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Dennis_Churaev [7]
2 years ago
11

A box slides down a frictionless plane inclined at an angle θ ¸ above the horizontal. The gravitational force on the box is dire

cted
a)vertically.
b)perpendicular to the plane.
c)at an angle θ ¸ below the inclined plane.
d)parallel to the plane in the same direction as the movement of the box.
e)parallel to the plane in the opposite direction as the movement of the box.
Physics
1 answer:
anzhelika [568]2 years ago
4 0

Answer:

vertically

Explanation:

The gravitational force on any object will always be vertically. (at an angle 90-θ below the inclined plane).

<em>The force made by the inclinated plane on the box and the force made by the box on the plane will be perpendicular to the plane.</em>

<em>The force made by the inclinated plane on the box and the force made by the box on the plane will be perpendicular to the plane.The acceleration of the box will be parallel to the plane in the same direction as the movement of the box, and the fricction force will be eparallel to the plane in the opposite direction as the movement of the box.</em>

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Answer:

The velocity of the Mr. miles is 17.14 m/s.

Explanation:

It is given that,

Mr. Miles zips down a water-slide starting at 15 m vertical distance up the scaffolding, h = 15 m

We need to find the velocity of the Mr. Miles at the bottom of the slide. It is a case of conservation of energy which states that the total energy of the system remains conserved. Let v is the velocity of the Mr. miles. So,

v=\sqrt{2gh}

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v=\sqrt{2\times 9.8\times 15}

v = 17.14 m/s

So, the velocity of the Mr. miles is 17.14 m/s. Hence, this is the required solution.

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3 years ago
A baseball player friend of yours wants to determine his pitching speed. you have him stand on a ledge and throw the ball horizo
zhenek [66]

Answer:

The pitching speed of your friend is 33.20 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

Your friend throw the ball horizontally that means the vertical initial

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The height h=u_{y}t+\frac{1}{2}gt^{2}

<u><em>Remember:</em></u> the height is negative value because its below the point of

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<em>Substitute these values in the rule above</em>

⇒ 4=0-\frac{1}{2}(9.8)t^{2}

⇒ -4 = -4.9t² (multiply both sides by -1)

⇒ 4 = 4.9t² (divide both sides by 4.9)

⇒ 0.81633 = t² (take √ for both sides)

⇒ <em>t = 0.9035</em>

Then the time of the ball to land on the ground is 0.9035 seconds

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The range R=u_{x}*t, where u_{x} is the horizontal

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R = 30 meters and t = 0.9035

⇒ 30=u_{x}(0.9035) (divide both sides by 0.9035)

⇒ u_{x}=33.20 m/s

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Answer:

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Hope this helps

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Answer:

3.98V

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