Answer:
Explanation:
The speed of the water in the large section of the pipe is not stated
so i will assume 36m/s
(if its not the said speed, input the figure of your speed and you get it right)
Continuity equation is applicable for ideal, incompressible liquids
Q the flux of water that is Av with A the cross section area and v the velocity,
so,
the diameter decreases 86% so
Thus, speed in smaller section is 48.6 m/s
Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
1.
For the first part, we just need to write the equation of the forces along two perpendicular directions.
We have actually only two forces acting on the car, if we want it to go around the track without friction:
- The weight of the car, mg, downward
- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)
By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:
(1)
(2)
where in the second equation, the term represents the centripetal force, with v being the speed of the car and R the radius of the track.
Dividing eq.(2) by eq.(1), we get the following expression:
2.
In this second situation, the cars moves around the track at a speed
This means that the centripetal force term
is now larger than before, and therefore, the horizontal component of the normal reaction, , is no longer enough to keep the car in circular motion.
This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes
And re-arranging for F,
(3)
But from eq.(2) in the previous part we know that
So, susbtituting into eq.(3),
Answer:
Option B, 93 cm
Explanation:
An diagram of the seed's motion is attached to this solution.
This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.
And this would be obtained from the equations of motion,
First of, the height of the plant is related to some quantities of the motion with this relation.
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)
(substituting the t = √(2H/g) derived from above
R = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2
R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.
QED!
