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Sliva [168]
2 years ago
10

While walking past a construction site, a person notices a pipe sticking out of a second floor window with water pouring out. As

the water flows to the ground, it speeds up due to the effect of gravity. How does the diameter of the flowing stream of water change as it descends? Assuming that the flow remains laminar, its diameter
Physics
1 answer:
tia_tia [17]2 years ago
4 0

Answer:

Its diameter increases as it flows down from the pipe. Assuming laminar flow for the water, then Bernoulli's equation can be applied.

P1-P2 + (rho)g(h1 - h2) + 1/2(rho)(v1² - v2²) = 0

Explanation:

P1 = P2 = atmospheric pressure so, P1 - P2 = 0

h1 is greater than h2 so h1-h2 is positive. Rearranging the equation above 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho = v2²

From the continuity equation for fluids

A1v1 = A2v2

v2 = A1v1/A2

Substituting into the equation above

(A1v1/A2)² = 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho

Making A2² the subject of the formula,

A2² = (A1v1)²× rho/(2{ (rho)g(h1-h2) + 1/2(rho)v1²}

The denominator will be greater than the numerator and as a result the diameter of the flowing stream decreases.

Thank you for reading.

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In the diagram below, a 10-kilogram ball is fired with a
Tcecarenko [31]

Answer:

A) 5m/s

Explanation:

Momentum of ball in one direction=momentum of cannon in other direction

mv = mv

(10)(500)=(1000)v

5000=1000v

∴v=5m/s

3 0
2 years ago
A horizontal force of 24 N is being applied to a box on a level surface. The box is accelerating at the rate of 2.0 against a fr
Leya [2.2K]

Answer:

For SI units the mass m = 5 kg.

Explanation:

The sum of all horizontal forces is must be equal to maₓ:

∑Fₓ = maₓ = 24N - 14N = 10N

maₓ = 10N

m = 10N / aₓ = 10N / 2 m/s² = 5 m/s

7 0
3 years ago
A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.62 and μk =
GREYUIT [131]

The frictional force is 218.6 N

Explanation:

The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.

There are two forces acting along this direction:

- The component of the weight parallel to the incline, downward along the plane, of magnitude

mg sin \theta

where

m = 46 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

\theta=29^{\circ} is the angle of the incline

- The (static) frictional force, acting upward, of magnitude F_f

Since the block is in equilibrium, we can write

mg sin \theta - F_f = 0

And substituting, we find the force of friction:

F_f = mg sin \theta = (46)(9.8)(sin 29^{\circ})=218.6 N

Learn more about frictional force along an inclined plane:

brainly.com/question/5884009

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7 0
2 years ago
10.
aliina [53]

The normal force is the supporting force that is exerted on an object that is in contact with another stable object.

Answer: Option C

<u>Explanation: </u>

Normal force is forward or upward pushing force acting on an object. Mostly the normal force acts as supporting force exerted on the object by the neighbouring stable object with which the object in question is in contact. So normal force falls under the category of contact forces.

Generally, normal force will be acting to support the weight of any object placed on another object. The best examples of normal forces are the weight of the book supported by table or by the pushing force of the wall on the person leaning on the wall.

5 0
2 years ago
Read 2 more answers
PLEASE HELP
Anvisha [2.4K]

The speed of the rock at 20 m is 34.3 m/s

Explanation:

We can solve this problem by using the law of conservation of energy: the mechanical energy of the rock, sum of its potential energy + its kinetic energy) must be conserved in absence of air resistance. So we can write:

U_i +K_i = U_f + K_f

where :

U_i is the initial potential energy

K_i is the initial kinetic energy

U_f is the final potential energy

K_f is the final kinetic energy

The equation can also be rewritten as  follows:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 100 kg is the mass of the rock

g=9.8 m/s^2 is the acceleration of gravity

h_i = 80 is the initial height

u = 0 is the initial speed  (the rock starts at rest)

h_f = 20 m is the final height of the rock

v is the final speed when h = 20 m

And solving for v, we find:

v=\sqrt{2g(h_i-h_f)}=\sqrt{2(9.8)(80-20)}=34.3 m/s

Learn more about kinetic energy and potential energy here:

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5 0
3 years ago
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