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Harlamova29_29 [7]
3 years ago
11

How many moles are in 73.4 grams of Phosphorus?

Physics
1 answer:
mylen [45]3 years ago
4 0
23.2


hope this helped
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If a board with an area of 3m2 has a 12 N force exerted on it, what is the pressure on the board?
Cerrena [4.2K]
4 N/cm^2 is the answer. the way you get this is the formula for pressure is P=F/A. So you would plug in the numbers which would make the equation P=12/3 which you know equals 4. The units for pressure is N/cm^2 or N/m^2

I hope this helps 
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3 years ago
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Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin
Sergio [31]

Answer:

Explanation:

Potential energy on the surface of the earth

= - GMm/ R

Potential at height h

=  - GMm/ (R+h)

Potential difference

= GMm/ R -  GMm/ (R+h)

= GMm ( 1/R - 1/ R+h )

= GMmh / R (R +h)

This will be the energy needed  to launch an object from the surface of Earth to a height h above the surface.

Extra  energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

= 1/2 x potential energy at height h

= 1/2 x GMm / ( R + h)

8 0
3 years ago
An electron is pushed into an electric field where it acquires a 1-v electrical potential. suppose instead that two electrons ar
Lapatulllka [165]
1V electrical potential 
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3 years ago
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The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete
oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\

make shear stress (τ) the subject of the formula

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of  0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2

Part (b) shear stress at a distance of  0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0}{2*12} =0

3 0
3 years ago
Describe the relationship between motion<br> and a reference point.
steposvetlana [31]

Answer:

An object is in motion when its distance from another object is changing. ... A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point. You assume that the reference point is stationary, or not moving.

Explanation:

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3 years ago
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