Answer:

Explanation:
As we know that there is no external torque on the system of two disc
then the angular momentum of the system will remains conserved
So we will have

now we have

also we have

now from above equation we have

now we have


Answer:
x=4.06m
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
Answer:
When scientists have a question, they form a hypothesis, <em>which</em><em> </em><em>is</em><em> </em><em>an</em><em> </em><em>idea</em><em> </em><em>that</em><em> </em><em>may</em><em> </em><em>be</em><em> </em><em>proved</em><em> </em><em>or</em><em> </em><em>disproved</em><em> </em><em>by</em><em> </em><em>an</em><em> </em><em>experiment</em><em>.</em>
Answer:
1387908 lbm/h
Explanation:
Air flowing into jet engine = 70 lbm/s
ρ = Exhaust gas density = 0.1 lbm/ft³
r = Radius of exit with a circular cross section = 1 ft
v = Exhaust gas velocity = 1450 ft/s
Exhaust gas mass (flow rate)= Air flowing into jet engine + Fuel
Q = (70+x) lbm/s
Area of exit with a circular cross section = π×r² = π×1²= π m²
Now from energy balance
Q = ρ×A×v
⇒70+x = 0.1×π×1450
⇒70+x = 455.53
⇒ x = 455.53-70
⇒ x = 385.53 lbm/s
∴ Mass of fuel which is supplied to the engine each minute is 1387908 lbm/h