How much work is needed to stop a 20g bullet moving with a speed of 150ms-1
m-20g-0.02kg
v-150ms/1
k.e-1/2mv2
k.e-1/2*0.02*150*150
1/2 to decimal-
0.5*0.02*150*150
225joules
I don't know i am the one asking the question
How much work is needed to stop a 20kg bullet with a speed of 150 minutesHow much work is needed to stop a 20kg bullet with a speed of 150 minutes.
K. E =1/2MV²
MASS= 20g =20/1000
=0.02kg
V = 150ms-¹
K. E = 0.02*150*150/2
=225j
225j
225J
225j
W=ΔKE= mv²/2 =0.02•150²/2=…
Show the formula
225j
Since the bullet is in motion =k.e.ke=1/2*m*v^2.first convert the 20g to its unit kg.divide 20 divide 1000 =0.02.1/2*0.2*150^2=225 joules
How much work is needed to stop a 20kg bullet with a speed of 150 minutes
225j
W=f×s
=20×150
=3000
K.e+p.e
A loaded sack of rice of total Mass 100kg falls down from the floor of a lorry owned by mis Nancy if the hight of the lorry is 2cm calculate the work done by gravity on the load (g=10m/s)
A loaded sack of rice of total Mass 10kg falls down from the floor of a lorry owned by mis Nancy if the hight of the lorry is 2m calculate the work done by gravity on the load (g=10m/s)
W= f x s
W= 20 x 150
=3000j.
667j
m=20g
v=150ms
1/2 to decimal
0.5*0.02*150*150=
150joules
W=f×s
W=20×150=3000joules