The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table.
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2 M = 7/5*m
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
Answer:
Explanation:
Given that,
Basket ball is drop from height
H=10m
It is dropped on planet mass
And the acceleration due to gravity on Mars is given as
g= 3.7m/s²
Time taken for the ball to reach the ground
Initial velocity of the body is zero
u=0m/s
Using equation of motion: free fall
H = ut + ½gt²
10 = 0•t + ½ × 3.7 ×t²
10 = 0 + 1.85t²
10 = 1.85t²
Then, t² =10/1.85
t² = 5.405
t = √ 5.405
t = 2.325seconds
So the time the ball spend on the air before reaching the ground is 2.325 seconds
<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>
Answer:
1331.84 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity = 0
s = Displacement = 490 km
a = Acceleration
g = Acceleration due to gravity = 1.81 m/s² = a
From equation of linear motion
The speed of the material must be 1331.84 m/s in order to reach the height of 490 km
The ball is against the vector of gravity. Then, the gravity will be negative.
The ball will stop in the air after approx. 4.72 seconds. And will take the same time to hit the ground.
It will stay approx. 9.44 seconds in the air.
