Answer:
The vehicle travels 56.25 metres in the interval during which body decelerates .
Explanation:
- Initial velocity of vehicle, u = 32 m/s
- Final velocity of vehicle, v = 22 m/s
- Rate of acceleration, a = - 4.8 m/
Let the distance travelled be s .
We have to determine the distance travelled by the vehicle during this time.
The equation of motion is given by
s =
<u>s = 56.25 metres</u>
The vehicle travels 56.25 metres in the interval during which body decelerates .
Answer:
I =1.8 kgm^2
Explanation:
In order to calculate the moment of inertia of the door you use the following formula, which relates the torque applied to the door with its moment of inertia and angular acceleration:
(1)
τ: torque applied to the door
I: moment of inertia of the door
α: angular acceleration = 5 rad/s^2
The torque is also given by τ = Fd, where F is the force applied at a distance of d to the pivot of the door (hinge axis).
F = 10 N
d = 0.9 m
You replace the expression for τ, and solve for I:
The moment of inertia of the door is 1.8 kgm^2
Answer:
Time period, T = 1.98 seconds
Explanation:
It is given that,
Mass of the block, m = 300 g = 0.3 kg
Force constant of the spring, k = 3 N/m
Displacement in the block, x = 3 cm
Let T is the period of the motion of the block. The time period of the block is given by :
T = 1.98 seconds
So, the period of the motion of the block is 1.98 seconds. Hence, this is the required solution.
Answer:
Rotation around the front-to-back axis is called roll. Rotation around the side-to-side axis is called pitch. Rotation around the vertical axis is called yaw.
Explanation:
Answer:
4
Explanation:
friction
weight
normal reaction
force to overcome inertia