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s344n2d4d5 [400]
2 years ago
5

You wish to lift a 12,000 lb stone by a vertical distance of 15 ft. Unfortunately, you can only generate a maximum pushing force

of 2,000 lb. What is the amount of work you must input in order to move the stone
Physics
1 answer:
DochEvi [55]2 years ago
6 0

Answer:

5,760,000 lb-ft²/s²

Explanation:

Work done, W = mgy where m = mass of stone = 12,000 lb, g = 32 ft/s² and y = vertical distance = 15 ft.

So, W = mgy = 12,000 lb × 32 ft/s² × 15 ft. = 5,760,000 lb-ft²/s²

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At present most of the world's energy needs are supplied by what kind of energy
antiseptic1488 [7]

Fossil fuels . . . coal, oil, natural gas

Among primitive cultures, wood is an important source.
4 0
3 years ago
Help meh in this question plzzz <br>​
iragen [17]

The Moment of Inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

I = I_{D} - I_{H} (1)

Where:

  • I_{D} - Moment of inertia of the Disk.
  • I_{H} - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right) (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole (m):

\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}

m = \frac{1}{2}\cdot M

And the resulting equation is:

I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)

I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}

I = \frac{15}{32}\cdot M\cdot R^{2}

The moment of inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

5 0
2 years ago
A compound has a melting point of 812°C. what type of bonding is in the compound?
AURORKA [14]

Answer:

Ionic compound.

Explanation:

812° C is a very high melting point. Such high melting points are generally ionic compound. Ionic compounds are have very strong bond between the elements ( electrostatic bond). In order to break this bond, large amount of heat energies are needed. So, they have high melting point. Also, Ionic compound are very good conductors of electricity.

7 0
2 years ago
In which situation is chemical energy being converted to another form of energy?
otez555 [7]

Answer:

A burning candle. (chemical energy into energy of heat and light, i.e. thermal and wave)

Explanation:

5 0
2 years ago
A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 385 kV. The secondary of
enot [183]

Answer:

a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new output is 86% of the old output

c) The losses in the new line are 74% the losses in the old line.

Explanation:

a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:

\frac{n_1}{n_2}=\frac{V_1}{V_2}

In the old transformer the ratio of voltages was:

\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{385} =0.03117\\\\n_2=n_1/0.03117=32.1n_1

In the new transformer the ratio of voltages is:

\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1

In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new current ratio is

\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1

If the old current output was 425 kV, the ratio of current was:

\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{425}= 0.028\\\\I_2=0.028I_1

Then, the ratio of the new output over the old output is:

\frac{I_{2new}}{I_{2old}} =\frac{0.024\cdot I_1}{0.028\cdot I_1}= 0.86

The new output is 86% of the old output (smaller output currents lower the losses on the transmission line).

c) The power loss is expressed as:

P_L=I^2\cdot R

Then, the ratio of losses is (R is constant for both power losses):

\frac{P_n}{P_o} =\frac{I_n^2R}{I_o^2R} =(\frac{I_n}{I_o} )^2=0.86^2=0.74

The losses in the new line are 74% the losses in the old line.

7 0
2 years ago
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