A body of mass 2kg is at rest. A force of 20N makes it move with a velocity of 5m/s, the impulse of the body would be 10 Kgm/s
<h3>What is impulse?</h3>
The product of the average applied force and the time for which it is exerted is known as an impulse. The unit for the impulse of force is the same as that of momentum which is kg*m/s
The mathematical relation for impulse is
Impulse = F * Δt =change in momentum
As given in the problem body of mass of 2 kg is at rest. A force of 20N makes it move with a velocity of 5m/s
Change in the momentum of the body
Δm = m(v₂-v₁)
As the body was initially at rest v₁ =0
Δm = 2×5
Δm = 10 Kgm/s
As we know that change in the momentum represents the impulse of the force
Thus, the impulse of the body would be 10 Kgm/s
Learn more about impulse, here
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Answer:
<h2>f=a×m</h2>
m=1800kg
1800000g×10N/kg
18000000N force is required to life the car
Answer:
V = 0.896 m/s
Explanation:
This is a typical problem of momentum conservation, whic states the following:
m₁V₁ + m₂V₂ = m₁V₃ + m₂V₄ (1)
In this case V₃ and V₄ would be the final velocity of the trucks after the collision.
With the given data let's see what we have:
m₁ = 5.5x10⁵ kg
m₂ = 2.3x10⁵ kg
V₁ = 5 m/s
V₂ = -5 m/s because it's going to the left (-x axis)
V₄ = 9.1 m/s to the right (Meaning is positive)
V₃ = ??
So to calculate V₃ we just need to replace the data into (1) and solve for V₃:
(5.5x10⁵ * 5) - (2.3x10⁵ * 5) = 5.5x10⁵V₃ + (2.3x10⁵ * 9.1)
2.75x10⁶ + 1.15x10⁶ = 5.5x10⁵V₃ + 2.093x10⁶
V₃ = 2.75x10⁶ - 1.15x10⁶ - 2.093x10⁶ / 5.5x10⁵
V₃ = -0.493x10⁶ / 5.5x10⁵
V₃ = -0.896 m/s
With this sign, it means that is going in the same sense of the other truck, but it's going to the left so this would be positive:
<h2>
V₃ = 0.896 m/s</h2>
Hope this helps
Answer
given,
mass of ball, m = 57.5 g = 0.0575 kg
velocity of ball northward,v = 26.7 m/s
mass of racket, M = 331 g = 0.331 Kg
velocity of the ball after collision,v' = 29.5 m/s
a) momentum of ball before collision
P₁ = m v
P₁ = 0.0575 x 26.7
P₁ = 1.535 kg.m/s
b) momentum of ball after collision
P₂ = m v'
P₂ = 0.0575 x (-29.5)
P₂ = -1.696 kg.m/s
c) change in momentum
Δ P = P₂ - P₁
Δ P = -1.696 -1.535
Δ P = -3.231 kg.m/s
d) using conservation of momentum
initial speed of racket = 0 m/s
M u + m v = Mu' + m v
M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)
0.331 u' = 3.232
u' = 9.76 m/s
change in velocity of the racket is equal to 9.76 m/s
Concept:
Frequency- It is defined as the number of oscillations occur in one second.
Its SI unit is Hertz (Hz)
Given: Produced sound vibrations is 18,500 cycles in 0.75 seconds
∵ In 0.75 second, produced sound has oscillations = 18,500 cycles
∴ In 1.0 second, produced sound has oscillations = (18,500 ÷ 0.75) Hz
The frequency of the sound will be ≈ 24,667 Hz
From the study of the given graph, only the animals (c) Cats, (b) Moths and (a) Bats can hear the produced sound because their upper audible frequency range is greater than 24,667 Hz.