Question

6.
A body of mass 0.25 kg moving with a velocity of 12 ms' is stopped by applying a force of 0.6
IN! The impulse of this force
3 Ns
b) – 9 Ns
c) - 12 Ns a ) - 15 Ns
Kirahvusen vertically with a speed of smys and rebounds with the​

Answer 1

Answer:

Answers

chaitanyakrishn1

chaitanyakrishn1 Ace

MASS IS 0.25 KG

INITIAL VELOCITY IS 12m/s

F = 0.6 N

We have to calculate the time.

So here it is

Final velocity v will be 0

F = ma = m (v-u)/t = 0.25 ×-12/ t

Since force is opposite in direction to velocity therefore

F = -0.6 N = -3t

Hence t = 6/30 = 1/5 s = 0.2 seconds

Hope it helps

Thank u★★★

Answer 2

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Impulse and forces

Now basically, An Impulse is large magnitude force applied for a very short period of time.

Thus it is denoted by "J" and it's formula is

J = F. dt

thus units of the impulse are,

Newton-seconds or Ns

now here,

⭐J = F. dt

since we know that, F = m*a

so we get as,

⭐J = m*dv/dt*dt

here, dt gets cancelled, so we left with,

=> J = m*dv

=> J = 0.25*12

=> J = 3 Ns

⭐thus the impulse on the body is 3 Ns.