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erik [133]
3 years ago
12

The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on

, the initial current decreases until the filament reaches its steady-state temperature. The temperature coefficient of resistivity of the filament is 43 times 10^-3 K^-1. The final current through the filament is one- eighth the initial current. What is the change in temperature of the filament?
a.30100 K
b.378 K
c.1627 K
d.814 K
e.1162 K
Physics
1 answer:
Neporo4naja [7]3 years ago
5 0

Answer:

162.8 K

Explanation:

initial current = io

final current, i = io/8

Let the potential difference is V.

coefficient of resistivity, α = 43 x 10^-3 /K

Let the resistance is R and the final resistance is Ro.

The resistance varies with temperature

R = Ro ( 1 + α ΔT)

V/i = V/io (1 + α ΔT )

8 = 1 + 43 x 10^-3 x ΔT

7 = 43 x 10^-3 x ΔT

ΔT = 162.8 K

Thus, the rise in temperature is 162.8 K.

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Answer:

The generator produces electrical energy at a rate of 1378125000 J per second.

Explanation:

volume of water falling each second is 1250 m^{3}

height through which it falls, h is 150 m

mass of 1 m^{3} of water is 1000 kg

⇒mass of 1250 m^{3} of water, m = 1250×1000 = 1250000 kg

acceleration due to gravity, g = 9.8 \frac{m}{sec^{2} }

in falling through 150 m in each second, by Work-Energy Theorem:

Kinetic Energy(KE) gained by it = Potential Energy(PE) lost by it

⇒KE = mgh

        = 1250000×9.8×150 J

        = 1837500000 J

Electrical Energy = \frac{3}{4}(KE)

                            = \frac{3}{4}×1837500000

                            = <u>1378125000 J per second</u>

8 0
3 years ago
A parallel-plate capacitor is disconnected from a battery, and the plates are pulled a small distance farther apart. Does Q incr
kolezko [41]

Answer:

Q stay the same

Explanation:

Charging of capacitor is done by battery . If battery is disconnected , charging will stop . There will not be any discharging as plates are separate . So pulling the plates apart will not affect the charge lying on the capacitor . It will decrease its capacity and increase its potential , keeping its charge constant.

4 0
3 years ago
One end of a horizontal spring with force constant 130.0 Ni'm is attached to a vertical wall. A 4.00-kg block sitting on the flo
tekilochka [14]

Answer:

Explanation:

When the spring is compressed by .80 m , restoring force by spring on block

= 130 x .80

= 104 N , acting away from wall

External force = 82 N , acting towards wall

Force of friction acting towards wall = μmg

= .4 x 4 x 9.8

= 15.68 N

Net force away from wall

= 104 -15.68 - 82

= 6.32 N

Acceleration

= 6.32 / 4

= 1.58 m / s²

It will be away from wall

Energy released by compressed spring = 1/2 k x²

= .5 x 130 x .8²

= 41.6 J

Energy lost in friction

= μmg x  .8

= .4 x 4 x 9.8 x .8

= 12.544 J

Energy available to block

= 41.6 - 12.544 J

= 29 J

Kinetic energy of block = 29

1/2 x 4 x v² = 29

v = 3.8 m / s

This will b speed of block as soon as spring relaxes. (x = 0 )

4 0
3 years ago
How many molecules are there in 39 grams of water
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Answer:

39 g H2O contains 1.3 ×1024molecules H2O.

Explanation:

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3 years ago
Determine the field strength, E, experienced by a test charge, q, if a charge of 7.0 × 10-5 coulombs is placed on q and a force
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Formula for feild strength= F/q
q=7.0^10-5 coulombs
F=5.2 N
E=5.2 / 7.0^10-5
E=
7 0
3 years ago
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