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3 years ago

12

An what is any device that makes work easier by changing the input force.

1 answer:

klio [65]3 years ago

6 0

Answer:

A machine

Explanation:

A machine is any device that makes work easier. Machines make work easier by changing the strength or direction of a force. Machines do not decrease the amount of work that needs to be done, they just make work easier by changing the way it is done.

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explain why the EMF of a dry cell drops if a large current is drawn for a short time and then recovers if allowed to rest

Delvig [45]

Answer:

Manganese (iv )oxide is a slow depolarizer and polarization occurs with a large current, on resting, the depolarization returns the p.d of the dry cell.

8 0

3 years ago

Magnetic field lines curve out from one pole and return to the same pole.<br>t/f

geniusboy [140]

That statement is a big fat prevarication.
Magnetic field lines start at one Pole and end at the OTHER one.

6 0

3 years ago

Daniel [21]

Can't say for sure but I think it's a) or c)

3 0

4 years ago

A system in which only one particle can move has the potential energy shown in (figure 1). Suppose u1 = 60 j. What is the y-comp

marta [7]

The y-component of the force on the particle at the given position is 120 N.

<h3>Electric force on the particle</h3>

The electric force on the particle is determined by applying Coulomb's law and work-energy theorem as shown below;

Fd = W

Where;

F is the applied force
d is the distance
W is potential

F = W/d

F = 60/0.5

F = 120 N

Thus, the y-component of the force on the particle at the given position is 120 N.

Learn more about electric force here: brainly.com/question/20880591

3 0

2 years ago

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.250 A/s,

sdas [7]

Explanation:

Given that,

Induced emf $\epsilon= 1.65\times10^{-3}\ V$

Rate of current = 0.250 A/s

Number of turns =30

(a). We need to calculate the mutual inductance of the pair of coils

Using formula of the mutual inductance

$M=\dfrac{\epsilon}{|\dfrac{\Delta i}{\Delta t}|}$

$M=\dfrac{1.65\times10^{-3}}{0.250}$

$M=0.0066=6.6\times10^{-3}\ Hz$

$M=6.6\ mHz$

The mutual inductance of the pair of coils is 6.6 mHz.

(b). We need to calculate the flux through each turn

Using formula of flux

$\phi_{B}=\dfrac{Mi}{N}$

Put the value into the formula

$\phi_{B}=\dfrac{6.6\times10^{-3}\times1.25}{30}$

$\phi_{B}=0.000275 =2.75\times10^{-4}\ Wb$

The flux through each turn is $2.75\times10^{-4}\ Wb$

(c). We need to calculate the magnitude of the induced emf in the first coil

Using formula of induced emf

$\epsilon=M|\dfrac{\Delta i_{2}}{\Delta t}|$

$\epsilon=6.6\times10^{-3}\times0.3$

$\epsilon=0.00198 =1.98\times10^{-3}\ mV$

The magnitude of the induced emf in the first coil is $1.98\times10^{-3}\ mV$

Hence, This is the required solution.

8 0

3 years ago

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