Answer:
<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>Explanation:
Part a)
As we know that drag force is given as
so we have
So acceleration of the ball is
Part B)
As per kinematics we know that
Answer:
a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve is
.
Explanation:
a) The work, measured in joules, is a physical variable represented by the following integral:
Where
,
- Initial and final position, respectively, measured in meters.
- Force as a function of position, measured in newtons.
Given that and the fact that
when
, the spring constant (
), measured in newtons per meter, is:
Now, the work function is obtained:
The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.
b) Let be . The area of the region enclosed by one loop of the curve is given by the following integral:
By using trigonometrical identities, the integral is further simplified:
The area of the region enclosed by one loop of the curve is
.
Answer:
4 m/s
Explanation:
Momentum is defined as:
where
m is the mass of the object
v is its velocity
For the object in this problem, we know:
p = 200 kg m/s is the momentum
m = 50 kg is the mass
Solving for the velocity, we find: