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Yuki888 [10]
3 years ago
15

A block of mass mm starts from rest and slides down from the top of a wedge of height hh and length dd. The surface of the wedge

forms an angle of \thetaθ with respect to the horizontal direction. The force of kinetic friction between the block and the wedge is \vec{f} f ​ . How fast is the block traveling when it reaches the bottom of the wedge?
Physics
1 answer:
lutik1710 [3]3 years ago
6 0

Answer:

v=\sqrt{ 2 (g sin\theta- f/m) d}

Explanation:

Given that

Mass = m

Height = h

Length = d

Angle = θ

Friction force = f

When block slides downward then friction force will act upwards

The force due to gravity along inline plane

Fg= mg sinθ

Lets 'a' is the acceleration

From Newtons law

mg sinθ  - f = m a

a= (g sinθ  - f/m)

Given that initial speed u= o ,lets v is final speed

v²=u²+ 2 a s

v²=2 x (g sinθ  - f/m) x d

v=\sqrt{ 2 (g sin\theta- f/m) d}

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The maximum height reached by the ball is 99.2 m

Explanation:

When the ball is thrown straight up, it follows a free fall motion, which is a uniformly accelerated motion with constant acceleration (g=9.8 m/s^2 towards the ground). Therefore, we can use the following suvat equation:

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In this problem, we have:

u = 44.1 m/s is the initial vertical velocity of the ball

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a=-g=-9.8 m/s^2 is the acceleration of gravity (downward, so negative)

Solving for s, we find the maximum height reached by the ball:

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Learn more about free fall:

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3 years ago
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astra-53 [7]

Answer:

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Explanation:

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7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

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Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

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Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

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Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
MArishka [77]

Complete Question:

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167•F which is c


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