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
Answer:
The water is flowing at the rate of 28.04 m/s.
Explanation:
Given;
Height of sea water, z₁ = 10.5 m
gauge pressure,
= 2.95 atm
Atmospheric pressure,
= 101325 Pa
To determine the speed of the water, apply Bernoulli's equation;

where;
P₁ = 
P₂ = 
v₁ = 0
z₂ = 0
Substitute in these values and the Bernoulli's equation will reduce to;

where;
is the density of seawater = 1030 kg/m³

Therefore, the water is flowing at the rate of 28.04 m/s.
Hi there!
We can use Newton's Second Law:

∑F = net force (N)
m = mass (kg)
a = acceleration (m/s²)
We are given the mass and acceleration, so:
∑F = 20 · 2 = <u>40 N</u>
Answer
D. move a small magnet back and forth within a section of the coiled wire.
Explanation:
i put that for the test and i got it right