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Blababa [14]
3 years ago
5

A ski starts from rest and slides down a 22 incline 75 m long. if the coefficient of friction is 0.090, what is the ski's speed

at the base of the incline?
Physics
1 answer:
miskamm [114]3 years ago
8 0

L = length of the incline = 75 m

θ = angle of incline = 22 deg

h = height of skier at the top of incline = L Sinθ = (75) Sin22 = 28.1 m

μ = Coefficient of friction = 0.090

N = normal force by the surface of incline

mg Cosθ = Component of weight of skier normal to the surface of incline opposite to normal force N

normal force "N" balances the component of weight opposite to it hence we get

N = mg Cosθ

frictional force acting on the skier is given as

f = μN

f = μmg Cosθ

v = speed of skier at the bottom of incline

Using conservation of energy

potential energy at the top of incline = kinetic energy at the bottom + work done by frictional force

mgh = f L + (0.5) m v²

mgh = μmg Cosθ L + (0.5) m v²

gh = μg Cosθ L + (0.5) v²

(9.8 x 28.1) = (0.09 x 9.8 x 75) Cos22 + (0.5) v²

v = 20.7 m/s

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3 years ago
While finding the spring constant, if X1 = 12 cm, X2 = 15 cm, and hanging mass = 22 grams, the value of spring constant K would
Pavel [41]

Answer:

If x₁=12 cm then k=1.7985 N/m

If x₂=15 cm then k=1.4388 N/m

Explanation:

Hanging mass= 22 g=0.022 kg

Acceleration due to gravity g=9.81 m/s²

If x₁=displacement= 12 cm=0.12 m

k= spring constant

F=ma\\\Rightarrow F=0.022\times 9.81\\\Rightarrow F=0.21582\ N

\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.012\\\Rightarrow k=1.7985\ N/m\\

∴k = 1.7985 N/m

If x₂=15 cm=0.15 m

Force of the hanging mass is same however the spring constant will change

\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.015\\\Rightarrow k=1.4388\ N/m\\

∴k = 1.4388 N/m

As the mass is not changing the spring constant has to change. That means that here there are two spring one with k=1.7985 N/m and the other with k= 1.4388 N/m

4 0
3 years ago
A baseball has a mass of 145 g. A bat exerts a force of 18,400 N on the ball. What is the acceleration of the ball?
amid [387]

The correct formula to use for the situation given above is: F = MA, where F is the applied force, M is the mass of the object and A is the acceleration.

From the details given in the question, we are told that:

F = 18, 400N

M = 145 g = 145 / 1000 = 0.145 kg

A = ?

From the equation F = MA

A = F / M

A = 18,400 / 0.145 = 126,896.55 = 1.27 *10^5.

Therefore, the correct option is C.

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The volume of gas in a flexible container at a depth of 10 m/33 ft will expand to double its original volume if taken to the surface.

  • What is volume:

The amount of space that a substance or object occupies.

Volume is denoted with letter V

  • What is a gas molecule?

Generally, a gas molecule is simply defined as a combination of numbers of atoms that are connected one to another.

here, 10 m/33 feet of sea water is taken to exerts a gauge pressure that is the same pressure as the atmosphere.

This means that we add one atmosphere/bar pressure for every 10 m/33 feet you descend.

The effect of this is that,

at 10 m/33 ft absolute pressure is two ata/bar

Adding another 10 m/33 ft to this depth to make it 20 m/66 ft, puts you under under three ata/bar of pressure and so on.

Descending to a depth it means that you have double the ata/bar pressure acting on you. If you decide to resurface, the pressure on the flexible container will reduce by a factor of two.

Since volume is inversely proportional to the pressure,

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3 0
1 year ago
In the 25 ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int
Aleonysh [2.5K]

Answer:

a) <em>8.33 x 10^-6 Pa</em>

b) <em>8.23 x 10^-11 atm</em>

c) <em>1.67 x 10^-5 Pa</em>

d) <em>1.65 x 10^-10 atm</em>

<em></em>

Explanation:

Intensity of the light I = 2500 W/m^2

speed of light c<u> </u>= 3 x 10^8 m/s

a) we know that the pressure for for a totally absorbing surface is given as

P_{abs} = I/c = 2500/(3 x 10^8) = <em>8.33 x 10^-6 Pa</em>

b) 1 atm = 101325 Pa

P_{abs} = (8.33 x 10^-6)/101325 = <em>8.23 x 10^-11 atm</em>

c) for a totally reflecting surface

P_{ref} = 2I/c = twice the value for totally absorbing

P_{ref}  = 2 x 8.33 x 10^-6 = <em>1.67 x 10^-5 Pa</em>

d)  1 atm = 101325 Pa

P_{ref} = 2 x 8.23 x 10^-11  = <em>1.65 x 10^-10 atm</em>

8 0
3 years ago
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