Explanation:
because the boy has larger surface area due to which he offers the larger air resistance which decreases the acceleration so, he will fall towards the earth's surface approximately with constant velocity.
(a) The velocity of the first ball before the collision with the second ball is 11.18 m/s.
(b) The final velocity of the two balls after the collision is determined as 5.59 m/s.
<h3>
Speed of the block when pushed by the spring</h3>
The speed of the block when pushed by the spring is calculated as follows;
K.E = Ux
¹/₂mv² = ¹/₂kx²
mv² = kx²
v² = kx²/m
v² = (25 x 0.5²)/0.05
v² = 125
v = 11.18 m/s
<h3>Final velocity of the two balls after the collision</h3>
The velocity of the two balls after the collision is calculated as follows;
Pi = Pf
where;
- Pi is initial momentum
- Pf is final momentum
m1u1 + m2u2 = v(m1 + m2)
0.05(11.18) + 0.05(0) = v(0.05 + 0.05)
0.559 = 0.1v
v = 5.59 m/s
Learn more about velocity here: brainly.com/question/4931057
#SPJ1
Rocket thrust equation
= ( mass flow rate of fuel burnt ) X (Velocity of gas ejected ) + ( Exit Pressure - Outdoor Pressure ) X ( Area of exhaust )
In this case, we can assume the exit pressure = outdoor pressure and since area of exhaust is not given, it can be assumed to be negligible.
In this case, by Newton 3rd’s law,
Force exerted by gas on rocket
= Force exerted by rocket on gas
= (10kg/s) X (5 x 10^3 m/s)
= 5 x 10^4 N
The answer would be false