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3 years ago

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.

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Normal atmospheric pressure at sea level is ?

iris [78.8K]

1 bar or 100 000 pascales. Or 1020 hPa. It kinda differentiates.

5 0

3 years ago

4) Block A has a mass of 3kg and velocity of 13m/s, catching up with a second block B that has a mass of 3kg and is moving with

serious [3.7K]

Answer:

Option A is the correct answer.

Explanation:

Here momentum is conserved.

That is $\left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}$

Substituting values

$3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s$

Speed of block A after collision = 10 m/s

Option A is the correct answer.

5 0

3 years ago

The condition required to work to be done

Dvinal [7]

Answer:

work=f(costheta)

Explanation:

work is done when a force acts on a body and displaces it on the direction of force

6 0

3 years ago

The body went 450 meters within 30 seconds of starting the movement. In what time did the first 50 meters go?

Andrew [12]

3.33 seconds.

<u>Explanation:</u>

We can find the speed of the body using the formula,

Speed = Distance traveled in meters / time taken in seconds

= 450 m / 30 seconds

= 15 m/s

So per second the distance traveled by the body is 15 m.

So time needed to travel 50 m can be found as,

time = distance/speed

= 50 m / 15 m /s

= 3.33 s

8 0

3 years ago

In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted?

agasfer [191]

Answer:

The frequency of the photon is $3.069\times10^{14}\ Hz$ .

Explanation:

Given that,

Energy $E=4.07\times10^{-19}\ J$

We need to calculate the energy

Using relation of energy

$E_{4}-E_{2}=\Delta E$

Where, $\Delta E$ = energy spacing

$4h\nu-2h\nu=4.07\times10^{-19}$

$\nu=\dfrac{4.07\times10^{-19}}{2h}$

Put the value of h into the formula

$\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}$

$\nu=3.069\times10^{14}\ Hz$

Hence, The frequency of the photon is $3.069\times10^{14}\ Hz$ .

4 0

3 years ago