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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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A dolphin swims 56 meters in 8 seconds and a walrus swims 30 meters in 6 seconds. Which is one has the faster speed, the dophin
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Answer:

dolphin= 7 meters/1 second      walrus= 5 meters/1 second

Explanation:

56 divided by 8 is 7

30 diviided by 6 is 5

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A 2 kg ball going 10 m/s follows a path that is perpendicular to the surface of a wall. It impacts the wall and looses 20 % of i
Ostrovityanka [42]

Answer:

If two objects make a head on collision, they can bounce and move along the same direction they approached from (i.e. only a single dimension). However, if two objects make a glancing collision, they'll move off in two dimensions after the collision (like a glancing collision between two billiard balls).

For a collision where objects will be moving in 2 dimensions (e.g. x and y), the momentum will be conserved in each direction independently (as long as there's no external impulse in that direction).

In other words, the total momentum in the x direction will be the same before and after the collision.

\Large \Sigma p_{xi}=\Sigma p_{xf}Σp

xi

​

=Σp

xf

​

\Sigma, p, start subscript, x, i, end subscript, equals, \Sigma, p, start subscript, x, f, end subscript

Also, the total momentum in the y direction will be the same before and after the collision.

\Large \Sigma p_{yi}=\Sigma p_{yf}Σp

yi

​

=Σp

yf

​

\Sigma, p, start subscript, y, i, end subscript, equals, \Sigma, p, start subscript, y, f, end subscript

In solving 2 dimensional collision problems, a good approach usually follows a general procedure:

Identify all the bodies in the system. Assign clear symbols to each and draw a simple diagram if necessary.

Write down all the values you know and decide exactly what you need to find out to solve the problem.

Select a coordinate system. If many of the forces and velocities fall along a particular direction, it is advisable to use this direction as your x or y axis to simplify calculation; even if it makes your axes not parallel to the page in your diagram.

Explanation:

i think this what your aswer is let me know if you got  it right if not i fix it or i will look in my answer book

please rate me in say thanks

4 0
3 years ago
What is 7 to the power of 5
LekaFEV [45]
7⁵ = 7 × 7× 7× 7× 7

     = 16807
5 0
3 years ago
Read 2 more answers
If a certain brand of solar panels is rated at a value of 1.50 KW/m2 , and a person needed to generate 2.50 MJ in an hour, what
alisha [4.7K]

Answer:

A=0.462\ m^2

Explanation:

Power rating of a solar panel is 1.50 KW/m²

It generates 2.50 MJ in an hour.

We need to find the area of this type of solar panel would be needed. The power pertaining to generate this energy is given by :

P=\dfrac{2.5\times 10^6}{1\ h}\\\\P=\dfrac{2.5\times 10^6}{3600\ s}\\\\P=694.44\ W

Let A be the area of the solar panel. It is calculated as follows :

\dfrac{P}{A}=1.5\times 10^3\\\\A=\dfrac{694.44}{1.5\times 10^3}\\A=0.462\ m^2

So, the required area of the solar panel is 0.462\ m^2.

4 0
3 years ago
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