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zhuklara [117]
3 years ago
11

PLEASE HELPPPPP!!!!!!!!!

Physics
1 answer:
Sholpan [36]3 years ago
3 0
The answer is 35
The unit is ohms
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A 0.25-kg ball sits on the roof of a building that is 10 meters tall. Find the GPE. (Gravity = 9.8 on Earth)
Tasya [4]

Gravitational potential energy = mgh or mass times acceleration due to gravity times the height

Here the mass is 0.25kg, the height is 10m, and gravity is 9.8m/s^2 so...

GPE = (0.25)(10)(9.8)

GPE = 24.5 J

7 0
4 years ago
Power is the rate at which...........is done or the rate at which........... is converted from one form to another .
34kurt

Answer:

the answer is c I thought

8 0
3 years ago
A mass moves back and forth in simple harmonic motion with amplitude A and period T.
Sever21 [200]

a. 0.5 T

- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position

- The period T is the time the system takes to complete one oscillation

During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

1 T : 4 A = t : 2 A

and solving for t we find

t=\frac{(1T)(2 A)}{4A}=0.5 T

b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

- the mass takes a time of 1 T to cover a distance of 4A

we can set the following proportion:

1 T : 4 A = t : 5 A

And by solving for t, we find

t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T

6 0
3 years ago
A guitarist finds that the pitch of one of her strings is slightly flat—the frequency is a bit too low. Should she increase or d
Yuri [45]

Answer:

The guitarist should increase the tension of the string.

Explanation:

The frequency of a vibrating string is determined by fn=(n/(2L))√T/μ. So if the tension in the string increased, the rate at which it vibrates (the frequency) will also increase.

Therefore it is advisable that she increase the tension of the string.

I hope it helps, please give brainliest if it does

6 0
3 years ago
You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh
lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

F=\frac{kQq}{r^2}

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}

And the force on the charge due to the charge on the north is,

\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of 45^\circ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

#SPJ4

3 0
2 years ago
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