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vlada-n [284]
3 years ago
11

What is an ellipse?

Physics
1 answer:
Rom4ik [11]3 years ago
6 0

Answer:

i think it's C thx correct me if wrong

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A ski lift carries people along a 220-meter cable up the side of a mountain. Riders are lifted a total of 110 meters in elevatio
jeka94

The ideal mechanical advantage (IMA) can be determined by the following equation:

 IMA= Input distance/Output distance

 The Input distance and Output distance are:

 Input distance=220 meters

 Output distance=110 meters

 When you substitute in the equation of the ideal mechanical advantage (IMA), you obtain:

 IMA= Input distance/Output distance

 IMA= 220 meters/110 meters

 IMA=2

3 0
3 years ago
Gravity is a <br><br> A.pushing force<br> B. not a force at all<br> C. pulling force
melamori03 [73]

Answer:

C

Explanation:

gravity is a pulling force according to Newton

4 0
3 years ago
A roller of radius 12.5 cm turns at 14 revolutions per second. What is the linear velocity of the roller in meters per second?
Firdavs [7]

12.5 times 14 and convert to meters its 1.75 meters per second

7 0
3 years ago
Read 2 more answers
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
4 years ago
What is happening in this picture?
Alexandra [31]
The answer is b no problem
7 0
3 years ago
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