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.
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?
27 answers
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.
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.
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.
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.
50.0 rey * (π/200) radians/rey = 0.785 radians
Therefore, 50.0 rey is equal to 0.785 radians.
50.0 rev to radians.
To convert from revolutions (rev) to radians, we need to multiply by the conversion factor of 2π radians/rev:
50.0 rev * 2π radians/rev = 100π radians
Therefore, 50.0 revolutions is equal to 100π radians. This can also be simplified to 314.16 radians if we use an approximate value of π.
50.0 rev * 2π radians/rev = 100π radians
Therefore, 50.0 revolutions is equal to 100π radians. This can also be simplified to 314.16 radians if we use an approximate value of π.
487 rad to revolutions.
To convert from radians to revolutions, we need to divide by the conversion factor of 2π radians/rev:
487 rad ÷ (2π radians/rev) = 77.39 rev
Therefore, 487 rad is equal to 77.39 revolutions when rounded to two decimal places.
487 rad ÷ (2π radians/rev) = 77.39 rev
Therefore, 487 rad is equal to 77.39 revolutions when rounded to two decimal places.
487π rad to revolutions
To convert from radians to revolutions, we need to divide by the conversion factor of 2π radians/rev:
487π rad ÷ (2π radians/rev) = 243.50 rev
Therefore, 487π rad is equal to 243.50 revolutions when rounded to two decimal places.
487π rad ÷ (2π radians/rev) = 243.50 rev
Therefore, 487π rad is equal to 243.50 revolutions when rounded to two decimal places.
72.0 rps to rad/s
To convert from revolutions per second (rps) to radians per second, we need to multiply by the conversion factor of 2π radians/rev:
72.0 rps * 2π radians/rev = 452.39 rad/s
Therefore, 72.0 rps is equal to 452.39 rad/s when rounded to two decimal places.
72.0 rps * 2π radians/rev = 452.39 rad/s
Therefore, 72.0 rps is equal to 452.39 rad/s when rounded to two decimal places.
1.50 x 10cube rpm to rad/s
First, we need to convert revolutions per minute (rpm) to revolutions per second (rps) by dividing by 60:
1.50 x 10^3 rpm / 60 = 25 rps
Next, we can convert from revolutions per second to radians per second by multiplying by the conversion factor of 2π radians/rev:
25 rps * 2π radians/rev = 157.08 rad/s
Therefore, 1.50 x 10^3 rpm is equal to 157.08 rad/s when rounded to two decimal places.
1.50 x 10^3 rpm / 60 = 25 rps
Next, we can convert from revolutions per second to radians per second by multiplying by the conversion factor of 2π radians/rev:
25 rps * 2π radians/rev = 157.08 rad/s
Therefore, 1.50 x 10^3 rpm is equal to 157.08 rad/s when rounded to two decimal places.
22.0 rad/s to rpm
To convert from radians per second (rad/s) to revolutions per minute (rpm), we need to use two conversion factors:
- 60 seconds/minute to convert from seconds to minutes
- 1 revolution = 2π radians to convert from radians to revolutions
22.0 rad/s * (60 s/min) * (1 rev/2π rad) ≈ 209.44 rpm
Therefore, 22.0 rad/s is approximately equal to 209.44 rpm when rounded to two decimal places.
- 60 seconds/minute to convert from seconds to minutes
- 1 revolution = 2π radians to convert from radians to revolutions
22.0 rad/s * (60 s/min) * (1 rev/2π rad) ≈ 209.44 rpm
Therefore, 22.0 rad/s is approximately equal to 209.44 rpm when rounded to two decimal places.
2.000 rad/s to deg/s
To convert from radians per second (rad/s) to degrees per second (deg/s), we need to multiply by the conversion factor of 180/π degrees/radian:
2.000 rad/s * (180/π) deg/s/rad ≈ 114.59 deg/s
Therefore, 2.000 rad/s is approximately equal to 114.59 deg/s when rounded to two decimal places.
2.000 rad/s * (180/π) deg/s/rad ≈ 114.59 deg/s
Therefore, 2.000 rad/s is approximately equal to 114.59 deg/s when rounded to two decimal places.
Express 40.0 deg/s in (a) rev/s, (b) rev/min, and (c) rad/s.
(a) To convert from degrees per second (deg/s) to revolutions per second (rev/s), we need to use the conversion factor of 1 rev = 360 degrees:
40.0 deg/s * (1 rev/360 degrees) = 0.111 rev/s (rounded to three decimal places)
Therefore, 40.0 deg/s is equal to 0.111 rev/s when rounded to three decimal places.
(b) To convert from revolutions per second (rev/s) to revolutions per minute (rpm), we need to multiply by the conversion factor of 60 seconds/minute:
0.111 rev/s * 60 s/min = 6.67 rev/min (rounded to two decimal places)
Therefore, 40.0 deg/s is equal to 6.67 rev/min when rounded to two decimal places.
(c) To convert from degrees per second (deg/s) to radians per second (rad/s), we need to use the conversion factor of π/180 radians/degree:
40.0 deg/s * (π/180 radians/degree) ≈ 0.698 rad/s
Therefore, 40.0 deg/s is approximately equal to 0.698 rad/s.
40.0 deg/s * (1 rev/360 degrees) = 0.111 rev/s (rounded to three decimal places)
Therefore, 40.0 deg/s is equal to 0.111 rev/s when rounded to three decimal places.
(b) To convert from revolutions per second (rev/s) to revolutions per minute (rpm), we need to multiply by the conversion factor of 60 seconds/minute:
0.111 rev/s * 60 s/min = 6.67 rev/min (rounded to two decimal places)
Therefore, 40.0 deg/s is equal to 6.67 rev/min when rounded to two decimal places.
(c) To convert from degrees per second (deg/s) to radians per second (rad/s), we need to use the conversion factor of π/180 radians/degree:
40.0 deg/s * (π/180 radians/degree) ≈ 0.698 rad/s
Therefore, 40.0 deg/s is approximately equal to 0.698 rad/s.
A flywheel turns at 480 rpm. Compute the angular speed at any point on the wheel and the tangential speed
30.0 cm from the center.
30.0 cm from the center.
The angular speed at any point on the wheel can be found using the formula:
ω = (2πN)/60
where ω is the angular speed in radians per second, N is the speed in revolutions per minute (rpm), and the factor of 60 is used to convert from seconds to minutes.
Plugging in the values, we get:
ω = (2π * 480 rpm)/60 = 50.27 rad/s
Therefore, the angular speed at any point on the wheel is 50.27 rad/s.
The tangential speed at a distance 30.0 cm from the center can be found using the formula:
v = ωr
where v is the tangential speed in meters per second, ω is the angular speed in radians per second (which we just calculated), and r is the distance from the center in meters.
We need to convert 30.0 cm to meters by dividing by 100:
r = 30.0 cm / 100 = 0.3 m
Plugging in the values, we get:
v = 50.27 rad/s * 0.3 m = 15.08 m/s
Therefore, the tangential speed 30.0 cm from the center is 15.08 m/s.
ω = (2πN)/60
where ω is the angular speed in radians per second, N is the speed in revolutions per minute (rpm), and the factor of 60 is used to convert from seconds to minutes.
Plugging in the values, we get:
ω = (2π * 480 rpm)/60 = 50.27 rad/s
Therefore, the angular speed at any point on the wheel is 50.27 rad/s.
The tangential speed at a distance 30.0 cm from the center can be found using the formula:
v = ωr
where v is the tangential speed in meters per second, ω is the angular speed in radians per second (which we just calculated), and r is the distance from the center in meters.
We need to convert 30.0 cm to meters by dividing by 100:
r = 30.0 cm / 100 = 0.3 m
Plugging in the values, we get:
v = 50.27 rad/s * 0.3 m = 15.08 m/s
Therefore, the tangential speed 30.0 cm from the center is 15.08 m/s.
3 answers