Question

find the value of the constant froce that is applied for 1 second on a 10 kg to increase the velocity of the object from 5m/s to 10m/s​

Answer 1

Answer:

50N

Explanation:

a=v-u/t

a=10-5/1

a=5m/s²

F=ma

F=10×5

Force =50N

I hope I helped you

Answer 2

$\bold{Given :-}$

$\bold{mass \: = \: 10kg}$

$\bold{time \: = \: 1sec}$

$\bold{initial \: velocity \: (u) = 5m/s}$

$\bold{final \: velocity (v) \: = 10m/s}$

$\bold \red{acceleration \: = \frac{v - u}{t}}$

$\bold{a \: = \: \frac{10 - 5}{1} }$

$\bold{a \: = \: 5m/s²}$

$\bold \red{force \: = \: m \times a}$

$\bold{force \: = \: 10kg \: \times 5m/s²}$

$\bold \green{force \: = 50N.... \: \: \: ans}$