Using second law of motion, derive the relation between force and acceleration. A bullet of 10 g strikes a sand-bag at a speed of 103 m s-1 and gets embedded after travelling 5 cm. Calculate (i) the resistive force exerted by the sand on the bullet PLS GIVE THE EXPLANATION THAT WHY HAVE YOU USED POSITIVE OR NEGATIVE SIGN. MENTION THE SIGN OF THE FORCE, You will get 41 points for this
Answers
Explanation:
Answer:
Explanation:
Given
A bullet of 10 g strike the sand bag at a speed of 1000 m/s and gets embedded after travelling 5 cm. calculate
- (a)Resistive force exerted by sand on bullet.
- (b)Time taken by bullet to come to rest.
- Given mass of bullet m = 10 g = 0.01 kg
- Speed v = 1000 m/s
- Distance s = 5 cm = 0.05 m
We know that F = ma
- v^2 = u^2 – 2as
- a = v^2 – u^2 / 2s
Now F = m (v^2 – u^2 / 2s)
= 0.01 x (1000^2 – 0 / 2 x 0.05)
Resistive force will be F = 10^5 N
Now time taken to come to rest will be
S = (u + v / 2) t
So t = 2s / u + v
= 2 x 0.05 / 1000
t = 10 ^-4 secs
Given
A bullet of 10 g strike the sand bag at a speed of 1000 m/s and gets embedded after travelling 5 cm. calculate
(a)Resistive force exerted by sand on bullet.
(b)Time taken by bullet to come to rest.
Given mass of bullet m = 10 g = 0.01 kg
Speed v = 1000 m/s
Distance s = 5 cm = 0.05 m
We know that F = ma
v^2 = u^2 – 2as
a = v^2 – u^2 / 2s
Now F = m (v^2 – u^2 / 2s)
= 0.01 x (1000^2 – 0 / 2 x 0.05)
Resistive force will be F = 10^5 N
Now time taken to come to rest will be
S = (u + v / 2) t
So t = 2s / u + v
= 2 x 0.05 / 1000
t = 10 ^-4 secs
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