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Gre4nikov [31]
3 years ago
6

Water runs through a plumbing with a flow of 0.750m3/s and arrives to every exit of a fountain. At what speed will the water com

e out of the holes if these have a diameter of 40mm? At what speed would the water come out if the diameter is 120 mm?
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
7 0

Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:

diameter = 40 mm   ==>   area = <em>π</em> (0.04 m)² ≈ 0.00503 m²

diameter = 120 mm   ==>   area = <em>π</em> (0.12 m)² ≈ 0.0452 m²

Then the speed at the end of the 40 mm pipe is

(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s

(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s

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Lerok [7]

Answer:

Option A

Explanation:

The Equation represents the displacement of the object which is represented by x

x=t-t^2

so, x_0  means when time is zero so we replace t with zero in the equation,

x_0=(0)-(0)^2\\x_0=0

now for v which is velocity we need to differentiate the function as the formula for velocity is rate of change of displacement over time so we derivate the equation once and get,

v=1-2t\\

now for  v_0  we insert t = 0 and get

v_0=1-2(0)\\v_0=1

now for a which is acceleration the formula of acceleration is rate of change of velocity over time, so we differentiate the the equation of v(velocity) once or the equation of x(displacement) twice so now we get,

a=-2

so Option A is your answer.

Remember derivative of a constant is always zero because a constant value has no rate of change has its a constant hence the derivative is 0

5 0
3 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Sever21 [200]

Answer:2.89\approx 2.9^{\circ}C/s

Explanation:

Given

Power\left ( P\right )=150 MW

mass of core\left ( m\right )=1.60\times 10^5 kg

Average specific heat \left ( C\right )=0.3349 KJ/kg^{\circ}C

And rate of increase of temperature =\frac{\mathrm{d}T}{\mathrm{d} t}

Now

P=mc\frac{\mathrm{d}T}{\mathrm{d} t}

150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}

Thus \frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}

\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s

6 0
4 years ago
A 35.8 kg box initially at rest is pushed 2.38 m along a rough, horizontal floor with a constant applied horizontal force of 108
tiny-mole [99]

Answer:

The work done by the applied force is 259.22 J.

Explanation:

The work done by the applied force is given by:

W = F*d

Where:

F: is the applied horizontal force = 108.915 N

d: is the distance = 2.38 m  

Hence, the work is:

W = F*d = 108.915 N*2.38 m = 259.22 J

Therefore, the work done by the applied force is 259.22 J.

I hope it helps you!                                                

6 0
3 years ago
Which type of rock is made of tiny bits of animal shells? A. gabbro B. coquina C. clastic sandstone D. limestone
marysya [2.9K]
Hello,



The answer is <span>C. clastic sandstone


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8 0
3 years ago
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Humans first traveled to the Moon in a spacecraft in 1969. What happened as the spacecraft traveled farther from Earth?
horsena [70]

Answer:

The answer is C)The force of gravity from Earth acting on the spacecraft decreased because the distance from Earth increased.

Explanation:

Gravity, a force, is dependent on the mass of the object exerting the gravity and the distance of an outside object from that object. The larger the object, the more gravity it will exert on an outside object. This force decreases as you move away from the object, but it will always still exist and never be equal to 0.

4 0
3 years ago
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