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:
Explanation:
Let T be the tension in the cord.
Impulse by cord = change in momentum of block A .
T x 5s = 10 ( 2 -0) = 20
T = 4 poundal .
acceleration of block B = 2 / 5 = 0.4 m /s²
Net force applied on A = m ( g + a ) where m is mass of block B , a is acceleration of block B .
= 8 ( 32 + .4 ) = 259.2 poundal
Frictional force on block A = 259.2 - 4 = 255.2 poundal
μ x 10 x 32 = 255.2
320μ = 255.2
μ =0 .8 .
Explanation:
It is given that,
Mass of the ball, m = 0.06 kg
Initial speed of the ball, u = 50.4 m/s
Final speed of the ball, v = -37 m/s (As it returns)
(a) Let J is the magnitude of the impulse delivered to the ball by the racket. It can be calculated as the change in momentum as :
J = -5.24 kg-m/s
(b) Let W is the work done by the racket on the ball. It can be calculated as the change in kinetic energy of the object.
W = -35.1348 Joules
Hence, this is the required solution.
Answer:
8.3 x 10⁻⁷ C
Explanation:
Electric flux will enter the face at x=0 and exit at face x= 25 m
On the other faces , field lines are parallel so no flux will enter or exit .
Flux entering the face at x = 0
= electric field x face area
= 560 x 25 x 25 = 350000 weber
Flux exiting the face at x = 25
= 410 x 25 x25
= 256250 weber
Net flux exiting from cube ( closed face )
350000 - 256250 = 93750 web
Apply gauss'es theorem
Q / ε = Flux coming out
Q is charge inside the closed cube
Q / ε = 93750
Q = 8.85 x 10⁻¹² x 93750
= 8.3 x 10⁻⁷ C
Answer:
Explanation:
Ignoring air resistance
Initial vertical velocity is 30sin35 = 17.2 m/s
Gravity reduces this velocity to zero in a time of
t = v/g =17.2 / 9.8 = 1.755 s
it takes the same time to come back down to ground level for a total flight time of 2(1.755) = 3.51 s
The horizontal velocity is 30cos35 = 24.57 m/s
the distance traveled horizontally is
d = vt = 24.57(3.51) = 86.298... = 86 m
