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Kisachek [45]
3 years ago
7

A train travels 55 km, south along a straight track in 34 minutes.

Physics
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

96.5 km/h south

Explanation:

The average velocity is given by:

v=\frac{d}{t}

where

d is the displacement

t is the time

We have:

d = 55 km

t = 34 min \cdot \frac{1}{60}=0.57 h

Therefore, the average velocity is

v=\frac{55}{0.57}=96.5 km/h

And the direction is the same as the  displacement (south)

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1. 23°

2. 65°

Explanation:

1.

The law of reflection provides that for the same medium, the angle of reflection equals the angle of incidence.

Similarly, here, angle of incidence and reflection will be 46°÷2=23°

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3 years ago
A compressed vertical spring stores 40 J of potential energy. The spring has a 0.1-kg stoneresting on it. The spring is released
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(A) 40 J

Explanation:

The initial potential energy stored in the spring is:

U=40 J

this energy is stored in the spring when the spring is compressed by a certain amount \Delta x, such that the elastic potential energy of the spring is

U=\frac{1}{2}k(\Delta x)^2

where k is the spring constant. On the contrary, when it is at rest, the kinetic energy of the stone is zero:

K=\frac{1}{2}mv^2 = 0

because the speed is zero (v=0).

When the spring is released, the spring returns to its equilibrium position, so that

\Delta x = 0

and

U=0

so the elastic potential energy becomes zero: so the total energy must be conserved, this means that all the potential energy has been converted into kinetic energy of the spring, so 40 J.

(B) 40 J

When the stone starts its motion, its kinetic energy is 40 J:

K = 40 J

While its gravitational potential energy is zero:

U = mgh = 0

where m is the mass of the stone, g is the gravitational acceleration, and h=0 is the height when the stone is thrown up.

As the stone goes up, its gravitational potential energy increases, since h in the formula increases; this means that the kinetic energy decreases, since the total energy must be constant.

When the stone reaches its maximum height, its speed becomes zero:

v = 0

This means that

K = 0

And so all the kinetic energy has been converted into gravitational potential energy, therefore

U = 40 J

(C) 40.8 m

At the maximum height of the trajectory of the stone, we have that the gravitational potential energy is

U=mgh = 40 J

where

m = 0.1 kg is the mass of the stone

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

h is the maximum height

Solving the formula for h, we find:

h=\frac{U}{mg}=\frac{40 J}{(0.1 kg)(9.8 m/s^2)}=40.8 m

(D) The initial compression of the spring must be increased by a factor \sqrt{2}

Here we want to double the maximum height reached by the stone:

h' = 2h

In order to do that, we must double its gravitational potential energy:

U' = 2U

This means that the initial potential energy stored in the spring must also be doubled, so

U' = 80 J

The elastic potential energy of the spring is

U' = \frac{1}{2}k(\Delta x)^2

We see that the compression of the spring can be rewritten as

\Delta x = \sqrt{\frac{2U'}{k}}

And we see that \Delta x is proportional to the square root of the energy: therefore, if the energy has doubled, the compression must increase by a factor \sqrt{2}.

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Answer:

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