Explain how a step-down transformer makes electricity safe for use in homes.

Physics

2 answers:

notka56 [123]2 years ago

5 0

Answer:

Sample Response: High-voltage current enters the primary winding, which induces a magnetic field in the core. The magnetic field induces a current in the secondary winding. The voltage in the secondary winding is lower because it has fewer coils than the primary winding.

Explanation

Savatey [412]2 years ago

3 0

Answer:

it lowers the voltage in homes

Explanation:

this is due to safety reasons as a human cant survive power line voltages which are between 150k and 760k and if you plugged anything short of a piece of 1 inch thick wire not counting insulation it would immediately start smoking and burn the cable.

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Now the elevator is moving downward with a velocity of v = -2.8 m/s but accelerating upward with an acceleration of a = 5.5 m/s2

borishaifa [10]

Answer:

160.75 N

Explanation:

The downward velocity has no effect on the force situation, it is only changes in velocity (plus, of course, gravity, which is always there) that require a force. At constant velocity, the bottom spring s_3 is supporting its mass m_3 to balance gravity.

As the elevator slows, though, it also ends up slowing down the spring arrangement, too. However, because the stretching takes time, it means that some damped harmonic motion will be set up in the spring chain.

When the motion has finally damped out, the net force the bottom spring s3 exerts on m3 has two components--that of gravity and of the deceleration of the elevator:

F_3net = m3 * (g + a) = 10.5×(9.81+5.5)= 10.5×15.31= 160.75 N

5 0

3 years ago

A straight segment of a current-carrying wire has a current element IL where I = 2.70 A and L = 3.20 cm i + 4.30 cm j. The segme

myrzilka [38]

The component of the force in negative z-direction is -0.144 N.

The given parameters;

current in the wire, I = 2.7 A
length of the wire, L = (3.2 i + 4.3j) cm
magnetic filed, B = 1.24 i

The force on the segment of the wire is calculated as follows;

$F = ILBsin(\theta)$

where;

θ is the angle wire and magnetic field

The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.

The angle between the wire and the magnetic field is calculated as follows;

$\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0$