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
Boyle's law<span> talks about the relationship </span>between<span> pressure and volume (high pressure = low volume, and vice-versa), while </span>Charles's law<span> talks about the relationship </span>between<span> volume and temperature (high temperature = high volume, and vice-versa).</span>
Answer:
120,000J
Corrected question;
In one hour, coal supplies 500 000 J of energy. The wasted energy amounts to 380 000 J. How much useful energy is produced in one hour?
Explanation:
Given;
Total energy Et = 500,000 J
Wasted Energy Ew = 380,000J
The amount useful energy is the amount of energy that is available for supply.
This can be derived by subtracting the wasted energy from the total energy.
Useful energy = Total Energy - wasted energy
Eu = Et - Ew
Substituting the given values;
Eu = 500,000J - 380,000
Eu = 120,000 J
The amount of useful energy produced in one hour is 120,000 J
Answer:
B meet A 0.01 km east of flagpole
Explanation:
given data
distance A = 5.7 km west
velocity V1 = 8.9 km/h
distance B = 4.5 km east
velocity V2 = 7 km/h
to find out
How far runners from the flagpole, when paths cross
solution
we know A and B are 5.7 + 4.5 = 10.2 km apart
and we consider here B will run distance x km for meet
so time will be for B is
time B = distance / velocity
time B = x / 7 ...................1
and
for A distance for meet = ( 10.2 - x ) km
so time A = distance / velocity
time A = ( 10.2 - x ) / 8.9 .............2
now equating equation 1 and 2
time A = time B
x / 7 = ( 10.2 - x ) / 8.9
x = 4.490
so distance of B run for meet is 4.490 km
so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km
so B meet A 0.01 km east of flagpole
F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
