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valentina_108 [34]
3 years ago
10

A block is placed on an inclined plane with an angle of inclination θ (in degrees) with respect to horizontal. The coefficient o

f static friction between the block and the inclined plane is 0.4. For what maximum value of θ will the block remain stationary on the inclined surface?
Physics
1 answer:
hjlf3 years ago
6 0

Answer:

The maximum value of θ that will cause the block to remain stationary on the inclined surface is 21.8°

Explanation:

Given;

coefficient of static friction, μ = 0.4

for the block to remain stationary on the inclined plane, force pushing the block upward must be equal to the force acting downwards.

μR = mgsinθ

μmgcosθ = mgsinθ

μcosθ = sinθ

μ = sinθ/cosθ

μ = tanθ

θ = tan⁻¹(0.4) = 21.8°

Therefore, the maximum value of θ that will cause the block to remain stationary on the inclined surface is 21.8°

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Answer:

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

m_{1} v_{1}=m_{2} v_{2}

\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }

\mathrm{v}_{1} \text { velocity before the collision }

\mathrm{v}_{2} \text { velocity after the collision }

\mathrm{m}_{1}=600 \mathrm{kg}

\mathrm{m}_{2}=900 \mathrm{kg}

\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}

600 \times 5=900 \times v_{2}

3000=900 \times v_{2}

\mathrm{v}_{2}=\frac{3000}{900}

\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

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What is the momentum of a 248 g rubber ball traveling at 30.0 m/s?
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<span>The answer is, 7.44 kg*m/s</span>
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Titanium reacts less with oxygen than most metals do. This is a___.
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this is a physical property

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Which statement correctly explains molecular motion in different states of matter using the kinetic theory?
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You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce
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The time needed for the 7th car to pass is 13.2 s

Explanation:

The motion of the train is a uniformly accelerated motion, therefore we can use suvat equations.

We start by analzying the motion of the first car, by using the equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered by the first car in a time t, which corresponds to the length of one car

u = 0 is the initial velocity

a is the acceleration

t = 5.0 s is the time

The equation can be rewritten as

a=\frac{2s}{t^2}=\frac{2L}{(5.0)^2}=0.08L[m/s^2]

where L is the length of one car.

The same equation can be written considering the first 7 cars:

7L = ut+\frac{1}{2}at'^2

where

7L is the distance covered by the 7 cars

t' is the time needed

We still have

u = 0

And the acceleration is constant so it is

a=0.08L

Substituting into the equation, we can find t':

7L = \frac{1}{2}(0.08L)t'^2\\7=0.04t'^2\\t'=\sqrt{\frac{7}{0.04}}=13.2 s

In attachment the graph of the distance covered versus the time taken: since the motion is uniformly accelerated, the relationship between the two variables is quadratical.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
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