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2 years ago

Why is the fuse used in a circuit called safety fuse?

Physics

2 answers:

Mazyrski [523]2 years ago

6 0

Answer:

The guy above me is right

Explanation:

Sunny_sXe [5.5K]2 years ago

4 0

Answer:

The maximum current which can flow through a fuse without melting it, is called its rating. ... If current higher than 8 A flows through the fuse, it would melt and circuit gets broken. ... Hence, fuse acts as a safety device

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A magnetic field of magnitude 0.550 T is directed parallel to the plane of a circular loop of radius 43.0 cm. A current of 5.80

Assoli18 [71]

Answer:

$\mu = 3.36\times 10^{-3}\ A-m^2$

Explanation:

Given that,

The magnitude of magnetic field, B = 0.55 T

The radus of the loop, r = 43 cm = 0.43 m

The current in the loop, I = 5.8 mA = 0.0058 A

We need to find the magnetic moment of the loop. It is given by the relation as follows :

$\mu = AI\\\\\mu=\pi r^2\times I$

Put all the values,

$\mu=\pi \times (0.43)^2\times 0.0058\\\\=3.36\times 10^{-3}\ A-m^2$

So, the magnetic moment of the loop is equal to $3.36\times 10^{-3}\ A-m^2$ .

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2 years ago

What is kinetic energy

mixas84 [53]

<span>(symbol K)</span><span> Energy that an object possesses because it is in motion. It is the energy given to an object to set it in motion; it depends on the mass (</span>m) of the object and its velocity (v<span>), according to the equation K = 1/2 </span>mv2<span>. On impact, it is converted into other forms of energy such as heat, sound and light.</span>

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2 years ago

What is the working principle of an electric motor?

slava [35]

The answer B homie trust

5 0

2 years ago

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Which medium caused the most bending of light as it passed through?

babunello [35]

I wasn't there observing the experiment while you and your class
performed it, so I don't really know how it was set up, or what
happened.

But I can tell you this: Light doesn't bend while passing through
any medium. It only bends at the boundary where one medium
meets a different one.

4 0

3 years ago

A)Calculate the effective value of g, the acceleration of gravity, at 7900 m above the Earth's surface.

arsen [322]

The solution for the acceleration of gravity is given as

$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$$

$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$$

This is further explained below.

<h3>What is the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.?</h3>

Generally,

Mass of earth $M=5.97 \times 10^{24} \mathrm{~kg}$

Radius of earth $R=6371 \mathrm{~km}$

Gravitational const. $G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$

height $h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$$$

$R+h_{1}=6371+7.9 &\\\\R+h_{1}=6378.9 \mathrm{~km} \\\\&R+h_{1}=6378.9 \times 10^{3} \mathrm{~m}\end{aligned}$

In conclusion, acceleration due to gravity at this point will be

$g_{1}=\frac{G M}{\left(\bar{R}+\overline{h_{1}}\right)^{2}}$\\\\$g_{1}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(6378.9 \times 10^{3}\right)^{2}}$\\\\$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$$

for $h_{2}=7900 \mathrm{~km}$

$R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\$g_{2}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(14271 \times 10^{3}\right)^{2}}$\\\\$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$$

Read more about acceleration due to gravity

brainly.com/question/13860566

#SPJ1

5 0

1 year ago

Why is the fuse used in a circuit called safety fuse?​

Why is the fuse used in a circuit called safety fuse?