Can you tell Which body part does not help in the perception and production of sound in humans?

A horizontal force, F, acts on a crate weighing 8 N, which moves on a horizontal floor at constant speed. The coefficient of fri

RoseWind [281]

The answer is 24 J

F K =.25*8 N

= 2N

F = f k = 2 N

Since a = 0

W = f * s

2 N * 12 m = 24 J

The coefficient of friction is a ratio used to quantify the friction force among two gadgets when it comes to the everyday pressure this is keeping them collectively. The coefficient of friction is critical attention at some stage in material selection and floor requirement determination.

For instance, ice on steel has a low coefficient of friction – the 2 materials slide past each different without problems – whilst rubber on the pavement has an excessive coefficient of friction – the substances no longer slide past each other without difficulty.

The coefficient of friction is dimensionless and it does not have any unit. it is a scalar, meaning the direction of the force does not have an effect on the physical quantity. The coefficient of friction depends on the gadgets that are causing friction.

Learn more about the coefficient of friction here brainly.com/question/20241845

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7 0

1 year ago

The superheroine Xanaxa, who has a mass of 68.1 kg , is pursuing the 75.3 kg archvillain Lexlax. She leaps from the ground to th

Alexandra [31]

The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as

$U = mg\Delta h$

Here,

$\Delta h$ = Change in height

m = mass of super heroine

g = Acceleration due to gravity

The change in height will be,

$\Delta h = h_f - h_i$

The final position of the heroin is below the ground level,

$h_f = -16.1m$

The initial height will be the zero point of our system of reference,

$\Delta h = -16.1m-0m$

$\Delta h = -16.1m$

Replacing all this values we have,

$U = mg\Delta h$

$U = (68.1kg)(9.8m/s^2)(-16.1m)$

$U = -10744.81J$

Since the final position of the heroine is located below the ground, there will net loss of gravitational potential energy of 10744.81J

4 0

3 years ago

What is the power output of a pump which can raise 60kg of water height to height of 10m every minute

sattari [20]

Answer:

<h3>Power = Work Done/time</h3>

=> Power = 60×10×10/60

=> Power = 6000/60

=> Power = 100 Watt

Hence the power output of a pump is 100 Watts.

8 0

2 years ago

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w

Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s$

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

$\theta= \omega t$

and substituting t = 75 seconds, we find

$\theta= (0.20 rad/s)(75 s)=15 rad$

In degrees, it is

$15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}$

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

$v=\omega r$

where we have

$\omega=0.20 rad/s$

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

$v=(0.20 rad/s)(13.5 m)=2.7 m/s$

6 0

3 years ago

A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba

Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

The mass of the rubber ball is $m_r = 2 \ kg$

The initial speed of the rubber ball is $u = 3 \ m/s$

The final speed at which it bounces bank $v - 3 \ m/s$

The mass of the clay ball is $m_c = 2 \ kg$

The initial speed of the clay ball is $u = 3 \ m/s$

The final speed of the clay ball is $v = 0 \ m/s$

Generally Impulse is mathematically represented as

$I = \Delta p$

where $\Delta p$ is the change in the linear momentum so

$I = m(v-u)$

For the rubber is

$I_r = 2(-3 -3)$

$I_r = -12\ kg \cdot m/s$

=> $|I_r| = 12\ kg \cdot m/s$

For the clay ball

$I_c = 2(0-3)$

$I_c = -6 \ kg\cdot \ m/s$

=> $| I_c| = 6 \ kg\cdot \ m/s$

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

8 0

3 years ago