RC circuit condenser charge 63 percent of the full charge from initial time to constant time

Now, the

63% that is equal to 0.63 which is full equilibrium charge

Therefore, the time required to maintain will be Equal to time (t) constant that is 2.7s

So, the correct answer is 2.7s

8 0
3 years ago
What happens to the particles in water as the water is heated and turns to vapor? (2 points)
Naddik [55]

Answer:

The particles will more likely to move faster since they are converted from a liquid to gas.

Rules for States of Matter:

1. Solid particles always are packed close together and don't have much space to move.

2. Liquid particles have space to move around but are still packed together, but not as close as solid.

3. Gas particles are moving freely, in fact they are in the air! Gas particles are free to move wherever. For example, the air has gas particles that are constantly bumping into each other.

Let me know if I am right =)

4 0
3 years ago
1. A student lifts a box of books that weighs 185 N. The box is
aksik [14]

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

W = \Delta U = (mg) \Delta h

where

mg is the weight of the object

\Delta h is the change in height

For the box in this problem,

mg = 185 N

\Delta h = 0.800 m

Substituting into the equation, we find:

W=(185)(0.800)=148 J

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

W=Fd

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

W=(825)(35)=28875 J

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

W=Fd

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

W' = 2(28875)=57750 J

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

F = mg

where

m = 0.180 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

Solving the equation,

F=(0.180)(9.8)=1.76 N

Now we find the work done by gravity using the same formula applied before:

W=Fd

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

W=(1.76)(2.5)=4.4 J

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

\Delta h = 1.2 m

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

W=mg \Delta h

Solving for m, we find

m=\frac{W}{g \Delta h}

And substituting the numerical values, we find the mass of the box:

m=\frac{7000}{(9.8)(1.2)}=595.2 kg

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

W=mg \Delta h

Where \Delta h is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of \Delta h is the same for both of them, so the work they have done is exactly the same.

5 0
3 years ago
What color is seen when the red light is on
Dennis_Churaev [7]

Answer:

Red is seen

Explanation:

8 0
3 years ago
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