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erica [24]
2 years ago
13

An ice-skater with a mass of 80kg is holding a bowling ball with a mass of 8 kg. Suppose that the skater tosses the bowling ball

forward with a speed of 6 m/s. What is the skater's reactive velocity? Show all work.
Physics
1 answer:
musickatia [10]2 years ago
8 0

Answer:

0.6 m/s

Explanation:

The details of the masses and velocities are;

The mass of the ice skater, m₁ = 80 kg

The mass of the ball, m₂ = 8 kg

The speed with which the skater tosses the ball forward, v₂ = 6 m/s

Therefore;

According to the principle of conservation of linear momentum, we have;

m₁·v₁ = m₂·v₂

Where;

v₁ = The skater's reactive velocity

Therefore, we get;

80 kg × v₁ = 8 kg × 6 m/s

v₁ = 8 kg × 6 m/s/(80 kg) = 0.6 m/s

The skater's reactive velocity, v₁ = 0.6 m/s.

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How long will it take you to travel 20 miles on a bus that drives 60 miles/h?
ValentinkaMS [17]

Answer:

<u>20 Minutes</u>

<u></u>

Explanation:

Well we know Mph (Miles per hour) is distance over time : \frac{distance}{time} \\

R (rate) = 60

d (distance) = 20

t (time) = Unknown

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 R = \frac{d}{t}  

   ↓

60 = \frac{20}{t}

   ↓

 t = \frac{20}{60}

    ↓

 t = \frac{1}{3}  or 0.3333

<em>So basically it would take one third of an hour. Lets change these units to minutes.</em>

60 * 0.333333 = 20

<em>So it would take you </em><u><em>20 minutes</em></u><em> to drive 20 miles on a bus that drives 60 mph</em>

<em />

Hope that helps

<em>~Siascon~</em>

8 0
3 years ago
What is a short circuit?
Marizza181 [45]

Answer:

A short circuit is an electrical circuit that allows a current to travel along an unintended path. It has no or very low electrical impedances. The opposite of a short circuit

Explanation:

3 0
3 years ago
Read 2 more answers
A stuffed toy with a mass of 0.900 kilograms sits on the edge of a bed at a height of 0.830 if the toy falls off the bed what wi
Olenka [21]
Mechanical energy (ME) is the sum of potential energy (PE) and kinetic energy (KE). When the toy falls, energy is converted from PE to KE, but by conservation of energy, ME (and therefore PE+KE) will remain the same.

Therefore, ME at 0.500 m is the same as ME at 0.830 m (the starting point). It's easier to calculate ME at the starting point because its just PE we need to worry about (but if we wanted to we could calculate the instantaneous PE and KE at 0.500 m too and add them to get the same answer).

At the start:

ME = PE = mgh
ME = 0.900 (9.8) (0.830)
ME = 7.32 J
8 0
3 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

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3 years ago
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