Answer:
= -32.53 m / s
this velocity is directed downwards
Explanation:
This is a free fall exercise, let's use the expression
= v_{oy}^{2} + 2 g (y -yo)
where we are assuming that there is friction with the air, as the body falls its initial velocity is zero
v_{oy} = √ 2g (y - y₀)
let's calculate
v_{y} = √ (2 9.8 (0-54.0))
= -32.53 m / s
this velocity is directed downwards
Answer:
The angle through which the wheel turned is 947.7 rad.
Explanation:
initial angular velocity, = 33.3 rad/s
angular acceleration, α = 2.15 rad/s²
final angular velocity, = 72 rad/s
angle the wheel turned, θ = ?
The angle through which the wheel turned can be calculated by applying the following kinematic equation;
Therefore, the angle through which the wheel turned is 947.7 rad.
Answer: Acceleration will have 2 components, vertical and horizontal.
Net-vertical component can be positive, zero or negative depending upon the magnitude of the upward component of the applied acceleration.
Net-horizontal acceleration will be equal to the horizontal component of the applied acceleration.
Explanation:
Since acceleration is a vector quantity and the cart is being pushed up the ramp, the ramp would be at some angle to the horizontal and hence there will be vertical and horizontal components of acceleration.
<u>For vertical acceleration:</u>
If the magnitude of the upward component of the applied acceleration is greater than the value of the acceleration due to gravity then the net vertical acceleration will be upward because it will overtake the value of acceleration due to gravity.
In case the upward component of the applied acceleration is lesser than the value of the acceleration due to gravity then the net vertical acceleration will be downward.
<u>For horizontal acceleration:</u>
This component remains unaffected and is equal to the horizontal component of the applied acceleration because there is no other acceleration acting in the horizontal direction.
But the net acceleration will not be solely in the vertical or horizontal direction because the block has to move forward on the inclined ramp so there will always exist a horizontal and a vertical component making the net acceleration to parallel to the ramp in upward direction if the body is going up the ramp.
