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Anarel [89]
3 years ago
6

A machine has an efficiency of 80% how much work must be done on the machine so to make it do 50000 J of outfit work

Physics
1 answer:
katrin [286]3 years ago
6 0
With efficiency of 80%, the machine outputs 50000 J. The input to the machine should represent 100%.

If
80% --- 50000 J
100% --- ?? J

Solution:
Input work = (100/80)*500000 J = 62500 J
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If it attractive it has opposite pole and if it repulsive it has same pole
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You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is
katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

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brainly.com/question/16492221

5 0
2 years ago
A process with a negative change in enthalpy and a negative change in entropy will generally be:________
valentina_108 [34]

A process with a negative change in enthalpy and a negative change in entropy will generally be: <u>spontaneous</u>.

<h3>Gibbs free energy:</h3>

Since the Gibbs free energy is a parameter that tells us whether a chemical reaction is spontaneous (Gibbs free energy less than 0) or nonspontaneous (Gibbs free energy greater than 0) in this situation, we can describe it mathematically as:

                                      ΔG = ΔH - TΔS

Therefore, any process with a negative change in enthalpy and a positive change in entropy will be spontaneous. If the enthalpy and the entropy are both negative, the subtraction becomes always negative, for which the Gibbs free energy is also negative.

One of the most crucial thermodynamic functions for the characterization of a system is the Gibbs free energy. It influences results like the voltage of an electrochemical cell and the equilibrium constant for a reversible reaction, among others.

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3 0
1 year ago
Which is an example of transfroming potential energy to kinetic​
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Answer:

A ball being dropped to the ground

6 0
3 years ago
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On the kelvin scale what is the freezing point of water
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Freezing point of the water is known as 273 K

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