The tangental acceleration of the flywheel of radius 2.12 cm with a final angular velocity of 1820 rev/min and an angular acceleration of 16.2 rad/s is 0.34 m/s².
<h3>What is tangental acceleration?</h3>
This is the rate of change of tangental velocity of an object in circular motion.
To calculate the tangental acceleration, we use the formula below.
Formula:
- a = αr............. Equation 1
Where:
- a = Tangental acceleration
- r = radius of the flywheel
- α = Angular acceleration of the flywheel.
From the question,
Given:
- r = 2.12 cm = 0.0212 m
- α = 16.2 rad/s²
Subsitute these values into equation 1
- a = (0.0212×16.2)
- a = 0.34 m/s²
Hence, The tangental acceleration of the flywheel of radius 2.12 cm with a final angular velocity of 1820 rev/min and an angular acceleration of 16.2 rad/s is 0.34 m/s².
Learn more about tangental acceleration here: brainly.com/question/11476496
Answer:
0 m/s
Explanation:
At the instant the ball reaches the top of its flight, it isn't moving up or down. So the vertical component of the instantaneous velocity is 0 m/s.
Answer:
sec
Explanation:
Consider the down direction as negative and upward direction as positive.
= mass of the bullet = 0.00728 kg
= mass of the wooden block = 2.75 kg
= velocity of the bullet before striking the block = 641 ms⁻¹
= velocity of the wooden block before collision
= velocity of the wooden block + bullet after collision = V
Using conservation of momentum
ms⁻¹
Consider the motion of the wooden block in free fall after being dropped:
= Initial velocity of the wooden block = 0 ms⁻¹
= acceleration of the wooden block = - 9.8 ms⁻²
= time of travel
= Final velocity of the wooden block = - 0.85 ms⁻¹
One the basis of above set of data , we can use the kinematics equation
sec
Answer:
1375J
Explanation:
The gravitational potential/potential energy of the at the top of the tree which is the energy by virtue of its position.
P.E = mgh
mass = m
Acceleration due to gravity = g
height = h
At the top of the tree, the value of h (height) is high resulting in the gravitational potential. When the cat lands on the ground, the value of h is zero, the the gravitational potential would be zero and all the potential energy have been converted to other forms of energy.
Therefore, the total gravitational potential store is equal to the maximum amount of energy that can be transferred which is equal to 1375J.
Answer:5m
Explanation:a=-10m/s (since it moved upwards against gravity)
V=0
U=10m/s
v2=u2+2as
0= 100-20s
20s=100
s=5m