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stich3 [128]
2 years ago
12

Two figure skaters, one weighing 625 N and the other 725 N, push off against each other on frictionless ice. If the heavier skat

er travels at 1.5 m/s, how fast will the lighter one travel?
Physics
1 answer:
lora16 [44]2 years ago
3 0

The lighter one travel with the speed of 0.574 m/sec. Force is defined as the product of mass and acceleration. Its unit is Newton.

<h3>What is force?</h3>

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

The two skaters push off against each other on frictionless ice, then torque act by the one skater on the other is equal.

P₁₂=P₂₁

F₁₂V₁ =F₂₁V₂

625 ×V₁= 725 N×  1.5 m/s

V₁=0.574 m/sec

Hence, the lighter one travel with the speed of 0.574 m/sec.

To learn more about the force, refer to the link;

brainly.com/question/26115859

#SPJ1

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Blaine steps onto a ski slope with an angle of 25°. There is a coefficient of kinetic friction of 0.15 between him and the groun
Ymorist [56]

θ = angle of the incline surface from the horizontal surface = 25⁰

μ = Coefficient of friction = 0.15

m = mass of the person = 65 kg

f_{k} = kinetic frictional force acting on the person as he slides down

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the free body diagram showing the forces acting on the person is given as


3 0
2 years ago
The width of the central maximum is defined as the distance between the two minima closest to the center of the diffraction patt
Gnom [1K]

Answer:

The value of the angle is \bf{ \sin^{-1}[h/am_{e}v]}.

Explanation:

Given:

The condition for diffraction minima is

a \sin \theta = m \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

where, a is the slit-width, \theta is the angle of incidence, m is the order number and \lambda is the wavelength of the light.

The wavelength of an electron traveling through a medium is governed by de Broglie's hypothesis.

According to de Broglie's hypothesis

\lambda &=& \dfrac{h}{p}\\               &=& \dfrac{h}{m_{e}v}

Here, h is Planck's constant, m_{e} is the mass of the electron and v is the velocity of the electron.

For first minimum m = 1.

From equation (1), we have

&& a \sin \theta = \dfrac{h}{m_{e}v}\\&or,& \theta = \sin^{-1}[\dfrac{h}{am_{e}v}]

6 0
3 years ago
A ball is launched from ground level and hits the ground again after an elapsed time of 4 seconds and after traveling a horizont
Mumz [18]

Answer:

a. It is constant the whole time the ball is in free-fall

Explanation:

If we divide the movement on its vertical and horizontal components, and we concentrate on the vertical component, let's call x-component, and analyze Newton's second's law:

\sum\overrightarrow{F_{x}}=m\overrightarrow{a_{x}}

with \sum\overrightarrow{F_{x}}, a_{x} the acceleration on horizontal direction and m the mass of the ball, because the only force acting on the object is gravity that is always vertical, there're not forces on the horizontal direction that means \sum\overrightarrow{F_{x}}=0 and by (1) that implies a_{x}=0 there's not acceleration on horizontal direction.

Because acceleration is the rate at what velocity changes and there's no acceleration, there's no change in velocity, in other words velocity is constant on horizontal direction.

6 0
2 years ago
Qualitative studies draw conclusions.<br> True<br> False
Marina86 [1]

Answer:

Hey there!

This is false. A qualitative study is about how a thing looks, not based on any mathematical or scientific data. Quantitative studies, on the other hand, draw conclusions.

Let me know if this helps :)

3 0
3 years ago
Assume you built a really big machine that could launch the projectile a “significant” distance; for instance, several hundred m
Gwar [14]
<span>First, a problem would be the sheer amount of wind resistance. If an object travels as far as even just one hundred miles it could encounter different wind patterns that could change the trajectory of the object. Second would be the size of the projectile. This creates a problem because the bigger it is, the more momentum it could potentially pick up, given that it is not too big to complete the distance. This is another problem with size, how far the projectile can actually travel. You would have to actually calculate the ideal size of said object to make sure it could actually make the distance you're looking for. hope it helps:)</span>
3 0
2 years ago
Read 2 more answers
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