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Elina [12.6K]
2 years ago
5

Describe and explain how energy is transferred in the bulbs. Use scientific ideas to support your answer

Physics
1 answer:
Lelechka [254]2 years ago
6 0

Answer:

In the light bulb, the flow of charge through the filament heats it up and causes it to glow. In this way, the light bulb converts electrical energy to heat energy and light energy.

Explanation:

In the light bulb, the flow of charge through the filament heats it up and causes it to glow. In this way, the light bulb converts electrical energy to heat energy and light energy.

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Where must an object be placed to form an image 30.0 cm from a diverging lens with a focal length of 43.0 cm?
Schach [20]
Using lens equation;

1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)

Substituting;

1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm

Therefore, the object should be place 99.23 cm from the lens.
6 0
3 years ago
The half-life of iodine-131 is 8.04 days. It’s decay constant equals
Svetllana [295]

Answer:

Explanation:

we know that half life of an element is

T=0.693/λ

where λ is decay constant in order to find decay constant

λ=0.693/T

λ=0.693/8.04

λ=0.086

7 0
3 years ago
Single Replacement: HCI + Zn
Kay [80]

During a single replacement, the Zn and H will switch places. That means the product formed will be ZnCl and H2.

4 0
3 years ago
A 2-kW electric heater takes 15 min to boil a quantity of water. If this is done once a day and power costs 10 cents per kWh, wh
frozen [14]

Answer:

<h2> $1.50</h2>

Explanation:

Given data

power P= 2 kW

time t= 15 min to hours = 15/60= 1/4 h

cost of power consumption per kWh= 10 cent = $0.1

We are expected to compute the cost of operating the heater for 30 days

but let us computer the energy consumption for one day

Energy of heater  for one day= 2* 1/4 = 0.5 kWh

the cost of operating the heater for 30 days= 0.5*0.1*30= $1.50

<u><em>Hence it will cost  $1.50 for 30 days operation</em></u>

4 0
3 years ago
A power plant produces 1000 MW to suply a city 40Km away.Current flows from the power plant on a single wire of resistance0.050
Westkost [7]

Answer:

The current in wire resistance 2Ω

a). 8696 A

b). fraction power 15.1% a 115kV

Explanation:

Resistance

R=0.05Ω/Km*40km

R=2Ω

P=1000 MW

a).

P=V*I\\I=\frac{P}{V}=\frac{1000x10^{6}W}{115x10^{3}k }  =8696.65A

Using law ohm

b).

V=I*R\\I=\frac{V}{R}

P=I*I*R\\P=I^{2} *R\\P=8696.65^{2}*2\\P=151.228 x10^{6}  W

e=\frac{151.228x10^{6} }{1000x10^{6} }*100= 15.12%

8 0
3 years ago
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