PHYSICS COLLEGE

A 0.26-kg stone is held 1.2 m above the top edge of a water well and then dropped into it. The well has a depth of 4.5 m. (a) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone−Earth system before the stone is released?

Answers

Answer 1

Answer:

Complete Question

The

Answer:

$E_r = 3.058 \ J$

$E_b = -11.466 \ J$

$\Delta E_n = -14.524 \ J$

Explanation:

From the question we are told that

The mass of the stone is $m_s = 0.26 \ kg$

The height above the top of the water is $h = 1.2 \ m$

The depth of the well is $d = 4.5 \ m$

The gravitational potential of the stone before it was released is

$E_r = mgh$

substituting values

$E_r = 0.26 * 9.8 * 1.2$

$E_r = 3.058 \ J$

The gravitation potential of the stone when it reaches the bottom of the well is

$E_b = mg(- d)$

The negative shows that the potential energy of the stone as compared to the earth is reducing

substituting values

$E_b = 0.26 * 9.8 *(- 4.5)$

$E_b = -11.466 \ J$

The change in the systems gravitational potential is

$\Delta E_n = E_b - E_r$

substituting values

$\Delta E_n = -11.466 - 3.058$

$\Delta E_n = -14.524 \ J$

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Observer 1 rides in a car and drops a ball from rest straight downward, relative to the interior of the car. The car moves horizontally with a constant speed of 3.80 m/s relative to observer 2 standing on the sidewalk.a) What is the speed of the ball 1.00 s after it is released, as measured by observer 2?b) What is the direction of travel of the ball 1.00 s after it is released, as measured relative to the horizontal by observer 2?

A 81.0 kg diver falls from rest into a swimming pool from a height of 4.70 m. It takes 1.84 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.

Answers

Explanation:

The given data is as follows.

height (h) = 4.70 m, mass = 81.0 kg

t = 1.84 s

As formula to calculate the velocity is as follows.

$\nu$ = 2gh

= $2 * 9.8 m/s^(2) * 4.70 m$

= 92.12 $s^(2)$

As relation between force, time and velocity is as follows.

F = $(m * \nu)/(t)$

Hence, putting the given values into the above formula as follows.

F = $(m * \nu)/(t)$

= $(81.0 kg * 92.12 s^(2))/(1.84 s)$

= 4055.28 N

Thus, we can conclude that the magnitude of the average force exerted on the diver during that time is 4055.28 N.

A grandfather clock keeps time using a pendulum consisting of a light rod connected to a small heavy mass. With a rod of length L, the period of oscillation is 2.00 s. What should the length of the rod be for the period of the oscillations to be 1.00 s?

Answers

Answer:

The length of the rod should be

$(L)/(4) \n$

Explanation:

Period of simple pendulum is given by

$T=2\pi\sqrt{(l)/(g)} \n$

We have

$(T_1^2)/(T_2^2)=(l_1)/(l_2)\n\n(2^2)/(1^2)=(L)/(l_2)\n\nl_2=(L)/(4) \n$

The length of the rod should be

$(L)/(4) \n$

A frictionless pendulum is made with a bob of mass 12.6 kg. The bob is held at height = 0.650 meter above the bottom of its trajectory, and then pushedforward with an initial speed of 4.22 m/s. What amount of mechanical energy does the bob have when it reaches the bottom?

Answers

The answer to your question is 55

g Adjacent rows in the first part of the experiment are found to have potentials of 3.66 V and 4.22 V. If the distance between rows is found to be 0.4 cm, what is the magnitude of the electric field at the location between the rows

Answers

Answer:

$E=140V/m$

Explanation:

If the electric field is uniform, the electric field between two points at potentials $V_1$ and $V_2$ which are separated by a distance d will be given by the formula:

$E=(\Delta V)/(d)$

So in our case, we have $E=(4.22V-3.66V)/(0.004m)=140V/m$

Help me with my physics, please

Answers

The right answer would be

-20t+ 80

Which of the following types of waves is not part of the electromagnetic spectrum? A) microwaves
B) gamma rays
C) ultraviolet radiation
D) radio waves
E) sound waves

Answers

Answer: Sound Waves

Explanation:

Sound waves are the only waves on this list that are not part of the electromagnetic spectrum. This is because sound waves require a medium to travel (molecules to transmit the sound waves), while waves on the electromagnetic spectrum do not require a medium. They are able to travel through space for example, while sound would not be able to.

Sound waves (E) are not electromagnetic at all.

Microwaves, gamma rays, ultraviolet waves, and radio waves all are.

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