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taurus [48]
2 years ago
11

PLEASE HELP I HAVE NO TIME PLEASEEE DONT SKIP

Physics
1 answer:
Triss [41]2 years ago
7 0
2m/s




it has to be 20 charecters just ignore this your answer is up there
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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
How do u answer this?
gayaneshka [121]

Answer:

food

Explanation:

did you get a chance to look at the maximum number of devices allowed by

4 0
3 years ago
A sled of mass 15 kg slides across the ground with a friction force of 80.85 N. What is k between the sled and the ground?
Dvinal [7]
The expression for the frictional force between the sled and the ground is:
F=k m g
where k is the coefficient of friction, m is the mass of the object and g=9.81 m/s^2 is the gravitational acceleration.

The friction force in our problem is F=80.85 N. The mass of the object is m=15 kg. Re-arranging the formula, we can find the value of k:
k= \frac{F}{mg}= \frac{80.85 N}{(15 kg)(9.81 m/s^2)}=0.55
8 0
3 years ago
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Which of the followijg is an example of kinetic energy
Ipatiy [6.2K]

Answer:

D

Explanation:

D. A swing moving back and forth

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Un alambre de hierro se tendió entre dos paredes fijas resistentes, estando la temperatura a 150 ºC ¿A qué temperatura se romper
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Answer:

Explanation:

porfa respondan

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