<span>Discharge is the volume of water moving down a stream or river per unit of time, commonly expressed in cubic feet per second or gallons per day. In general, river discharge is computed by multiplying the area of water in a channel cross section by the average velocity of the water in that cross section: discharge = area * velocity. In this case, the answer is 0.2 m/s.</span>
Answer:
see explanation below
Explanation:
Given that,
500°C
= 25°C
d = 0.2m
L = 10mm = 0.01m
U₀ = 2m/s
Calculate average temperature

262.5 + 273
= 535.5K
From properties of air table A-4 corresponding to
= 535.5K 
k = 43.9 × 10⁻³W/m.k
v = 47.57 × 10⁻⁶ m²/s

A)
Number for the first strips is equal to


Calculating heat transfer coefficient from the first strip


The rate of convection heat transfer from the first strip is

The rate of convection heat transfer from the fifth trip is equal to


Calculating 

The rate of convection heat transfer from the tenth strip is


Calculating

Calculating the rate of convection heat transfer from the tenth strip

The rate of convection heat transfer from 25th strip is equal to

Calculating 

Calculating 

Calculating the rate of convection heat transfer from the tenth strip

Answer:

Explanation:
Since volume of all the liquid is always conserved
so here we know that




now we know that total surface area is given as



Explanation:
Given that,
Force with which a child hits a ball is 350 N
Time of contact is 0.12 s
We need to find the impulse received by the ball. The impulse delivered is given by :

So, the impulse is 42 N-m..
We know that he change in momentum is also equal to the impulse delivered.
So, impulse = 42 N-m and change in momentum =42 N-m.
Answer:
The ball was in air for 3.896 s
Explanation:
given,
g = 9.8 m/s², acceleration due to gravity,
If the launch angle is 45°, the horizontal range will be maximum.
The horizontal and vertical launch velocities are equal, and each is equal to
v_h = v cos θ
v_h = 27 × cos 45°
= 19.09 m/s.
The time to attain maximum height is one half of the time of flight.
v = u + at ∵ v = 0 (max. height)
19.09 - 9.8 t₁ = 0
t₁ = 1.948 s
The time of flight is twice of the maximum height time
2 t₁ = 3.896 s
The horizontal distance traveled is
D = v × t
D = 3.896×19.09
= 74.375 m
The ball was in air for 3.896 s