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Scilla [17]
3 years ago
13

A skier of mass 82.9 kg starts from rest at the top of a frictionless incline of height 20 m. At the bottom of the incline, the

skier encounters a horizontal surface. The skier travels on the horizontal surface a distance of 95.2 meters before coming to a rest. What is the magnitude of the (constant) friction force (in N) that the rough horizontal surface exerts on the skier
Physics
1 answer:
ehidna [41]3 years ago
7 0

Answer: 170.67 N

Explanation:

Given

Mass of skier is m=82.9\ kg

Height of the inclination is h=20\ m

Here, the potential energy of the skier is converted into kinetic energy which is consumed by the friction force by applying a constant force that does work to stop the skier.

\Rightarrow mgh=F\cdot x\quad \quad [\text{F=constant friction force}]\\\\\Rightarrow 82.9\times 9.8\times 20=F\cdot 95.2\\\\\Rightarrow F=\dfrac{16,248.4}{95.2}\\\\\Rightarrow F=170.67\ N

Thus, the horizontal friction force is 170.67 N.

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A spherical raindrop 2.5 mm in diameter falls through a vertical distance of 3900 m. Take the cross-sectional area of a raindrop
Karolina [17]

Answer:

276.62 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)

Equation of motion

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 3900+0^2}\\\Rightarrow v=276.62\ m/s

<u>Neglecting air drag</u> the velocity of the spherical drop would be 276.62 m/s

5 0
3 years ago
A motor does a total of 480 joules of work in 5.0 seconds to lift a 12-kilogram block to the top of a rampThe average power deve
Flauer [41]

Answer:

there it is fella

we neglect the mass data

3 0
3 years ago
a physics student throws a stone horizontally off a cliff. one second later, he throws a second identical stone in exactly the s
Virty [35]

The second stone hits the ground exactly one second after the first.

The distance traveled by each stone down the cliff is calculated using second kinematic equation;

h = v_0_yt + \frac{1}{2} gt^2

where;

  • <em>t is the time of motion </em>
  • <em />v_0_y<em> is the initial vertical velocity of the stone = 0</em>

h = \frac{1}{2} gt^2

The time taken by the first stone to hit the ground is calculated as;

t_1 = \sqrt{\frac{2h}{g} }

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

t_2 = \sqrt{\frac{2h}{g} } + 1

t_2 = t_1 + 1

Thus, we can conclude that the second stone hits the ground exactly one second after the first.

"<em>Your question is not complete, it seems be missing the following information;"</em>

A. The second stone hits the ground exactly one second after the first.

B. The second stone hits the ground less than one second after the first

C. The second stone hits the ground more than one second after the first.

D. The second stone hits the ground at the same time as the first.

Learn more here:brainly.com/question/16793944

8 0
3 years ago
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Alexxx [7]
1.)
Velocity is in m/s, and acceleration is in m/s^2 like you said. Because of this, we can calculate this by dividing the speed by the time it took to get to that speed.
(20 meters/second) / 10 seconds = 2 meters/ second^2

2.)
Same thing with the first one.
(100 meters/second) / 4 seconds = 25 meters / seconds^2
7 0
3 years ago
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Thermal energy transfers throughout the water in the beaker. Describe and explain how this happens?
ziro4ka [17]
The transfer of heat between the bottom surface of the beaker and water inside it is due to Convection phenomenon. When a fluid, such as air or a liquid, is heated and then travels away from the source, it carries the thermal energy along. ... The fluid above a hot surface expands, becomes less dense, and rises
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3 years ago
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