A boy is exerting a force of 70 N at a 50-degree angle on a lawn mower. He is accelerating at 1.8 m/s2. Round the answers to the
2 answers:
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Refer to the figure shown below.
m₁ = 1100 kg, the mass of the car
m₂ = 700 kg, the mass of the trailer and boat
F = 1900 N, the driving force acting on the car
N₁ = m₁g, the normal reaction on the car
N₂ = m₂g, the normal force on the trailer and boat
μN₁ and μN₂ are frictional forces, where = kinetic coefficient of friction
T = the force in the hitch between the car and trailer.
Part (a)
Let R₁ = the total force that resists the motion of the car, boat, and trailer.
Because the acceleration is 0.550m/s², therefore
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100+700 kg)*(0.55 m/s²) = (1900 - R₁) N
990 = 1900 - R
R₁ = 910 N
Answer: The resistive force is 910 N
Part (b)
80% of the resistive forces are experienced by the boat and trailer.
Let the resistive force be R₂.
Then
R₂ = 0.8*R₁ = 728 N
If the tension in the hitch between the car and the trailer is T, then
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²)
T - 728 = 385
T = 1113 N
Answer: The force in the hitch is 1113 N
m₁ = 1100 kg, the mass of the car
m₂ = 700 kg, the mass of the trailer and boat
F = 1900 N, the driving force acting on the car
N₁ = m₁g, the normal reaction on the car
N₂ = m₂g, the normal force on the trailer and boat
μN₁ and μN₂ are frictional forces, where = kinetic coefficient of friction
T = the force in the hitch between the car and trailer.
Part (a)
Let R₁ = the total force that resists the motion of the car, boat, and trailer.
Because the acceleration is 0.550m/s², therefore
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100+700 kg)*(0.55 m/s²) = (1900 - R₁) N
990 = 1900 - R
R₁ = 910 N
Answer: The resistive force is 910 N
Part (b)
80% of the resistive forces are experienced by the boat and trailer.
Let the resistive force be R₂.
Then
R₂ = 0.8*R₁ = 728 N
If the tension in the hitch between the car and the trailer is T, then
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²)
T - 728 = 385
T = 1113 N
Answer: The force in the hitch is 1113 N
a) 4F0
b) Speed of planet B is the same as speed of planet A
Speed of planet C is twice the speed of planet A
Explanation:
a)
The magnitude of the gravitational force between two objects is given by the formula
where
G is the gravitational constant
m1, m2 are the masses of the 2 objects
r is the separation between the objects
For the system planet A - Star A, we have:
So the force is
For the system planet B - Star B, we have:
So the force is
So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.
For the system planet C - Star C, we have:
So the force is
So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.
b)
The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:
where
m is the mass of the planet
M is the mass of the star
v is the tangential speed
We can re-arrange the equation solving for v, and we find an expression for the speed:
For System A,
So the tangential speed is
For system B,
So the tangential speed is
So, the speed of planet B is the same as planet A.
For system C,
So the tangential speed is
So, the speed of planet C is twice the speed of planet A.