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tatiyna
3 years ago
11

A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom o

f a gully. Its momentum upon hitting the ground is
Physics
1 answer:
Digiron [165]3 years ago
6 0

Answer:

2940 kg m/s

Explanation:

First of all, we need to find the final velocity of the motorcycle when it hits the ground. The horizontal component of the velocity is zero, so we can just find the vertical component, which is given by:

v_y = v_0+gt

where

v0 =0 is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration due to gravity

t = 3 s is the time of the fall

Substituting into the formula, we find

v_y=0+(9.8 m/s^2)(3 s)=29.4 m/s

And so, we can now find the momentum of the motorcycle upon hitting the ground, which is given by the product between the mass and the velocity:

p=mv=(100 kg)(29.4 m/s)=2940 kg m/s

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3 years ago
Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The ga
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So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

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P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion  P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄    

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K  

T₃ = 90.1 K

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a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

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