Answer :
556.59 m/s.
Explanation:
Given that,
An airplane is traveling at an altitude of 15,490 meters.
A box of supplies is dropped from its cargo hold. We need to find the velocity of cargo when it hits the ground.
The initial velocity of the box is 0 as it as at rest. Let v is the velocity of cargo when it hits the ground. We can find it using third equation of motion as follows :
Put u = 0 and a = g
So, when it hits the ground its velocity is 556.59 m/s.
2. Is right..... if not then go with 1
The closest answer is
Alpha - mass of 4 and charge of +2; beta - no mass and charge of -1; gamma - no mass and no charge (consists of energy)
It’s not exactly correct because a beta particle has the (small) mass of an electron (also the positron). All other choices are way off, I’d go with this one.
Answer:
V = 0.32 m /s
Explanation:
Momentum of the man on the sledge along with sledge
= (80+8) x 10 = 880 kg. m/s When the man jumps off the sledge , the velocity of the sledge will remain intact at 10 m/s
When the sledge hits the boulder and bounces back the momentum of the sledge - boulder system will remain conserved .
change in momentum of sledge = 8 x (10 +6 )
= 128 kg m/s
Change in the momentum of boulder
= 400 V -0
= 400V
400V = 128
V = 0.32 m /s
