When a rocket lifts off what provides the momentum match of the upper lift
1 answer:
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The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.
Explanation:
When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.
As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.
Since, the inlet volume flow is measured as the product of velocity with the area.
Inlet volume flow=Inlet velocity*Area*time
And the mass flow rate is
Mass flow rate in the inlet=density*area*inlet velocity*time
Mass flow rate in the outlet=density*area*outlet velocity*time
Since, the time and area is constant, the inlet and outlet will be same as
(Mass inlet)/(density*inlet velocity)=Area*Time
(Mass outlet)/(density*outlet velocity)=Area*Time
As the ratio of mass to density is termed as specific volume, then
(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)
Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity
As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity
is
So, the inlet velocity is 1.4035 m/s.
Then the inlet volume will be
As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s
So, the inlet volume is 0.019 m³/s.
Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.
a) See free-body diagram in attachment
b) The book is stationary in the vertical direction
c) The net horizontal force is 35 N in the forward direction
d) The net force on the book is 35 N in the forward horizontal direction
e) The acceleration is in the forward direction
Explanation:
a)
The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.
From the diagram of this book, we see there are 4 forces acting on the book:
- The applied force, F = 50 N, pushing forward in the horizontal direction
- The frictional force, , pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)
- The weight of the book, , where m is the mass of the book and
is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:
- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book
b)
The book is in equilibrium in the vertical direction, therefore there is no motion.
In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is
where a is the acceleration; however, since N = W, this becomes
And since the book is initially at rest on the desk, this means that there is no motion.
c)
We said there are two forces acting in the horizontal direction:
- The applied force, F = 50 N, forward
- The frictional force, , backward
Since they act along the same line, we can calculate their resultant as
and therefore the net force is 35 N in the forward direction.
d)
The net force is obtained as the resultant of the net forces in the horizontal and vertical direction. However, we have:
- The net force in the horizontal direction is 35 N
- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction
Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.
e)
The acceleration of the book can be calculated by using Newton's second law:
where
is the net force
m is the mass
a is the acceleration
Here we have:
(in the forward direction)
m = 4 kg
Therefore, the acceleration is
(forward)
Learn more about forces, weight and Newton's second law:
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Answer:
3.7 m/s
Explanation:
M = 444 kg
U = 5 m/s
m = 344 kg
u = - 5 m/s
Let the velocity of train is V and the car s v after the collision.
As the collision is elastic
By use of conservation of momentum
MU + mu = MV + mv
444 x 5 - 344 x 5 = 444 V + 344 v
500 = 444 V + 344 v
125 = 111 V + 86 v .... (1)
By using the formula of coefficient of restitution ( e = 1 for elastic collision)
-5 - 5 = V - v
V - v = - 10
v = V + 10
Substitute the value of v in equation (1)
125 = 111 V + 86 (V + 10)
125 = 197 V + 860
197 V = - 735
V = - 3.7 m/s
Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.