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Lady bird [3.3K]
3 years ago
12

Water from a fire hose is directed horizontally against a wall at a rate of 50.0 kg/s and aspeed of 42.0 m/s. Calculate the forc

e exerted on the wall, assuming the water’s horizontalmomentum is reduced to zero.
Physics
2 answers:
xxMikexx [17]3 years ago
6 0

Answer:

Explanation:

initial velocity, u = 42 m/s

mass per unit time, m = 50 kg/s

final velocity, v = 0

Force is equal to the rate of change of momentum

F = m ( v - u) / t

F = 50 x 42 = 2100 N

Ghella [55]3 years ago
4 0

Answer:

magnitude of the force exerted on the wall is 2100 N.

Explanation:

given data

rate = 50.0 kg/s

speed = 42.0 m/s

solution

momentum is reduced = zero

so final speed v = 0

so relation between the force and the momentum is express as

Force =  \dfrac{p}{t}    ...............1

Force = \dfrac{mv}{t}    

put here value and we get

Force = 50 \times 42  

Force = 2100 N

so that magnitude of the force exerted on the wall is 2100 N.

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Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

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Length = L

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since 2 r = D

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Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D   ^2}}{ \dfrac{2\rho L }{\pi \ (D)   ^2}}

\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D   ^2}}* { \dfrac  {\pi \ (D)   ^2} {2\rho L}}

\dfrac{R_s}{R_L} =2

{R_s}=2{R_L}

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

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