Studied by 1 person
02/20/25
Question 1A: a car with mass 725kg is moving at +100 km/h. A) Find its momentum.
Momentum is calculated using the formula ( p = mv ), where ( p ) is momentum, ( m ) is mass, and ( v ) is velocity. For a car with mass 725 kg moving at 100 km/h, its momentum is ( 725 \times 27.78 ) kg·m/s, or ( 20145 ) kg·m/s.
Question 1B: a car with mass 725kg is moving at +100 km/h. B) At what velocity is the momentum of a larger car, mass 2175kg, equal to that of the smaller car?
To find the velocity of the larger car that results in equal momentum to the smaller car, use the momentum formula ( p = mv ). Set the momentum of the smaller car ( 20145 kg·m/s ) equal to the momentum of the larger car ( 2175 kg × v ). Solving for v gives v = 20145 / 2175, resulting in approximately 9.25 m/s.
Question 2A: a snowmobile has a mass of 2.50 × 10² kg. A constant force is exerted on it for 60.0 seconds. The snowmobile’s initial velocity is 6.00 m/s and his final velocity is 28.0 m/s. A) what is its change in momentum?
The change in momentum is calculated by finding the difference between the final and initial momentum. Using the formula ( \Delta p = m(v_f - v_i) ), where ( m ) is mass, ( v_f ) is final velocity, and ( v_i ) is initial velocity, the change in momentum is ( 2.50 \times 10^2 kg \times (28.0 m/s - 6.00 m/s) ), resulting in a change of approximately 5,500 kg·m/s.
Question 2B: a snowmobile has a mass of 2.50 × 10² kg. A constant force is exerted on it for 60.0 seconds. The snowmobile’s initial velocity is 6.00 m/s and his final velocity is 28.0 m/s. B) what is the magnitude of the force exerted on it?
The magnitude of the force can be calculated using Newton's second law, which states that force is equal to the change in momentum divided by the time interval. Using the formula ( F = {triangle p}{ triangle t} ), where ( triangle p ) is the change in momentum (5,500 kg·m/s) and ( triangle t ) is the time (60.0 s), the force exerted is approximately 91.67 N.
Question 3A: The brakes exert a 6.40 × 10² N force on a car weighing 15,680 N and moving at 20.0 m/s. The car finally stops. A) what is the car’s mass?
Question 3B: The brakes exert a 6.40 × 10² N force on a car weighing 15,680 N and moving at 20.0 m/s. The car finally stops. B) what is the car’s initial momentum?
Question 3C: The brakes exert a 6.40 × 10² N force on a car weighing 15,680 N and moving at 20.0 m/s. The car finally stops. C) what is the change in the car’s momentum?
Question 3D: The brakes exert a 6.40 × 10² N force on a car weighing 15,680 N and moving at 20.0 m/s. The car finally stops. D) how long does the braking force act on the car to bring it to a halt?
What is the momentum of a 100kg motorbike travelling at 10m/s?
p = mv = 100kg(10m/s) = 1000kg x m/s (aka 1.0 × 10³ m/s)
What is the mass of a plane that is travelling at 200km/h and has a momentum of 1.1 × 10^6 kg x m/s?
To find the mass, use the formula momentum (p) = mass (m) x velocity (v). Rearranging gives m = p/v. Thus, m = (1.1 × 10^6 kg x m/s) / (200 km/h converted to m/s). Converting 200 km/h to m/s gives 55.56 m/s, so m = (1.1 × 10^6 kg x m/s) / (55.56 m/s) = 19784.17 kg.
How fast does a 0.01kg bug have to fly to have a momentum of 0.25kg x m/s?
To find the velocity, use the formula momentum (p) = mass (m) x velocity (v). Rearranging gives v = p/m. Thus, v = (0.25 kg x m/s) / (0.01 kg) = 25 m/s.
What are the standard units for mass, displacement, and time?
Mass is measured in kilograms (kg), displacement in meters (m), and time in seconds (s).
fill in the formula: p = _____
p = mv
fill in the formula: v = _____
v = p/m
fill in the forumla: m = _____
m = p/v
fill in the blank: to convert from km/h —> m/s, you divide the number by _____.
3.6
what units are used in variable “p?”
kg x m/s
Question 5
6.7 × 10^-2 m/s
Question 6
Question 7
Question 18
Question 19
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Question 21
Question 24
U
Question 1
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U
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Question 5 a) and b)
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