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Complete Question:
a. A hawk flies in a horizontal arc of radius 11.3 m at a constant speed of 5.7 m/s. Find its centripetal acceleration.
Answer in units of m/s2
b. It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s2. Find the magnitude of acceleration under
these new conditions.
Answer in units of m/s2
Answer:
a. 2.875m/s²
b. 3.172m/s²
Explanation:
a. The formula for centripetal acceleration = (speed²) ÷ radius
Centripetal acceleration = (5.7m/s)²÷ 11.3m
Centripetal acceleration = 2.875m/s²
b. Magnitude of acceleration can be calculated by finding the sum of the vectors for the both the centripetal acceleration and the increase in the speed rate.
Centripetal acceleration ( acceleration x) = 2.875m/s²
Increase in the speed rate ( acceleration n) = 1.34m/s²
Magnitude of acceleration = √a²ₓ + a²ₙ
=√( 2.875m/s²)²+ (1.34m/s²)²
= √ 10.06m/s²
= 3.172m/s²
Answer:
b
Explanation:
determine the rotational kinetic energy about it's center of mass
Answer:

Explanation:
Mass of block=10 kg
Applied horizontal force =F=20 N
Friction force=f=10 N
We have to find the acceleration of block.
Net force=Applied horizontal force-friction force

Where F= Horizontal force
f=Friction force
m=Mass of object
a=Acceleration of object


Hence, the acceleration of the block=
Answer:
0.453 m/s
Explanation:
Assuming the handle has diameter of 0.4 m while inner part diameter is 0.1 m then the circumference of outer part is
where d is diameter and subscript h denote handle. By substituting 0.4 for the handle's diameter then cirxumference of outer part is 
The rate of rotation will then be 1.81/1.256=1.441 rev/s
Similarly, circumference of inner part will be
where subscript i represent inner. Substituting 0.1 for inner diameter then

The rate of rotation found for outer handle applies for inner hence speed will be 0.3142*1.441=0.453 m/s
Answer:
(A) V = 9.89m/s
(B) U = -2.50m/s
(C) ΔK.E = –377047J
(D) ΔK.E = –257750J
Explanation:
The full solution can be found in the attachment below. The east has been chosen as the direction for positivity.
This problem involves the principle of momentum conservation. This principle states that the total momentum before collision is equal to the total momentum after collision. This problem is an inelastic kind of collision for which the momentum is conserved but the kinetic energy is not. The kinetic energy after collision is always lesser than that before collision. The balance is converted into heat by friction, and also sound energy.
See attachment below for full solution.