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3 years ago

A student is using a 12.9 ft ramp to raise an object 6 ft above the ground.

Engineering

1 answer:

Hatshy [7]3 years ago

8 0

Answer:

2.150

Explanation:

The ideal mechanical advantage of the ramp is given as a ratio between the height of the ramp and the distance from the ground ie

Ideal mechanical advantage of the ramp= $\frac {12.9}{6}=2.150$

Therefore, the answer is 2.150 to the nearest thousandth

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Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the o

tresset_1 [31]

Answer:

Vx = 6.242 x 10raised to power 15

Vy = -6.242 x 10raised to power 15

Explanation:

from E = IVt

but V = IR from ohm's law and Q = It from faraday's first law

I = Q/t

E = Q/t x V x t = QV

hence, E =QV

V = E/Q

3 0

3 years ago

Read 2 more answers

EggzOG43199 :) it has to be more than 20 characters so I'm just saying this

lawyer [7]

Answer:

giberishgiberishgiberishgiberishgiberishgiberishgiberishgiberishgiberish

Explanation:

i have no idea what the question is but do u 4 real need help?

8 0

2 years ago

As Becky was driving "Old Betsy," the family station wagon, the engine finally quit, being worn out after 171,000 miles. It can

andriy [413]

Answer:

1) 4.361 x 10 raised to power 8 revolutions

2) 1.744 x 10 raised to power 9 firings

3) 2.18 x 10 raised to power 8 intake strokes

Explanation:

The step by step explanation is as shown in the attachment

8 0

3 years ago

The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,

babunello [35]

Answer:

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × $10^6$ Pa

viscosity n = 100 Pas

angle = 18°

so Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 $\times 10^{-3}$ )²× 9 $\times 10^{-3}$ × sin18 × cos18

Qd = 94.305 × $10^{-6}$ m³/s

and

Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2

Qb = 11 × $10^{6}$ × π × 85 $\times 10^{-3}$ × ( 9 $\times 10^{-3}$ )³ × sin²18 ÷ 12 × 100 × 2

Qb = 85.2 × $10^{-6}$ m³/s

so here

volume flow rate Qx = Qd - Qb ..............3

Qx = 94.305 × $10^{-6}$ - 85.2 × $10^{-6}$

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

8 0

3 years ago

In the fully developed region of flow in a circular pipe, does the velocity profile change in the flow direction?

taurus [48]

Answer:

No, the velocity profile does not change in the flow direction.

Explanation:

In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.

3 0

3 years ago