Answer:
Explanation:
Due to heat energy , metal expands . Formula for linear expansion is as follows .
L = l ( 1 + α Δt )
where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .
To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L . The linear coefficient of brass and steel are
20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .
For steel sphere ,
L = 25 ( 1 + 12 x 10⁻⁶ Δt )
For brass ring
L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )
25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )
1.004( 1 + 12 x 10⁻⁶ Δt ) = ( 1 + 20 x 10⁻⁶ Δt )
1.004 + 12.0482 x 10⁻⁶ Δt = 1 + 20 x 10⁻⁶ Δt
.004 = 7.9518 x 10⁻⁶ Δt
Δt = 4000 / 7.9518
= 503⁰C.
final temp = 503 + 15 = 518⁰C .
Answer:
a) 24.4 Ω
b) 4.92 A
c) 495.9 W
d)
c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.
Explanation:
b)
The formula for power is:
P = IV
where,
P = Power of heater = 590 W
V = Voltage it takes = 120 V
I = Current Drawn = ?
Therefore,
590 W = (I)(120 V)
I = 590 W/120 V
<u>I = 4.92 A</u>
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a)
From Ohm's Law:
V = IR
R = V/I
Therefore,
R = 120 V/4.92 A
<u>R = 24.4 Ω</u>
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c)
For constant resistance and 110 V the power becomes:
P = V²/R
Therefore,
P = (110 V)²/24.4 Ω
<u>P = 495.9 W</u>
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d)
If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:
<u>c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.</u>
so, 444 eggs would have been released in 37yrs
Answer:
E = 13.2 kWh
, Cost = $ 10.8
Explanation:
We can look for the consumed energy from the expression of the power
P = W / t
The work is equal to the variation of the kinetic energy, for which
P = E / t
E = P t
look for the energy consumed in one day and multiply by the days of the month in the month
E = 110 4 30
E = 13200 W h
E = 13.2 kWh
the cost of this energy is
Cost = 0.9 12
Cost = $ 10.8
