Answer:
v = 719.2 m / s and a = 83.33 m / s²
Explanation:
This is a rocket propulsion system where the system is made up of the rocket plus the ejected mass, where the final velocity is
v - v₀ =
ln (M₀ / M)
where v₀ is the initial velocity, v_{e} the velocity of the gases with respect to the rocket and M₀ and M the initial and final masses of the rocket
In this case, if fuel burns at 75 kg / s, we can calculate the fuel burned for the 10 s
m_fuel = 75 10
m_fuel = 750 kg
As the rocket initially had a mass of 3000 kg including 1000 kg of fuel, there are still 250 kg, so the mass of the rocket minus the fuel burned is
M = 3000 -750 = 2250 kg
let's calculate
v - 0 = 2500 ln (3000/2250)
v = 719.2 m / s
To calculate the acceleration, let's use the concept of the rocket thrust, which is the force of the gases on it. In the case of the rocket, it is
Push = v_{e} dM / dt
let's calculate
Push = 2500 75
Push = 187500 N
If we use Newton's second law
F = m a
a = F / m
let's calculate
a = 187500/2250
a = 83.33 m / s²
Answer:
v = 3.04 m/s
Explanation:
given,
mass of the block, M = 6.6 Kg
horizontal force, F = 12.2 N
distance, L = 2.5 m
initial speed = 0 m/s
speed of the block,v = ?
we now
Work done is equal to change in Kinetic energy.
Work done = Force x displacement
W = Δ K E
Δ K E = Force x displacement


3.3 v² = 30.5
v² = 9.242
v = 3.04 m/s
speed of the block is equal to 3.04 m/s
Answer:
the 70kg man
Explanation:
because he has more weight and is moving faster
Answer:
C. crust, mantle, core
Explanation:
density increases as you travel from the crust to the inner core
the crust is on top
next is the mantle
and then the core
Answer:
∆PE = 749.7 J
At 0.9 m high, PE = 793.8 J
At 1.75 m high, PE = 1543.5 J