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elena-14-01-66 [18.8K]
3 years ago
5

A ringing bell sends sound waves in all directions places sides

Physics
1 answer:
Strike441 [17]3 years ago
4 0

Answer:

direction

Explanation:

because particles surround the bell, so when the bell vibrates, it causes particles surrounding it to vibrate back and forth vigorously. as these particles vibrate they collide with the neighbouring particles, passing on the energy.

hope this is what you are asking, if not please report it so that someone else gets to try it.

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The kinetic energy of an object is directly proportional to its mass, and to the square of its velocity.

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Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa
anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

Explanation:

Given that,

First charge q_{1}= 2.9\mu C

Second chargeq_{2}= 2.9\mu C

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

E_{1}=\dfrac{kq}{r'^2}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}

E_{1}=52200=52.2\times10^{3}\ N/C

For E₂,

Using formula of electric field

E_{1}=\dfrac{kq}{r^2}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}

E_{2}=104400=104.4\times10^{3}\ N/C

We need to calculate the horizontal electric field

E_{x}=E_{1}\cos\theta

E_{x}=52.2\times10^{3}\times\cos45

E_{x}=36910.97=36.9\times10^{3}\ N/C

We need to calculate the vertical electric field

E_{y}=E_{2}+E_{1}\sin\theta

E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45

E_{y}=141310.97=141.3\times10^{3}\ N/C

We need to calculate the net electric field

E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

E_{net}=146038.69\ N/C

E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

\theta=75.36^{\circ}

Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

4 0
3 years ago
A child accidentally drops a toy from his apartment window. It falls for 1.05 seconds. What is the velocity of the toy just befo
AVprozaik [17]

Sorry!

This cannot be answered. We don't have weight, height, etc.

6 0
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On a playground, there is a merry‑go‑round. In order to get it moving, Bonnie applies a force of 31 N31 N . The merry‑go‑round m
nika2105 [10]

Answer:

The magnitude of the torque is 263.5 N.

Explanation:

Given that,

Applied force = 31 N

Distance from the axis = 8.5 m

She applies her force perpendicularly to a line drawn from the axis of rotation

So, The angle is 90°

We need to calculate the torque

Using formula of torque

\tau=Fd\sin\theta

Where, F = force

d = distance

Put the value into the formula

\tau=31\times8.5\sin90

\tau= 263.5\ N

Hence, The magnitude of the torque is 263.5 N.

4 0
2 years ago
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