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ohaa [14]
3 years ago
12

Prior Knowledge Questions (Do these BEFORE using the Gizmo.)A buoy is anchored to the ocean floor. A large wave approaches the b

uoy. How will the buoy move as the wave goes by?
Physics
1 answer:
dimaraw [331]3 years ago
8 0

Answer:

bounce up and down

Explanation:

Buoys are used for two main reasons, one is to let the people on land know of a big incoming wave, while the second reason is to generate electricity. When a big wave is approaching the buoy starts to bounce up and down with the strength of the smalled previous waves and then bounce very strongly up as the bigger wave passes by. This movement is combined with pistons within the buoy in order to conduct electricity.

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By what factor must we increase the amplitude of vibration of an object at the end of a spring in order to double its maximum sp
strojnjashka [21]

Answer:

A'=2A

Explanation:

According to the law of conservation of energy, the total energy of the system can be expresed as the sum of the potential energy and kinetic energy:

E=U+K=\frac{kA^2}{2}\\E=\frac{kx^2}{2}+\frac{mv^2}{2}=\frac{kA^2}{2}

When the spring is in its equilibrium position, that is x=0, the object speed its maximum. So, we have:

\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}

In order to double its maximum speed, that is v'{max}=2v_{max}. We have:

A'=\sqrt{\frac{m(v'_{max})^2}{k}}\\A'=\sqrt{\frac{m(2v_{max})^2}{k}}\\A'=\sqrt{\frac{4mv_{max}^2}{k}}\\A'=2\sqrt{\frac{mv_{max}^2}{k}}\\A'=2A

6 0
3 years ago
Hello people ~
lozanna [386]

Answer:

It’s called a conservative field.

Explanation:

I think it’s going to be the conservative field because in the question it talks about how it is able to become possible to define potential at a point in an electric field because electric field.

3 0
2 years ago
Read 2 more answers
A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (
Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants (v_{oy} = 0) zero, so let's use the equation

     y =v_{oy} t -1/2 g t²

     y= - ½ g t²

     t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

     x = vox t

     x = v₀ₓ √2y/g

c) Speeds before touching the ground

     vₓ = vox = constant

     v_{y} = v_{oy} - gt

     v_{y} = 0 - g √2y/g

    v_{y}  = - √2gy

    tan θ = Vy / vx

    θ = tan⁻¹ (vy / vx)

    θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0
3 years ago
A roller coaster is stopped on a track. When the engineer presses a launch button on the coaster, the coaster moves forward. Exp
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A roller coaster is stopped on a track. When the engineer presses a launch button on the coaster, the coaster moves forward. Explain this change in terms of balanced and unbalanced forces.
5 0
3 years ago
A roller coaster car has a mass of 290. kilograms. Starting from rest, the car acquires
Olegator [25]
Now I can actually edit my answer directly: I'm fairly sure I've got this wrong, and my mind has gone blank for how to do it, if someone could delete this that would be great and I'll think about it and see if I can figure it out!
4 0
3 years ago
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