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PtichkaEL [24]
3 years ago
10

Give reason: 1. Copper wires are used in domestic circuits. 2. Tungsten is used as filament in bulb. 3. Filament bulbs are repla

ced by CFL and LEDs
Physics
2 answers:
Vilka [71]3 years ago
6 0

Answer:

1.<em>Copper electrical wires are safer to use than wires made of most other conductive metals because they are resistant to heat. As you can see, copper is the preferred metal for electrical wires for several reasons. It has high electrical conductive; it's inexpensive; it's ductile; and it's thermal resistant.</em>

2.<em>Tungsten is used for making filament of an electric bulb due to the following reasons: Being an alloy it has a very high melting point. It has very high resistivity so it does not burn easily at room temperature. The lamp glows at high temperatures.</em>

3.<em>Confused about choosing between CFL bulbs vs. LED bulbs? Since 2014, government regulations have been pushing consumers to replace familiar incandescent light bulbs with more energy-efficient compact fluorescent lamp (CFL) and light-emitting diode (LED) bulbs.Advances in both CFL and LED technologies have expanded consumer choice — yet make evaluating your options a little more complex. Knowing how different kinds of light bulbs stack up against each other will help you not only get the right lighting for any given space but also save energy in the process.</em>

Karolina [17]3 years ago
3 0

Answer:

your mom

Explanation:

You might be interested in
A jogger hears a car alarm and decides to investigate. While running toward the car, she hears an alarm frequency of 872.10 Hz.
Nata [24]

Answer:

v = 4.18 m/s

Explanation:

given,

frequency of the alarm = 872.10 Hz

after passing car frequency she hear = 851.10 Hz

Speed of sound = 343 m/s

speed of the jogger = ?

speed of the

v_f = \dfrac{872.10-851.10}{2}

v_f =10.5\ Hz

v_o = 872.1 - 10.5

V_0 = 861.6\ Hz

The speed of jogger

v = \dfrac{v_1 \times 343}{v_0}-343

v = \dfrac{872.1 \times 343}{861.6}-343

v = 4.18 m/s

5 0
3 years ago
An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much
cupoosta [38]

Answer:

time will elapse before it return to  its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × 10^{7} m/s

uniform electric field E = 1.18 × 10^{4} N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × 10^{-31} kg ×a = 1.18 × 10^{4} × 1.602 × 10^{-19}

a = 20.75 × 10^{14} m/s²

so acceleration is 20.75 × 10^{14} m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × 10^{7} = 0 + 20.75 × 10^{14} (t)

t = 11.80 × 10^{-9} s

so time will elapse before it return to  its staring point is

time = 2t

time = 2 ×11.80 × 10^{-9}

time is 23.6 × 10^{-9} s

time will elapse before it return to  its staring point is 23.6 ns

7 0
3 years ago
Can you please help - the answer I got so far is 7.624 but when I enter it in my homework in Quest, it i stating that this answe
torisob [31]

Answer:

-7.04

Explanation:

9.8 multipled by -0.719 b

6 0
3 years ago
What energy transfer will a stretched rubber band have when let go
GarryVolchara [31]

Answer:

when the rubber band is realeased the potential energy is quickly converted to kinetic energy this is equal to one mass of the the rubber band multiplied by its velocity( in meters per second)

3 0
3 years ago
Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the
BlackZzzverrR [31]

relation between potential difference and electric field is given as

E . d = \Delta V

so here we know that

d = 3 cm

\Delta V = 30 V

E \times 0.03 = 30

E = 1000 N/C

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

5 0
3 years ago
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