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3 years ago

If a fuse melts, does it create an open circuit, a closed circuit, or a short circuit?

Physics

2 answers:

Gekata [30.6K]3 years ago

8 0

Short circuit ……Lynn chekdidbdid eid

Andrew [12]3 years ago

7 0

Answer:

short circuit

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An 800-g block of ice at 0.00°C is resting in a large bath of water at 0.00°C insulated from the environment. After an entropy c

Allisa [31]

Answer:

Unmeltedd ice = 308.109 g

Explanation:

Gibbs Free energy:

A systems Gibbs Free Energy is defined as the free energy of the product of the absolute temperature and the entropy change less than the enthalpy change.

Therefore, G = ΔH-TΔS

where G is Gibbs Free Energy

ΔH is enthalpy change

T is absolute temperature

ΔS is entropy change

Here since there is a phase change, therefore G will be 0.

∴ΔH = TΔS

Given: Temperature, T = 0°C = 273 K

Entropy change,ΔS = 600 J/K

Latent heat of fusion of water = 333 J/g

∴ΔH = TΔS

∴ΔH = 273 x 600

= 163800 J

So this is the amount of enthalpy that will be used into melting of ice.

∴ΔH = mass of ice melted x latent heat of fusion of water

Mass of ice melted = ΔH / latent heat of fusion of water

= 163800 / 333

= 491.891 g

This is the mass of ice melted.

And initial amount of ice is 800 g

Amount of ice left after melting = Initial amount of ice - amount of ice melted

= 800-491.891

= 308.109 g

Amount of ice remained after melting = 308.109 g

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3 years ago

One problem with digital data is that it can be vulnerable to hackers.

hoa [83]

Answer: c. A person who gains unauthorized access to digital data

Explanation:

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1 year ago

A 30° incline permanently sits on a 1.1 meter high table. Starting from rest a ball rolls off the incline with a velocity of 2m/

exis [7]

It would be a good game for you but if I get a pic I don’t want you can you come to my crib I just

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2 years ago

Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0 incline. The coeffi cient of kinetic friction between the bl

UkoKoshka [18]

Answer:

a) $W_g =61.25J$

b) $W_k = -46.25J$

c) $W_N = 0$

d) $W_g$ would be the same.

$W_k$ would decrease.

$W_N$ would be the same.

Explanation:

a) On an inclined plane the force of gravity is the sine component of the weight of the block.

$F_g = mg\sin(\theta) = 5(9.8)\sin(30^\circ)\\W_g = F_g x = 5(9.8)\sin(30^\circ)2.5 = 61.25J$

b) The friction force is equal to the normal force times coefficient of friction.

$F_k = -mg\cos(\theta)\mu_k = -5(9.8)cos(30^\circ)0.436 = -18.5 N\\W_k = -F_kx = -46.25J$

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.

d) The relation between the vertical height and the distance on the ramp is

$h = x\sin(\theta)$

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of $x\sin(\theta)$ .

The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.

The work done by the normal force would still be zero.

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3 years ago

Sometimes when we refer to carbon dioxide we write it as CO2 what is CO2

frez [133]

C stands for carbon. The O stands for oxygen. CO2 is one carbon atom and two oxygen atoms

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2 years ago