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Murljashka [212]
2 years ago
13

What object requires less force to overcome inertia B.80 kg piece of furniture or A.1000 kg vehicle

Physics
1 answer:
Tju [1.3M]2 years ago
8 0

Less force will be necessary to overcome inertia for the 80 kg piece of furniture.

Force is a factor that has the power to alter an object's motion. A massed object's velocity can be changed or accelerated by a force. A push or a pull is a straightforward method to explain forces.

The term "moment of inertia" refers to the quantity that describes how a body resists angular acceleration and is calculated by multiplying each particle's mass by its square of the distance from the rotational axis.

I = mr², where m is the mass of the object and r is the distance to the rotation axis.

Therefore, The inertia is directly proportional to the mass of the object.

So, as the mass increases the inertia increases.

Therefore, 80 kg piece of furniture will require less force to overcome inertia.

Learn more about inertia here:

brainly.com/question/14460640

#SPJ9

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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
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