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GrogVix [38]
3 years ago
10

A 2-kg cart, traveling on a horizontal air track with a speed of 3m/s, collides with a stationary 4-kg cart. The carts stick tog

ether. The impulse exerted by one cart on the other has a magnitude of
Physics
1 answer:
ladessa [460]3 years ago
8 0

Answer:

The impulse exerted by one cart on the other has a magnitude of 4 N.s.

Explanation:

Given;

mass of the first cart, m₁ = 2 kg

initial speed of the first car, u₁ = 3 m/s

mass of the second cart, m₂ = 4 kg

initial speed of the second cart, u₂ = 0

Let the final speed of both carts = v, since they stick together after collision.

Apply the principle of conservation of momentum to determine v

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 3 + 0 = v(2 + 4)

6 = 6v

v = 1 m/s

Impulse is given by;

I = ft = mΔv = m(

The impulse exerted by the first cart on the second cart is given;

I = 2 (3 -1 )

I = 4 N.s

The impulse exerted by the second cart on the first cart is given;

I = 4(0-1)

I = - 4 N.s (equal in magnitude but opposite in direction to the impulse exerted by the first).

Therefore, the impulse exerted by one cart on the other has a magnitude of 4 N.s.

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A block of mass 500g is pulled from rest on ahorizontal frictionless bench by a steady force F and travels 8m in 2s find the acc
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<u>We are Given:</u>

Mass of the block (m) = 500 grams or 0.5 Kg

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An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe
kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

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