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Flauer [41]
3 years ago
6

A boy is playing with a water hose, which has an exit area of

Physics
1 answer:
Nookie1986 [14]3 years ago
7 0

Answer:

The exit velocity of water is  B. 15 m/s.

Explanation:

According to equation of continuity, for a steady flow of water, the volume of liquid entering a pipe in 1 second is equal to the volume that leaves per second.

If the initial exit area of the pipe is A₁ and the speed of exit is v₁ and the final exit area is A₂ and its corresponding exit velocity  is v₂, then,

Rewrite the expression for v₂.

Substitute 10 cm² for A₁, 2 cm² for A₂ and 3 m/s for v₁.

The exit speed of water from the hose is 15 m/s.

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A 2.4 kg box has an initial velocity of 3.6 m/s upward along a plane inclined at 27◦ to the horizontal. The coefficient of kinet
Vika [28.1K]

Answer:

d= 1.18 m

Explanation:

In abscense of  friction, total mechanical energy must be constant, i.e.,

ΔK + ΔU = 0

As we are told that there exists a kinetic friction between the box and the plane, we need to take into account the work done by the friction force in the equation, as follows:

ΔK + ΔU = Wnc (1)

If we take as our zero gravitational potential energy reference, the height at which the box is sent upward, we can write the following equations for the different terms in (1):

ΔK = Kf- K₀ = 0 - 1/2*m*v₀² = -1/2*2.4kg* (3.6)²(m/s)² = -15.6 J

ΔU = Uf - U₀ = m*g*h = *m*g*d*sin θ = 2.4 kg*9.81 m/s²*d*0.454 = 10.7*d J

Wnc = Ff. d* cos (180º) = μk*N*d*cos(180º) (2)

The friction force always opposes to the displacement, so the angle between force and displacement is 180º.

The normal force, as is always perpendicular to the surface, takes the value needed to equilibrate the component of the weight perpendicular to the incline, as follows:

N = m*g*cosθ =  2.4 kg*9.81 m/s²*cos 27º = 21 N

Replacing in (2):

Wnc = 0.12*21*cos (180º) = -2.52*d J

Replacing in (1):

-15.6 J + 10.7*d J = -2.52*d J

Solving for d:

d = 1.18 m

 

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4 years ago
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Mamont248 [21]

Answer: 18.35 m/s

Explanation:

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