Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 11.7 revolutions/s. Find the linear speed v
1 answer:
Answer:
The linear speed of the car, v, is 24.26 m/s
Explanation:
Given;
radius of the car's tire, r = 0.330 m
angular speed of the car, ω = 11.7 revolutions/s
The angular speed of the car in radian per second:
The linear speed of the car, v, is calculated as;
v = ωr
v = 73.523 rad/s x 0.33 m
v = 24.26 m/s
Therefore, the linear speed of the car, v, is 24.26 m/s
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Answer:
3.46 seconds
Explanation:
Since the ball is moving in circular motion thus centripetal force will be acting there along the rope.
The equation for the centripetal force is as follows -
Where, is the mass of the ball,
is the speed and
is the radius of the circular path which will be equal to the length of the rope.
This centripetal force will be equal to the tension in the string and thus we can write,
and,
Thus, m/s.
Now, the total length of circular path = circumference of the circle
Thus, total path length = 2πr = 2 × 3.14 × 2 = 12.56 m
Time taken to complete one revolution = =
= 3.46 seconds.
Thus, the mass will complete one revolution in 3.46 seconds.
Answer:
Frequency of oscillation, f = 4 Hz
time period, T = 0.25 s
Angular frequency,
Given:
Time taken to make one oscillation, T = 0.25 s
Solution:
Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:
f =
f =
Frequency of oscillation, f = 4 Hz
The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.
Therefore, time period, T = 0.25 s
Angular frequency of oscillation is given by:
Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m