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3 years ago

8

An autographed baseball rolls off of a 0.91 m high desk and strikes the floor 0.84 m away from the desk. How fast was it rolling

on the desk before it fell off? The acceleration of gravity is 9.81 m/s2 .

1 answer:

Rudik [331]3 years ago

4 0

The initial velocity (its speed before falling off) is approximately 1.95 m/s

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Sliva [168]

The stopwatch will be the most useful in determining the kinetic energy of a 50 g battery- powered car traveling a distance of 10 m.

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8 0

2 years ago

A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w

amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

10 x 3 + 25 x 0 = v(10 + 25)

30 = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

$K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\$

the final kinetic energy of the package;

$K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J$

the fraction of the initial energy lost;

$= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71$

7 0

3 years ago

Starting from rest, a 6.79 kg block slides 2.82 m down a rough 20.7 ◦ incline. The coefficient of kinetic friction between the b

Veronika [31]

Answer:

23.52092 J

Explanation:

m = Mass of block = 6.79 kg

s = Sliding distance = 2.82 m

$\theta$ = Angle of slide = 20.7°

$\mu$ = Coefficient of kinetic friction = 0.425

g = Acceleration due to gravity = 9.8 m/s²

Work done by the force of gravity is given by

$W=mgsin\theta\\\Rightarrow W=6.79\times 9.8\times sin20.7\\\Rightarrow W=23.52092\ J$

The work done by the force of gravity is 23.52092 J

8 0

3 years ago