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lukranit [14]
3 years ago
12

When a liquid is cooled, the kinetic energy of the particles ?

Physics
1 answer:
vodka [1.7K]3 years ago
4 0
When a liquid is cooled, the kinetic energy of the particles, DECREASES..
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Hanging from a horizontal beam are nine simple pendulums of the following lengths:
LenaWriter [7]

Answer:

Options d and e

Explanation:

The pendulum which will be set in motion are those which their natural frequency is equal to the frequency of oscillation of the beam.

We can get the length of the pendulums likely to oscillate with the formula;

L =\frac{g}{w^{2} }

where g=9.8m/s

         ω= 2rad/s to 4rad/sec

when ω= 2rad/sec

L= \frac{9.8}{2^{2} }

L = 2.45m

when  ω= 4rad/sec

L=\frac{9.8}{4^{2} }

L = 9.8/16

L=0.6125m

L is between 0.6125m and 2.45m.

This means only pendulum lengths in this range will oscillate.Therefore pendulums with length 0.8m and 1.2m will be strongly set in motion.

Have a great day ahead

8 0
4 years ago
When objects are forced to vibrate ( like when dropped onto a hard surface), they will do so at their
Marat540 [252]

They'll vibrate at their characteristic resonant frequency. That depends on the material the object is made of and its shape.


3 0
3 years ago
A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

4 0
3 years ago
If you can't throw a football what do u need to work on more
NeX [460]
Work out your arms.                                                          
3 0
3 years ago
Read 2 more answers
A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding
sveta [45]

Answer:

F1= 588 N

F2= 784 N

Explanation:

Please see the attached file.

5 0
3 years ago
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