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IrinaK [193]
4 years ago
9

A red rubber ball rolls down a hill from rest with an acceleration of 7.8 m/s 2 . How fast is it moving after it has traveled 5

m?
Physics
1 answer:
Wittaler [7]4 years ago
5 0

Answer:

8.83m

Explanation:

Given parameters:

  Acceleration  = 7.8m/s²

   Initial velocity = 0m/s

  distance covered  = 5m

Unknown:

Final velocity  = ?

Solution:

The equation to solve this problem with is given as;

    V² = U² + 2aS

V is the final velocity

U is the initial velocity

a is the acceleration

S is the distance

 Input the parameters and solve;

    V² = 0² + 2 x 7.8 x 5 = 78

    V = √78 = 8.83m

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A golf ball strikes a hard, smooth floor at an angle of 28.2 ° and, as the drawing shows, rebounds at the same angle. The mass o
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Answer:

    I = 1.06886  N s

Explanation:

The expression for momentum is

          I = F t = Δp

therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor

Let's find the components of the initial velocity

          sin 28.2 = v_y / v

          cos 28.2=  vₓ / v

          v_y = v sin 282

          vₓ = v cos 28.2

          v_y = 42.8 sin 28.2 = 20.225 m / s

          vₓ = 42.8 cos 28.2 = 37.72 m / s

since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making

θ = -28.2

         v_y = -20.55 m / s

         v_x = 37.72 m / s

X axis

         Iₓ = Δpₓ = p_{fx} - p_{ox}

since the ball moves in the x-axis without changing the velocity, the change in moment must be zero

         Δpₓ = m v_{fx} - m v₀ₓ = 0

          v_{fx} = v₀ₓ

therefore

          Iₓ = 0

Y axis  

        I_y = Δp_y = p_{fy} -p_{oy}

when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards

         v_{fy} = - v_{oy}

         Δp_y = 2 m v_{oy}

         Δp_y = 2 0.0260 (20.55)

         \Delta p_{y} = 1.0686 N s

the total impulse is

          I = Iₓ i ^ + I_y j ^

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Answer:

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Explanation:

As we know that Normal force is the reaction force of two contact surfaces which always act perpendicular to the contact surfaces

Here we know that the rock is moving inside the bowl

So Normal force on the rock must perpendicular to the surface of the bowl which always passes through the center of the bowl.

Since the rock is moving in vertical plane so it must have two acceleration

1) Tangential acceleration which will increase the magnitude of the speed along the tangential path

2) Centripetal acceleration which will change the direction of the rock

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the normal force on the rock acts perpendicular to the bowl's surface.

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