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yuradex [85]
3 years ago
12

Which wave is carrying the most energy?

Physics
2 answers:
larisa86 [58]3 years ago
7 0

Answer : Wave 2

Explanation :

A disturbance created in a medium is termed as a wave. Energy is transported from one location to another.

A wave can be characterized by the following parameters i.e.

(1) The amplitude

(2) The Frequency

(3) The Wavelength etc.

The amplitude means the maximum displacement of the particles. It also shows the transportation of energy.

If the amplitude is higher, it shows the maximum energy carried by the wave.

And if the amplitude is smaller, it implies the less energy carried by the wave.

Wave 2 have maximum amplitude, the maximum energy is carried by it.

So, the correct option is (2)

Rainbow [258]3 years ago
3 0
The second wave because it has the most frequencies. The larger the wave, the more frequencies it has.
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A bowling ball with a mass of 7.0kg strikes a pin that had a mass of 2.0kg the pin flies forward with a velocity of 6.0m/s, and
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The conservation of momentum P states that the amount of momentum remains constant when there are not external forces.

We don't have external forces, so:

P_0 = P_1\\m_bv_{0b}+m_pv_{0p}=m_bv_{1b}+m_pv_{1p}\\

Where:

  • mb is the mass of the bowling ball
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  • v_{0b}\quad and\quad v_{0p} the initial velocities of the bowling ball and the pin.
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Solving for v0b:

v_{0b} =\dfrac{m_bv_{1b}+m_pv_{1p}- m_pv_{0p}}{m_{b}}\\\\v_{0b} =\dfrac{(7\;kg)(4\;m/s)+(2\;kg)(6\;m/s)- (2\;kg)(0 \;m/s)}{7\;kg}\\v_{0b}=\dfrac{40}{7}\;m/s\\\\\boxed{v_{0b}\approx5.71\;m/s}

<h2>R/ The original velocity of the ball was 5.71 m/s.</h2>
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A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c
den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, L_1=L-1

New time period of the pendulum is, T_1=2.81\ s

We know that, the time period of a simple pendulum of length 'L' is given as:

T=2\pi\sqrt{\frac{L}{g}}-------------- (1)

So, for the new length, the time period is given as:

T_1=2\pi\sqrt{\frac{L_1}{g}}------------ (2)

Squaring both the equations and then dividing them, we get:

\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1

Now, plug in the given values and calculate 'L'. This gives,

L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m

Therefore, the original length of the simple pendulum is 2.97 m

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