Answer:
I = 2 kgm^2
Explanation:
In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:
(1)
I: moment of inertia of the door
α: angular acceleration of the door = 2.00 rad/s^2
τ: torque exerted on the door
You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:
(2)
F: force = 5.00 N
d: distance to the hinges = 0.800 m
You replace the equation (2) into the equation (1), and you solve for α:

Finally, you replace the values of all parameters in the previous equation for I:

The moment of inertia of the door around the hinges is 2 kgm^2
Answer:
A title
Explanation:
Because this is middle school.
Answer:
ball clears the net
Explanation:
= initial speed of launch of the ball = 20 ms^{-1}
= angle of launch = 5 deg
Consider the motion of the ball along the horizontal direction
= initial velocity = 
= time of travel
= horizontal displacement of the ball to reach the net = 7 m
Since there is no acceleration along the horizontal direction, we have
Eq-1
Consider the motion of the ball along the vertical direction
= initial velocity = 
= time of travel
= Initial position of the ball at the time of launch = 2 m
= Final position of the ball at time "t"
= acceleration in down direction = - 9.8 ms⁻²
Along the vertical direction , position at any time is given as

Since Y > 1 m
hence the ball clears the net
Answer:
Explanation:
angular momentum of the putty about the point of rotation
= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .
= .045 x 4.23 x 2/3 x .95 cos46
= .0837 units
moment of inertia of rod = ml² / 3 , m is mass of rod and l is length
= 2.95 x .95² / 3
I₁ = .8874 units
moment of inertia of rod + putty
I₁ + mr²
m is mass of putty and r is distance where it sticks
I₂ = .8874 + .045 x (2 x .95 / 3)²
I₂ = .905
Applying conservation of angular momentum
angular momentum of putty = final angular momentum of rod+ putty
.0837 = .905 ω
ω is final angular velocity of rod + putty
ω = .092 rad /s .
Answer:
V is approximately = 23m/s
Explanation:
Kinetic energy = ½ mv²
Where m= mass = 0.450kg
V= velocity =?
K. E = 119J
Therefore
K. E = ½ mv²
Input values given
119= ½ × 0.450 × v²
Multiply both sides by 2
119 ×2 = 2 × 1/2 × 0.450 × v²
238= 0.450v²
Divide both sides by 0.450
238/0.450 = 0.450v²/0.450
v² = 528.89
Square root both sides
Sq rt v² = sq rt 528.89
V = 22.998m/s
V is approximately = 23m/s
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