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vredina [299]
4 years ago
12

What is the purpose of a circuit breaker?

Physics
1 answer:
sashaice [31]4 years ago
8 0
It’s designed to protect an electrical circuit from damage caused by overcurrent, usually resulting from an overload or short circuit. Its basic function is to interrupt current flow after a fault is detected.

That’s really just the basic purpose.
Happy to help!
~Brooke❤️
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How is air temperature related to high and low pressure air
vitfil [10]

Answer: So, I looked at it to see what was the correct one, and the correct answer is Cool air near surface forms high-pressure areas, warm air forms low pressure areas. I hope this helps :D :)

Explanation:

3 0
3 years ago
Um comentarista de futebol certa vez comentou:"A bola bateu na trave e voltou duas vezes mais forte". Sabendo que quando a bola
ryzh [129]

Answer:

Por ela ter batido na trave, não tem como voltar 2x mais forte, por que toda ação correspondente a uma reação de igual intensidade, mas que atua no sentido oposto

Explanation:

7 0
3 years ago
Which of these describes a real image?
Margaret [11]
Image from a far away object formed by a concave mirror

I have no idea but this is my best guess as a sophomore in college
8 0
3 years ago
A man climbs on to a wall that is 3.6m high and gains 2268J of potential energy. What is the mass of the man? *
Ilia_Sergeevich [38]
The mass of the man is 63 kilograms
There is it written down properly. Its hard to type this stuff in here

3 0
4 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
hodyreva [135]

Answer : The correct option is, (c) 3.7\times 10^2J/^oC

Explanation :

First we have to calculate the energy or heat.

Formula used :

E=V\times I\times t

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

E=(3.6V)\times (2.6A)\times (350s)

E=3276J

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}

where,

C = heat capacity of the calorimeter

T_{initial} = initial temperature = 20.3^oC

T_{final} = final temperature = 29.1^oC

Now put all the given values in this formula, we get:

C=\frac{3276J}{(29.1-20.3)^oC}

C=372.27J/^oC=3.7\times 10^2J/^oC

Therefore, the heat capacity of the calorimeter is, 3.7\times 10^2J/^oC

7 0
3 years ago
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