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victus00 [196]
2 years ago
8

An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of th

e jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?
A) We’ll see the bell move, but we won’t hear it ring.
B) We won’t see the bell move, but we’ll hear it ring.
C) We’ll see the bell move and hear it ring.
D) We won’t see the bell move or hear it ring.
E) We’ll see the sound waves exit the vacuum pump.

Physics
2 answers:
aksik [14]2 years ago
4 0
A) We'll see the bell move, but we won't hear it ring

Because, light can travel through vacuum and sound cannot travel through vacuum. Sound waves are vibrations of particles in any media. So Sound requires a medium to travel and cannot travel in vacuum as there are no atoms or molecules to vibrate
olchik [2.2K]2 years ago
3 0

Answer:

A

Explanation:

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Force of static friction between the two surfaces

Explanation:

When two surfaces come into contact, they exert a force that resist the sliding of the two surfaces. This force is called static friction.

This force is given by the relation

                                       F_{s}=\mu_{s}\eta

Where,

                             μ - coefficient of static friction

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According to the Big Bang theory how long ago did the universe started??
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True or False <br> Most magnets are made<br> from 100% aluminum
bazaltina [42]

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4 0
2 years ago
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A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
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