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tekilochka [14]
3 years ago
11

In a practical machine, the power output is smaller than the power input. Considering the law of energy conservation, what can e

xplain this occurrence?
A. Some of the energy is used to combat friction, and thus is transformed from mechanical energy to heat.

B. Some of the energy going into the machine is being created at a quicker rate than the energy coming out is being destroyed.

C. Some of the energy coming out of the machine is being destroyed.

D. Some of the energy going into the machine is being destroyed
Physics
2 answers:
TEA [102]3 years ago
7 0

PF STUDENTS:  Some of the energy is used to combat friction, and thus is transformed from mechanical energy to heat.

Grace [21]3 years ago
5 0

The law of energy conservation states that energy cannot be created or destroyed, so choices B to D are immediately invalid. Choice A can explain this occurrence: <u>A. Some of the energy is used to combat friction, and thus is transformed from mechanical energy to heat.</u>

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Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

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Answer:

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220V is applied to two different conductors made of the same material. One conductor is twice as long and twice as thick as the
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Answer:

Explanation:

For calculating resistance of a conductor , the formula is

R = ρ l / A , ρ is specific resistance , l is length and A is cross sectional area of wire.

For first wire length is l₁ , area is A₁ resistance is R₁, for second resistance is R₂ , length is l₂ and area is A₂

Given , l₁ = 2l₂ , A₁ = 4A₂ , area is proportional to square of thickness.

R₁ / R₂ = I₁A₂ / I₂A₁

= 2l₂ x A₁ / 4 I₂A₁

= 1 / 2

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Ratio of power = (V² / R₁) x (R₂ / V²)

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