All categories

3 years ago

Please Please Help!!, What happens when a current moves through a wire and the wire is placed near a magnet?

Physics

2 answers:

USPshnik [31]3 years ago

7 0

I think it is c not sure

LenaWriter [7]3 years ago

7 0

'd' is the correct choice. Don't make me type out the whole statement again.

You might be interested in

Two boxes (m1 = 43.0 kg and m2 = 39.5 kg) are connected by a light string that passes over a light, frictionless pulley. One box

wlad13 [49]

The net force on the system:
F = m₂g - m₁gsin(∅)
F = 39.5 x 9.81 - 43 x 9.81 x sin(30)
F = 176.58 N

Now, we use F = ma to find the acceleration on each mass.
F = m₁a₁
a₁ = 176.58 / 43
a₁ = 4.11 m/s²

F = m₂a₂
a₂ = 176.58 / 39.5
a₂ = 4.47 m/s²

5 0

3 years ago

What is the role of the neutral wire

lions [1.4K]

Answer:

The neutral wire is often confused with ground wire, but in reality, they serve two distinct purposes. Neutral wires carry currents back to power source to better control and regulate voltage. Its overall purpose is to serve as a path to return energy.

5 0

3 years ago

Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another

Tanya [424]

Answer: $I_2=0.618\times 10^{-7} W/m^2$

Explanation:

Given

Distance between source and receiver $d_1=3 m$

Sound Intensity $I_1=1.1\times 10^{-7} W/m^2$

Distance of of second observer $d_2=4 m$

Intensity varies as

$I\propto \frac{1}{d^2}$

using this

$I=\frac{k}{d^2}$

$\frac{I_1}{I_2}=\frac{d_2^2}{d_1^2}$

$\frac{1.1\times 10^{-7}}{I_2}=\frac{4^2}{3^2}$

$I_2=0.75^2\times 1.1\times 10^{-7}$

$I_2=0.618\times 10^{-7} W/m^2$

6 0

3 years ago

How many seconds will elapse between seeing lightning and hearing the thunder if the lightning strikes 1mi (5280 ft) away and th

alexandr1967 [171]

Answer:

t = 4.58 s

Explanation:

In this problem, we need to find the time elapse between seeing lightning and hearing the thunder if the lightning strikes 1mi (5280 ft) away and the air temperature is 90.0°F.

T = 90.0°F = 32.2 °C

The speed of sound at temperature T is given by :

v = (331.3 +0.6T)

Put T = 32.2°C

So,

v = (331.3 +0.6(32.2))

= 350.62 m/s

We have, distance, d = 1 mile = 1609.34

So,

$t=\dfrac{d}{v}\\\\t=4.58\ s$