Exercising and eating right
Answer:
The block is moving 0.564 m/s after the bullet emerges it.
Explanation:
<u>Step 1:</u> Data given
Mass of the bullet m = 4.86 grams = 0.00486 kg
Initial speed of the bullet u = 306 m/s
Mass of the wooden block m' = 1.37 kg
Final speed of the bullet v = 147 m/s
Initial speed of the block u'= 0 m/s
Final speed of the block = v'
<u>Step 2:</u> conservation law of momentum
mu + m'u' = mv + m'v'
0.00486*306 + 1.37 * 0 = 0.00486*147 + 1.37*v'
1.48716 = 0.71442 + 1.37 v'
0.77274 = 1.37 v'
v' = 0.564 m/s
The block is moving 0.564 m/s after the bullet emerges it.
Answer:
3.25 seconds
Explanation:
It is given that,
A person throws a baseball from height of 7 feet with an initial vertical velocity of 50 feet per second. The equation for his motion is as follows :
Where
s is the height in feet
For the given condition, the equation becomes:
When it hits the ground, h = 0
i.e.
It is a quadratic equation, we find the value of t,
t = 3.25 seconds and t = -0.134 s
Neglecting negative value
Hence, for 3.25 seconds the baseball is in the air before it hits the ground.
A=2.5 m/s²
m= 100 Kg
F=m*a
F=force
m=mass
a=acceleration
F=100 Kg*2.5 m/s²=250 N
answer: F=250 N