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3 years ago

If the force of a golf club on a golf ball is 200 N forward, what will the force of the ball on the club be?

Physics

2 answers:

AVprozaik [17]3 years ago

8 0

Answer:

it's b 200 n backward

Explanation:

Afina-wow [57]3 years ago

5 0

Newton's Third Law is every action has an equal and opposite reaction.

B. 200 N Backwards is the correct answer

The ball should put 200 N of force towards the golfer.

The ball is exerting 200 N of force towards the club as well, but the opposite reaction is that it flies away.

For more details: For each reaction there must be a opposite reaction(Which is the ball going 200 N backwards) and an equal reaction(The ball putting 200 N of force towards the golfer and the golfer)

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The most common type of mirage is an illusion that light from faraway objects seem to be reflected by a pool of water that is no

olga nikolaevna [1]

Answer:

Mirage is an optical illusion due to the refraction of light as it travels through different layers of air at different temperature.

Explanation:

On a clear hot afternoon, the sun heats up the ground, and the ground heats up the air immediately above it. The result is that the air directly above the ground is hotter than the air above it, leaving the air layers with different refractive indies. Since refractive index decreases with increase in temperature, the air immediately above the ground bends the light that travels through it from the sky, directly into our eyes. The image seen is perceived to be reflected from the ground, but in an actual sense is a direct image of the sky, giving one the impression of seeing a pool of water on the ground.

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4 years ago

Elaborate on the reason(s) that matter is said to move even as in a solid state.

Mariulka [41]

Answer:

The matter does not move in solid state but vibrates.

Explanation:

The atoms inside the matter cannot move or shift their positions without any external force but makes some small vibration movements. Generally in solids, the particles are bound by the attractive forces acting in between the atoms inside the matter.

The small vibrations that are happening inside the matter are because of the external factors like temperature. The increase in temperature raises the kinetic energy of the atoms inside and makes them move faster and this results in the vibration of the matter.

6 0

3 years ago

Read 2 more answers

How long does it take the lava bomb to reach its maximum height? Answer with three significant digits and the correct unit. A sm

wlad13 [49]

Answer:

The time taken to reach the maximum height is 3.20 seconds

Explanation:

The given parameters are;

The initial height from which the volcano erupts the lava bomb = 64.4 m

The initial upward velocity of the lava bomb = 31.4 m/s

The acceleration due to gravity, g = 9.8 m/s²

The time it takes the lava bomb to reach its maximum height, t, is given by the following kinematic equation as follows;

v = u - g·t

Where;

v = The final velocity = 0 m/s at maximum height

u = The initial velocity = 31.4 m/s

g = The acceleration due to gravity = 9.8 m/s²

t = The time taken to reach the maximum height

Substituting the values gives;

0 = 31.4 - 9.8 × t

∴ 31.4 = 9.8 × t

t = 31.4/9.8 ≈ 3.204

The time taken to reach the maximum height rounded to three significant figures = t ≈ 3.20 seconds

4 0

3 years ago

A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at halfway

$K_f$ is the final kinetic energy, at halfway

The equation can be rewritten as

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 2.7 kg is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0.54$ is the initial height

u = 0 is the initial speed

$h_f = 0.25 m$ is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

$v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s$

The total mechanical energy of the block can be calculated from the initial conditions, and it is

$E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J$

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

$E=E_e = \frac{1}{2}kx^2$

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m$

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$