Answer:
Thrust developed = 212.3373 kN
Explanation:
Assuming the ship is stationary
<u>Determine the Thrust developed</u>
power supplied to the propeller ( Punit ) = 1900 KW
Duct distance ( diameter ; D ) = 2.6 m
first step : <em>calculate the area of the duct </em>
A = π/4 * D^2
= π/4 * ( 2.6)^2 = 5.3092 m^2
<em>next : calculate the velocity of propeller</em>
Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0
∴V^3 = Punit * 2 / A*β
= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )
hence V2 = 8.9480 m/s
<em>Finally determine the thrust developed </em>
F = Punit / V2
= (1900 * 10^3) / ( 8.9480)
= 212.3373 kN
Answer:
Temperature of the mixture=8 °C
Explanation:
Let, Temperature of the mixture=t °C
Q₁ = Q₂
=> m₁SΔΘ₁ = m₂SΔΘ₂
=> 39 × (20+t) = 21 × (60-t)
∴ t = 8 °C
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components of the speed of the coin is given as




now the time taken by the coin to reach the plate is given by



now in order to find the height



so it is placed at 1.52 m height
Answer:
Explanation:
The angular position of first intensity maxima is given by the expression
= λ / d where λ is wave length of light and d is slit separation
putting the values given in the problem. ,
.02 = λ₁ / D
Now wavelength has been increased to 2λ₁
angular separation of first maxima
= 2λ₁ /D
= 2 x .02
= .04 radians.
The work done on the puck is 96 J
Explanation:
According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.
Mathematically:
where
is the final kinetic energy of the puck, with
m = 2 kg being the mass of the puck
v = 10 m/s is the final speed
is the initial kinetic energy of the puck, with
u = 2 m/s being the initial speed of the puck
Substituting numbers into the equation, we find the work done by the player on the puck:
Learn more about work and kinetic energy:
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