A parachutist with a camera, both descending at a speed of 10.8 m/s, releases that camera at an altitude of 50 m. In this proble
1 answer:
Answer:
The velocity of the camera is 33.11 m/s.
Explanation:
Given that,
Speed = 10.8 m/s
Altitude = 50 m
Suppose determine the velocity of the camera just before it hits the ground?
We need to calculate the velocity of the camera
Using equation of motion
Where, v = final velocity of camera
u = initial speed of camera
s = distance
Put the value into the formula
The direction will be downward so it is the negative velocity.
Hence, The velocity of the camera is 33.11 m/s.
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Answer:
The minimum speed must the car must be 13.13 m/s.
Explanation:
The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.
We know that, mg be the weight of car and rider, which is equal to the centripetal force.
So, the minimum speed must the car must be 13.13 m/s.
Answer:
v = √[gR (sin θ - μcos θ)]
Explanation:
The free body diagram for the car is presented in the attached image to this answer.
The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.
Resolving the weight into the axis parallel and perpendicular to the inclined plane,
N = mg cos θ
And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ
Frictional force = Fr = μN = μmg cos θ
Centripetal force responsible for keeping the car in circular motion = (mv²/R)
So, a force balance in the plane parallel to the inclined plane shows that
Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)
(mv²/R) = (mg sin θ - μmg cos θ)
v² = R(g sin θ - μg cos θ)
v² = gR (sin θ - μcos θ)
v = √[gR (sin θ - μcos θ)]
Hope this Helps!!!
