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Blababa [14]
2 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]2 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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4 0
3 years ago
g suppose a spring with spring constant of 50 N/m is hanging from the ceiling. You hang 2.0 kg mass from the spring. How far is
adelina 88 [10]

Answer:

The extension of the spring is 0.392 m.

Explanation:

Given;

spring constant, k = 50 N/m

mass attached to the spring, m = 2.0 kg

let the extension of the spring = x

The extension of the spring is calculated by applying Hook's law;

F = kx

mg = kx

x = \frac{mg}{k} \\\\x = \frac{2 \times 9.8}{50} \\\\x = 0.392 \ m

Therefore, the extension of the spring is 0.392 m.

8 0
3 years ago
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8 0
3 years ago
A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length ℓ and of negligible
mr Goodwill [35]

Answer:

a)

mv l

b)

\frac{M }{(M + m)}

Explanation:

Complete question statement is as follows :

A wooden block of mass M resting on a friction less, horizontal surface is attached to a rigid rod of length ℓ and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

(a) What is the angular momentum of the bullet–block system about a vertical axis through the pivot? (Use any variable or symbol stated above as necessary.)

(b) What fraction of the original kinetic energy of the bullet is converted into internal energy in the system during the collision? (Use any variable or symbol stated above as necessary.)

a)

m = mass of the bullet

v = velocity of the bullet before collision

r = distance of the line of motion of bullet from pivot = l

L = Angular momentum of the bullet-block system

Angular momentum of the bullet-block system is given as

L = m v r

L = mv l

b)

V = final velocity of bullet block combination

Using conservation of momentum

Angular momentum of bullet block combination = Angular momentum of bullet

(M + m) V l = m v l\\V =\frac{mv}{(M + m)}

K_{o} = Initial kinetic energy of the bullet

Initial kinetic energy of the bullet is given as

K_{o} = (0.5) m v^{2}

K_{f} = Final kinetic energy of bullet block combination

Final kinetic energy of bullet block combination is given as

K_{f} = (0.5) (M + m) V^{2}

Fraction of original kinetic energylost is given as

Fraction = \frac{(K_{o} - K_{f})}{K_{o}} = \frac{((0.5) m v^{2} - (0.5) (M + m) V^{2})}{(0.5) m v^{2}}

Fraction = \frac{(m v^{2} - (M + m) (\frac{mv}{(M + m)})^{2})}{m v^{2}} = \frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}

Fraction = \frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}\\ \frac{M }{(M + m)}

6 0
3 years ago
An accessory such as a coupling that performs a mechanical rather than an electrical function is called?
Assoli18 [71]
Fitting is the answer
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