To calculate the force of impact F, first lets calculate the acceleration a of the ball:
a=v/t where v is the velocity of the ball and t is time
a=32/0.8=40 m/s²
To get the force F we need the Newtons second law:
F=m*a where m is the mass of the ball and a is the acceleration.
F=m*a= 0.2*40 = 8 N
So the impact force is F= 8 N.
Explanation:
We have,
The initial position of an object is zero.
The starting velocity is 3 m/s and the final velocity was 10 m/s.
The object moves with constant acceleration..
The area covered under the velocity-time graph gives displacement of the object. The correct option is "the area of the rectangle plus the area of the triangle under the line".
Answer:
a) 0.138J
b) 3.58m/S
c) (1.52J)(I)
Explanation:
a) to find the increase in the translational kinetic energy you can use the relation

where Wp is the work done by the person and Wg is the work done by the gravitational force
By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

the change in the translational kinetic energy is 0.138J
b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

the new speed is 3.58m/s
c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

hence, the change in Er is about 1.52J times the initial rotational energy
Explanation:
The given data is as follows.
Velocity of bullet,
= 814.8 m/s
Observer distance from marksman, d = 24.7 m
Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.
t =
(velocity in air = 343 m/s)
= 0.072 sec
Now, before the observer hears the report the distance traveled by the bullet is as follows.

= 
= 58.66
= 59 (approx)
Thus, we can conclude that each bullet will travel a distance of 59 m.