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Vikki [24]
3 years ago
8

A pendulum is used in a large clock. The pendulum is raised to its maximum height of 0.3 m and has a mass of 5.2 KG. What is the

speed of the pendulum as it reaches its lowest point?
A. 4.2 m/s
B. 3.1 m/s
C. 1.9 m/s
D. 2.4 m/s
Physics
1 answer:
umka21 [38]3 years ago
8 0

Answer:

2.4m/s, This is what I got.

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A moving particle is subject to conservative forces only. When its kinetic energy decreases by 10 J, what happens to its mechani
satela [25.4K]

Answer:

Its mechanical energy is the same.

Explanation:

If forces are only conservative, the mechanical energy will be the same.

It can be different if energy get transformed in another kind of energy like elastic energy for example, although the amount of energy is always the same.

If we just have mechanical energy not geting transformed we have:

Em=K+U

Em: Mechanical energy

K: Kinetic energý

U: Potential energy

Then if Kinetic energy decreases 10J, Potential energy will grow up 10J to keep the same amount of mechanical energy.

8 0
3 years ago
What is the frequency of a wave that has a wavelength of 20 cm and a speed of 10 m/s?
anygoal [31]
If you look at the picture you will see how, but the answer is 0.5 Hz.

6 0
4 years ago
A 320-g ball and a 400-g ball are attached to the two ends of a string that goes over a pulley with a radius of 8.7 cm. Because
Gnom [1K]

To solve this problem, it is necessary to apply the concepts related to force described in Newton's second law, so that

F = ma

Where,

m = mass

a = Acceleration (Gravitational acceleration when there is action over the object of the earth)

Torque, as we know, is the force applied at a certain distance, that is,

\tau = F*d

Where

F= Force

d = Distance

Our values are given as,

m_1 = 0.32Kg

m_2 = 0.4Kg

d = 8.7*10^{-2}m

Since the system is in equilibrium the difference of the torques is the result of the total Torque applied, that is to say

\tau = T_2-T_1

\tau = F_2*d-F_1*d

\tau = m_2g*d-m_1*g*d

\tau = (m_2-m_1)g*d

\tau = (0.4-0.32)(9.8)(8.7*10^{-2})

\tau = 0.068N\cdot m

Therefore the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released is \tau = 0.068N\cdot m

8 0
3 years ago
If you wanted to increase the gravitational force between two objects what would you do
adelina 88 [10]
Double the force on the object
6 0
4 years ago
Read 2 more answers
A ball is thrown vertically downwards at speed vo from height h. Draw velocity vs. time & acceleration vs. time graphs. In t
yuradex [85]

Answer:

Explanation:

let the ball is thrown vertically downwards with velocity u.

So, initial velocity, = - u (downwards)

acceleration = - g (downwards)

let the velocity is v after time t.

use first equation of motion

v = u + at

- v = - u - gt

v = u + gt

So, it is a straight line having slope g and y intersept is u.

The graph I shows the velocity - time graph.

Now the value of acceleration remains constant and it is equal to - 9.8 m/s^2.

So, acceleration time graph is a starigh line parallel to time axis having slope zero.

the graph II shows the acceleration - time graph.

Use III equation of motion to find the final speed in terms height.

v^{2}=u^{2}+2gh

And the time is

v = u + gt

t=\frac{v-u}{g}

4 0
4 years ago
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