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gtnhenbr [62]
3 years ago
9

Why the efficiency of an engine cannot be 100%?

Physics
2 answers:
Marysya12 [62]3 years ago
6 0
<span>No machine can operate at 100 percent efficiency because some of the energy input will always be used to overcome the force of gravity and the effects of friction and air resistance.</span>
Alex73 [517]3 years ago
3 0
No, because the engine can't convert enough heat. 100%? That's way too much. That is theoretically impossible. Its important to consider the equation, this will give you the better picture as to why it can't be. In fact, the maximum efficiency CANNOT be more than the Carnot limit. Its the 2nd Law of Thermodynamics. 
nth ≤ 1 - (Tc/Th) 

I hope this helped.
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The sun is made mostly up of Hydrogen.
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A block of plastic in the shape of a rectangular solid that has height 8.00 cm and area A for its top and bottom surfaces is flo
kifflom [539]

Answer:

0.35 kg

Explanation:

8 cm = 0.08 m

For the block to stay balance, the buoyancy force must be the same as gravity that pulls it down.

Let mass of the block be M, then the gravity would be Mg

Let water density be \rho_w = 1000 kg/m^3, the buoyancy force would be the weight of water that is displaced by the submerged block.

For example, when there is no coin, block is h_0 = 0.0312m submerged. The weight of water displaced must be

W_0 = Ah_0\rho_wg = 0.0312A1000g = 31.2Ag

Which is also the weight of block, of Mg

Therefore M = 31.2A.    (1)

As coins are stacked on top of block, h increase, so as weight of water displaced and total weight of block and coins. Now let m be the total weight of coins. The gravity of block and weight must be (M+m)g. And the weight of water displaced is:

W = Ah\rho g = (M + m)g

h = \frac{M}{A\rho} + \frac{m}{A\rho}

Since the linear plot of h vs m has a slope of 0.089 m/kg, we can interpret it as

\frac{1}{A\rho} = 0.089

A = \frac{1}{0.089\rho} = \frac{1}{89} = 0.011 m^2

So from the eq. (1) we can solve for M = 31.2A = 0.35 kg

8 0
3 years ago
A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C
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Answer:

3.62 m  and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}

\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}

\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}

x = - 1.4 m

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4 years ago
Eating a banana enables s person to perform about 4×10⁴ J of work. To what vertical height does eating a banana enable a 50kg wo
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Answer:

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Explanation:

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p.e = 50 X 10 X h

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h = 80m

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