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Paraphin [41]
3 years ago
8

A person is able to pull a rope with a force of 700 N. What is the minimum number of fixed and moveable pulleys that the person

will require to lift up a 27,800 N elephant? Assume the pulleys themselves are weightless and the system has 100% efficiency.
Physics
1 answer:
noname [10]3 years ago
6 0

Each pulley reduces the extra effort compulsory to lift the elephant (light and 100% efficiency). If the elephant "weighs" 27,800 N and the person can apply 700 N,
27,800 / 700 = 39.7 which is about 40 N. 
The answer that offers a combined # of 40 pulleys is computed by:
40 / 2 = 20

Therefore the answer is 20 fixed and 20 moveable. 

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The amount of energy needed is 2093 J

Explanation:

The amount of energy needed to increase the temperature of a substance by \Delta T is given by the equation

Q=mC\Delta T

where

m is the mass of the substance

C is its specific heat capacity

\Delta T is the increase in temperature

For the water in this problem, we have

m = 50.0 g = 0.050 kg

C=4186 J/g^{\circ}C (specific heat capacity of water)

\Delta T=10.0^{\circ}C

Therefore, the amount of energy needed is

Q=(0.050)(4186)(10)=2093 J

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Which of the following statements are true? Check all that apply. Check all that apply. The gravitational force between two obje
AveGali [126]

<em>The gravitational force between two objects is inversely proportional to the square of the distance between the two objects.</em>

The gravitational force between two objects is proportional to the product of the masses of the two objects.

The gravitational force between two objects is proportional to the square of the distance between the two objects.  <em> no</em>

The gravitational force between two objects is inversely proportional to the distance between the two objects.  <em> no</em>

The gravitational force between two objects is proportional to the distance between the two objects.  <em> no</em>

The gravitational force between two objects is inversely proportional to the product of the masses of the two objects. <em> no</em>

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Which statement provides the best description of a computational model?
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Answer:

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Explanation:

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3 years ago
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
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