Ask question

Ask question

All categories

3 years ago

7

How to use a fuse and where do we use it ? Please help!!!!

1 answer:

UNO [17]3 years ago

4 0

Answer:

Explanation:

A fuse is an electrical safety device built around a conductive strip that is designed to melt and separate in the event of excessive current. Fuses are always connected in series with the component(s) to be protected from overcurrent, so that when the fuse blows (opens) it will open the entire circuit and stop current through the component(s). A fuse connected in one branch of a parallel circuit, of course, would not affect current through any of the other branches.

You might be interested in

A 0.46-kg cord is stretched between two supports, 7.2 m apart. When one support is struck by a hammer, a transverse wave travels

Kryger [21]

Answer:

T = 6.0 N

Explanation:

given,

mass of the cord = 0.46 Kg

length of the supports = 7.2 m

time taken to travel = 0.74 s

tension in the chord = ?

using formula for tension calculation

$T = \dfrac{v^2.m}{l}$

$v = \dfrac{l}{s}$

$v = \dfrac{7.2}{0.74}$

v = 9.73 m/s

now, calculation of tension

$T = \dfrac{9.73^2\times 0.46}{7.2}$

T = 6.0 N

The tension in the cord is equal to 6.0 N.

7 0

3 years ago

How much power is used in a machine that produces 15 Joules of work in 3 seconds? Use the formula P = W/t.

Marta_Voda [28]

P=w/t

w=15
t=3
therefore, 5 watts (b)

7 0

3 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

5 0

3 years ago

A planet's temperature depends on its distance from the Sun as well as the strength of its greenhouse effect. Let's remove the e

pishuonlain [190]

Answer:

Smallest Effect

Mars

Earth

Venus

Largest Effect

5 0

4 years ago

The melting point of a substance occurs at the same temperatures as it’s blank point

azamat

I think it's a.

Explanation:

melting point is boiling

4 0

4 years ago

Read 2 more answers