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Yuki888 [10]
2 years ago
15

If a system of pulleys results in a force of 25% of the load, how far will the rope need to move to pull the load a distance of

10 feet?
Engineering
1 answer:
GaryK [48]2 years ago
8 0

Answer:

  40 ft

Explanation:

Assuming no loss of energy in the system of pulleys, the work done is the same whether you move the load directly or through the pulleys.

  W = Fd . . . . . . . . work is the product of force and distance

  F(10 ft) = (0.25F)(d) . . . . . where d is the distance we want to find

  d = 10F/(0.25F) = 40

The rope will need to move 40 feet.

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topjm [15]

Answer: Thats alright bro keep yo head up

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7 0
2 years ago
Read 2 more answers
g A food department is kept at -12oC by a refrigerator in an environment at 30oC. The total heat gain to the food department is
boyakko [2]

Answer:

a) \dot W = 0.417\,kW, b) COP_{R} = 2.198, c) Irreversible.

Explanation:

a) The power input required by the refrigerator is:

\dot W = \dot Q_{H} - \dot Q_{L}

\dot W = \left(4800\,\frac{kJ}{h} - 3300\,\frac{kJ}{h}\right)\cdot \left(\frac{1}{3600} \,\frac{h}{s} \right)

\dot W = 0.417\,kW

b) The Coefficient of Performance of the refrigerator is:

COP_{R} = \frac{\dot Q_{L}}{\dot W}

COP_{R} = \frac{3300\,\frac{kJ}{h} }{(0.417\,kW)\cdot \left(3600\,\frac{s}{h} \right)}

COP_{R} = 2.198

c) The maximum ideal Coefficient of Performance of the refrigeration is given by the inverse Carnot's Cycle:

COP_{R,ideal} = \frac{T_{L}}{T_{H}-T_{L}}

COP_{R,ideal} = \frac{261.15\,K}{303.15\,K - 261.15\,K}

COP_{R,ideal} = 6.218

The refrigeration cycle is irreversible, as COP_{R} < COP_{R,ideal}.

3 0
3 years ago
What form of joining uses heat to create coalescence of the materials?
Allushta [10]
The answer is Soldering
3 0
3 years ago
WILL MARK BRAINLEST PLEASE HELP
TEA [102]

I put

People who pursue a career in the creative imaging fields have qualities like a good imagination, creativity, open minds,  good with ideas, and handling situations. If you enter that field, you need imagination to create things and an open mind to be open to all creations. You need good ideas to make good thing that will work.

please don't plagiarise tho, re-word it.

6 0
3 years ago
a. Determine R for a series RC high-pass filter with a cutoff frequency (fc) of 8 kHz. Use a 100 nF capacitor. b. Draw the schem
Readme [11.4K]

Answer:

a) 199.04 ohms

b) attached in image

c) -0.696dB

Explanation:

We are given:

Fc = 8Khz = 8000hz

C = 100nF = 100*10^-^9F

a)Using the formula:

F_c = \frac{1}{2pie*Rc}

8000= \frac{1}{2*3.14*R*100*10^-^9}

R =\frac{1}{2*3.14*100*10^-^9*8000}

R = 199.04 ohms

b) diagram is attached

c) H(w) = \frac{V_out(w)}{Vin(w)} = \frac{1}{1-j\frac{wc}{w}}

H(F) = \frac{1}{1-j\frac{fc}{f}}

At F = 20KHz and Fc= 8KHz we have:

H(F)= \frac{1}{1-j\frac{8}{20}} = \frac{1}{1-j(0.4)}

|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}

=0.923

|H(F)| in dB = 20log |H(F)|

=20log0.923

= -0.696dB

5 0
3 years ago
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