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lora16 [44]
3 years ago
15

PLZ HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!! What is a front?

Physics
2 answers:
Phoenix [80]3 years ago
7 0

It’s a set of type of one particular face and size

Shkiper50 [21]3 years ago
4 0

Answer:

in computer and writing context it is many styles of writing that you can make the text look like. I.E: <em>helllo  </em>hello

Explanation:

A font is the combination of typeface and other qualities, such as size, pitch, and spacing. For example, Times Roman is a typeface that defines the shape of each character. Within Times Roman, however, there are many fonts to choose from -- different sizes, italic, bold, and so on.

Mark me the brainliest plz

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One side of the moon always faces Earth because the time it takes the moon to spin on its axis is blank the time it takes the mo
Kamila [148]
The word to fill in the blank is "equal". Because the time taken to rotate (spin on its axis) is equal to the time of revolution (going around the earth), this means that both have the same rate of angular rotation. So for every bit that the moon goes around its orbit around earth, the moon itself rotates accordingly to present the exact same side to earth.
6 0
3 years ago
A water pump pumps 3 kg of water every second from the bottom of a well to the top of the well 50 m above if the pump used 2000
vesna_86 [32]

watts = work per second.

work is mgh = 3x10x50=1500

watts out = 1500

Watts used = 2000

eff=1500/2000=75%


8 0
3 years ago
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
Help it’s actually physical science but plz help
Butoxors [25]
The correct answer is B
3 0
3 years ago
1 pts
mezya [45]

Answer:

46 dgres is the answer

Explanation:

<em>plz</em><em> </em><em>mark</em><em> </em><em>me</em><em> brainliest</em><em> plz</em><em> Plz</em>

4 0
2 years ago
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