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3 years ago

6. A skier starts from rest at the top of a frictionless incline of height 20.0 m. At

Physics

1 answer:

n200080 [17]3 years ago

6 0

Explanation:

a. KE at bottom = PE at top

½ mv² = mgh

v = √(2gh)

v = √(2 × 9.8 m/s² × 20.0 m)

v = 19.8 m/s

b. Work by friction = PE at top

mgμ d = mgh

d = h / μ

d = 20.0 m / 0.210

d = 95.2 m

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Calculate the index of refraction for a medium in which the speed of light is 2.1x 108 m/s. The speed of light in vacuum is 3x10

strojnjashka [21]

Answer:

n = 1.42

Explanation:

The refractive index for a medium is given by the ratio of the speed of light in vacuum to the speed of light in a medium.

$n=\dfrac{c}{v}\\\\n=\dfrac{3\times 10^8}{2.1\times 10^8}\\\\n = 1.42$

So, the refractive index of the medium is 1.42.

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If an object is not accelerating, how many forces act on it?

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Answer:

If there are no forces acting upon the object, then there is no acceleration. If there is no acceleration, then the object will move with a constant velocity.

Explanation:

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2 years ago

Read 2 more answers

Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela

NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

x-component of the tension in the rope on the right is approximately 91.93 N

y-component of the tension in the rope on the right is approximately 71.135 N

x-component of the tension in the rope on the left is -80.0 N

y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately 11.93 N

3. The net acceleration of the wagon in the horizontal direction is approximately 0.1193 m/s².

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The x and y component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ 91.93 N

y-component = 120.0 N × sin(40.0°) ≈ 77.135 N

The x and y component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = -80.0 N

y-component = 80.0 N × sin(180°) = 0.0 N

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = 11.93 N

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

$\displaystyle a_x = \frac{F_x}{m} \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}$

The acceleration of the wagon in the horizontal direction, aₓ ≈ 0.1193 m/s².

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