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Oksanka [162]
3 years ago
12

a waves amplitude is 0.5 meters. If the amplitude is increased to 1 metro, how does its energy change

Physics
1 answer:
marysya [2.9K]3 years ago
5 0

Answer:

The energy becomes 4 times greater.

Explanation:

We know that the energy of a wave is proportional to the square of its amplitude

E ∝ Amplitude^2

Since the original amplitude = 0.5 m

and the new amplitude becomes = 1 m

We are doubling the amplitude. This means that the new energy will be affected by a factor of 4

E_new  ∝ (2*Amplitude)^2  =

E_new ∝ 4*(Amplitude)^2  

E_new  = 4*E

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olga_2 [115]

Answer:

The answer is D

The second law of thermodynamics

5 0
3 years ago
When the motion of one or both of the particles is at an angle to the line of impact, the impact is said to be ________
Nana76 [90]

Answer: Oblique impact

Explanation:

When the motion of one or both of the particles is at an angle to the line of impact, the impact is said to be oblique impact.

On the other hand, when the directions of motion of the two colliding particles are moving along a line of impact, then it's refered to as central impact.

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2 years ago
A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of
ololo11 [35]

Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

v^2=u^2+2gh

h =\dfrac{v^2}{2g}....(I)

For second half,

We need to calculate the time

Using equation of motion

h =ut+\dfrac{1}{2}gt_{2}^2

h=0+\dfrac{1}{2}gt_{2}^2

t_{2}=\sqrt{\dfrac{2h}{g}}

Put the value of h from equation (I)

t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}

t_{2}=\dfrac{v}{g}

According to question,

t_{1}+t_{2}=10

t_{1}=t_{2}

Put the value of t₁ and t₂

\dfrac{v}{g}+\dfrac{v}{g}=10

\dfrac{2v}{g}=10

v=\dfrac{10\times g}{2}

Here, g = 10

The initial velocity is

v=\dfrac{10\times10}{2}

v=50\ m/s

Hence, The initial velocity is 50 m/s.

3 0
2 years ago
A vocalist with a bass voice can sing as low as 92 Hz.
Inessa05 [86]

Answer:

  • 3.26 x 10 to the power of 6

Explanation:

c = lambda × frequency

5 0
3 years ago
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7 0
3 years ago
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