Answer:
Explanation:
Given that,
The mass of the bullet is
M= 28g = 0.028kg.
Thickness oof sack bag
d = 30cm = 0.3m
Initial velocity of bullet Vi = 55m/s
Velocity at which the bullet emerges the sack bag is Vf = 18m/s
We want to calculate the frictional force present in the sack bag.
Using conservation of energy.
Work done by friction in the sack is equal to change in kinetic energy.
Work done by friction is given as
Wf = Fr × d
The distance is the thickness of the bag
Wf = Fr × 0.3
Wf = 0.3 • Fr
Change in kinetic energy is calculated by
∆K.E = ½mVf² — ½ mVi²
∆K.E = ½m(Vf² — Vi²)
∆K.E = ½ × 0.028 ( 18²-55²)
∆K.E = 0.014 × -2701
∆K.E = -37.814 J
Since
Work done by friction = ∆K.E
0.3 Fr = -37.814
Fr = -37.814/0.3
Fr = -126.05 N
The, negative sign show that frictional force is opposing the motion.
Fr = 126.05 N
If a car travels 18,000 meters in 30 minutes, what is its average speed?
Answer:
Venus
Explanation:
basic 2nd grade smarts "no offense"
