Answer:
(C) T
The tension T at equilibrium will be equal to the Buoyant force.
The Buoyant force is given by:
Fb = density x acceleration due to gravity x volume displaced
The change in height doesn't affect the Buoyant force and hence the tension.
Note: The figure of question is added in the attachment
Answer: 0.56 m/s
Explanation:
hello, there is 25° inclination angle for the chute in the drawing. Thankfully, I know this problem. The conservation of momentum.
so there are X and Y components for the momentum in this problem. The Y component is not conserved as when the coal gets in the cart, the normal force exerted by the surface reduces it to 0.
Now, the X component is definitely conserved here.
so you have the momentum of the cart which is 440*0.5 added to the momentum of the chunk which is 150*0.8*cos(25°), that is the momentum before the coupling between the objects. Afterwards both objects will have the same velocity, so we write the equation like this:

Answer:

Explanation:
given data:
flow Q = 9 m^{3}/s
velocity = 8 m/s
density of air = 1.18 kg/m^{3}
minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

here
is mass flow rate and given as


Putting all value to get minimum power


Answer:
W = 172.5 J
Explanation:
given,
mass of the fruit crate = 14.5 kg
initial velocity to lift = 0.500 m/s
increase in the tension = 150 N
lift of crate = 1.15 m
work done by the tension = ?
work done = force x displacement
W = F s cos θ
θ = 0°
W = F s x cos 0
W = 150 x 1.15 x 1
W = 172.5 J
Work done on the crate by the tension force = W = 172.5 J
To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From them we will consider speed as the distance traveled per unit of time. Said unit of time will be cleared to find the total time taken to travel the given distance. Later with the calculated average times and distances, we will obtain the average speed.
PART A)
The time taken to travel a distance of 250km with a speed of 95km/h is



Time taken for the lunch is

The time taken travel a distance of 250km with a speed of 55km/h



The total time taken is



The average speed is the ratio of total distance and total time


PART B)
As the displacement is zero the average velocity is zero.