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2 years ago

2. Think about the energy that was needed to form the wave in the pool. Where did the energy come from?

Physics

1 answer:

patriot [66]2 years ago

6 0

Wave energy is actually a concentrated form of solar power generated by the action of the wind blowing across the surface of the oceans water which can then be used as a renewable source of energy. As the suns rays strike the Earth's atmosphere, they warm it up.

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A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

3 0

3 years ago

How can a mass be hot enough to flow but still act like a solid

Lunna [17]

The inner core is solid because it is made of very dense, or heavy, materials like iron and nickel. Even though it is very hot, these materials don't

6 0

2 years ago

A football punter wants to kick the ball so that it is in the air for 4.5 s and lands 50 m from where it was kicked. Assume that

irakobra [83]

Answer:

(a) The angle of projection is 63 degree.

(b) The velocity of projection is 24.5 m/s.

Explanation:

Height, h = 1 m

horizontal distance, d = 50 m

time, t = 4.5 s

Let the initial velocity is u and the angle is A.

(a) Horizontal distance = horizontal velocity x time

50 = u cos A x 4.5

u cos A = 11.1 .....(1)

Use second equation of motion in vertical direction

$h = u t + 0.5 gt^2\\\\- 1 = u sin A \times 4.5 - 0.5 \times 9.8\times 4.5^2\\\\u sin A = 21.8 ..... (2)$

Divide (2) by (1)

tan A = 1.97

A = 63 degree

(b) Substitute the value of A in equation (2)

u x sin 63 = 21.8

u = 24.5 m/s

7 0

3 years ago

These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the neg

Kay [80]

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

$\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C$

Calculating the common potential:

$\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\$

Calculating the charge after redistribution:

$When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2$

$\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C$

6 0

3 years ago

CAN SOMEONE PLEASE HELP ME WITH MY PHYSICS QUESTIONS? I NEED CORRECT ANSWERS ONLY!

BARSIC [14]

<h2>Right answer: acceleration due to gravity is always the same </h2><h2 />

According to the experiments done and currently verified, in vacuum (this means there is not air or any fluid), all objects in free fall experience the same acceleration, which is <u>the acceleration of gravity</u>.

Now, in this case we are on Earth, so the gravity value is $9.8\frac{m}{s^{2}}$

Note the objects experience the acceleration of gravity regardless of their mass.

Nevertheless, on Earth we have air, hence <u>air resistance</u>, so the afirmation <em>"Free fall is a situation in which the only force acting upon an object is gravity" </em>is not completely true on Earth, unless the following condition is fulfiled:

If the air resistance is <u>too small</u> that we can approximate it to <u>zero</u> in the calculations, then in free fall the objects will accelerate downwards at $9.8\frac{m}{s^{2}}$ and hit the ground at approximately the same time.

5 0

3 years ago

2. Think about the energy that was needed to form the wave in the pool. Where did the energy come from? ​

2. Think about the energy that was needed to form the wave in the pool. Where did the energy come from?