A 1200-kg car going 30 m/s applies its brakes and skids to rest. If the friction force between the sliding tires and the pavement is 6000 N, how far does the car skid before coming to rest?
We can use the kinematic equation:
v_f^2 = v_i^2 + 2aΔx
where v_f is the final velocity (0 m/s), v_i is the initial velocity (30 m/s), a is the acceleration, and Δx is the distance traveled.
We can solve for Δx:
Δx = (v_f^2 - v_i^2) / (2a)
Since the car comes to rest, the acceleration can be found using Newton's second law:
F = ma
where F is the friction force (6000 N) and m is the mass of the car (1200 kg). Rearranging, we get:
a = F/m
Plugging in the values, we get:
a = 6000 N / 1200 kg = 5 m/s^2
Now we can solve for Δx:
Δx = (0^2 - 30^2) / (2 * (-5)) = 450 meters
Therefore, the car skids 450 meters before coming to rest.
How large a force is required to accelerate a 1300-kg car from rest to a speed of 20 m/s in a distance of 80 m?
We can use the kinematic equation:
v_f^2 = v_i^2 + 2aΔx
where v_f is the final velocity (20 m/s), v_i is the initial velocity (0 m/s), a is the acceleration, and Δx is the distance traveled.
We can solve for a:
a = (v_f^2 - v_i^2) / (2Δx)
Plugging in the values, we get:
a = (20^2 - 0^2) / (2 * 80) = 5 m/s^2
Now we can use Newton's second law to find the force required:
F = ma
where F is the force, m is the mass of the car (1300 kg), and a is the acceleration we just calculated.
Plugging in the values, we get:
F = 1300 kg * 5 m/s^2 = 6500 N
Therefore, a force of 6500 N is required to accelerate the car from rest to 20 m/s in a distance of 80 m.
A 40000-kg freight car is coasting at a speed of 5.0 m/s along a straight track when it strikes a 30000-kg stationary freight car and couples to it. What will be their combined speed after impact?
We can use the conservation of momentum principle to solve the problem:
m1v1 + m2v2 = (m1 + m2)v
where m1 is the mass of the first freight car, v1 is its velocity before the collision, m2 is the mass of the second freight car, v2 is its velocity before the collision, and v is their combined velocity after the collision.
We can plug in the values:
(40000 kg)(5.0 m/s) + (30000 kg)(0 m/s) = (40000 kg + 30000 kg) v
Simplifying, we get:
200000 kg m/s = 70000 kg v
Dividing by 70000 kg, we get:
v = 2.86 m/s
Therefore, their combined speed after the impact is 2.86 m/s.
An empty 15000-kg coal car is coasting on a level track at 5.00 m/S. Suddenly 5000 kg of coal is dumped Into it from directly above it. The coal instally has zero honzontal velocity. Find the final speed of the Car
We can use the conservation of momentum principle to solve the problem:
m1v1 + m2v2 = (m1 + m2)v
where m1 is the mass of the empty coal car, v1 is its velocity before the coal is dumped, m2 is the mass of the dumped coal, v2 is its velocity before the collision (which is zero), and v is the final velocity of the car and the coal.
We can plug in the values:
(15000 kg)(5.00 m/s) + (5000 kg)(0 m/s) = (15000 kg + 5000 kg) v
Simplifying, we get:
75000 kg m/s = 20000 kg v
Dividing by 20000 kg, we get:
v = 3.75 m/s
Therefore, the final speed of the car is 3.75 m/s.
50.0 rey to radians
To convert from rey to radians, we need to multiply by the conversion factor of (π/200) radians/rey:
50.0 rey * (π/200) radians/rey = 0.785 radians
Therefore, 50.0 rey is equal to 0.785 radians.