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Elodia [21]
3 years ago
9

A sandbag motionless in outer space is hit by a three-times-as massive sandbag moving at 12 m/s. They stick together. This is an

example of ____________ collision and it loses ________ % of initial kinetic energy.
Physics
1 answer:
vekshin13 years ago
4 0

Answer:

This is an example of inelastic collision and it loses zero(0) % of initial kinetic energy

Explanation:

Let the mass of the motionless sandbag = m₁

Let the mass of the moving sandbag = m₂ = 3m₁

initial velocity of the motionless sandbag, u₁ = 0

initial velocity of the moving sandbag, u₂ = 12 m/s

Let their final velocity, = v

Collision between two particles can either be elastic or inelastic.

Since they stick together after the impact, then the collision is inelastic

Apply the principle of conservation of linear momentum;

Initial kinetic energy = final kinetic energy

¹/₂m₁u₁² + ¹/₂m₂u₂² = ¹/₂v²(m₁ + m₂)

¹/₂m₁(0)² + ¹/₂(3m₁)(12)² = ¹/₂v²(m₁+3m₁)

216m₁ = 2m₁ v²

v² = 108

v = √108

v = 10.392 m/s

Change in kinetic energy = Final kinetic energy - initial kinetic energy

Initial Kinetic energy, KE₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

                                   KE₁ = ¹/₂m₁(0)² + ¹/₂(3m₁)(12)²

                                   KE₁ = 216m₁ J

Final kinetic energy, KE₂ = ¹/₂v²(m₁ + m₂)

                                  KE₂ = ¹/₂(108)(m₁ + 3m₁)

                                  KE₂ = 216m₁ J

ΔKE = KE₂ - KE₁ =  216m₁ J -  216m₁ J = 0%

Therefore, this is an example of inelastic collision and it loses zero(0) % of initial kinetic energy

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Gemiola [76]

The Mass of the barter and the ball is 3.1kg

Data Given;

  • F = 1.1*10^-9N
  • r = 0.76m
  • G = 6.67*10^-11 m^3 kg^-1 s^-1

Assuming the barter and the ball have equal mass, we can apply gravitational force formula.

<h3>Gravitational Force</h3>

F = \frac{Gm_1m_2}{r^2}

This is product of the mass of the two bodies and divided by the square of their distance apart.

Substituting the values into the formula

F = \frac{Gm_1m_2}{r^2\\}\\m_1 = m_2 = m\\1.1*10^-^9=\frac{6.67*10^-^1^1*m^2}{0.76^2}  \\m^2 = \frac{1.1*10^-^9*0.76^2}{6.67*10^-^1^1} \\m^2 = 9.525\\m = \sqrt{9.525}\\m =3.086=3.1kg

From the calculations above, the mass of the barter and the ball is 3.1kg

Learn more on gravitational force here;

brainly.com/question/11359658

8 0
3 years ago
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kaheart [24]

Answer:

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Explanation:

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5 0
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4.5 cm3 of water is boiled at atmospheric pressure to become 4048.3 cm3 of steam, also at atmospheric pressure. Calculate the wo
Mrrafil [7]

1. 408.4 J

The work done by a gas is given by:

W=p\Delta V

where

p is the gas pressure

\Delta V is the change in volume of the gas

In this problem,

p=1.01\cdot 10^5 Pa (atmospheric pressure)

\Delta V=4048.3 cm^3 - 4.5 cm^3 =4043.8 cm^3 = 4043.8 \cdot 10^{-6}m^3 is the change in volume

So, the work done is

W=(1.01\cdot 10^5 Pa)(4043.8 \cdot 10^{-6}m^3)=408.4 J

2. 10170 J

The amount of heat added to the water to completely boil it is equal to the latent heat of vaporization:

Q = m \lambda_v

where

m is the mass of the water

\lambda_v = 2.26\cdot 10^6 J/kg is the specific latent heat of vaporization

The initial volume of water is

V_i = 4.5 cm^3 = 4.5\cdot 10^{-6}m^3

and the water density is

\rho = 1000 kg/m^3

So the water mass is

m=\rho V_i = (1000 kg/m^3)(4.5\cdot 10^{-6}m^3)=4.5\cdot 10^{-3}kg

So, the amount of heat added to the water is

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8 0
4 years ago
1. The head of a rattlesnake can accelerate 50.00 m/s2 in striking a victim. If a car
scoray [572]

Answer:

2.000 s

Explanation:

Given:

a = 50.00 m/s²

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v₀ = 0 m/s

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(100.0 m/s) = (50.00 m/s²) t + (0 m/s)

t = 2.000 s

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3 years ago
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