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3 years ago

What happens to the strength of the magnetic field as you come closer to the current carrying wire?

1 answer:

lina2011 [118]3 years ago

8 0

Electric energy is carried by current, which is the flow of electrons, which are the negatively charged sub-particles of atoms. This transfer of electrons from one place to another powers our lights, phones, appliances and many other things we use every day. Another interesting phenomenon of flowing current is that it produces its own magnetic field. Electricity and magnetism are very closely linked in that all closed loop currents create their own magnetic fields, and magnetic fields acting upon closed loop circuits can produce current or even change current diction.

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What is the initial vertical velocity?

Marat540 [252]

Answer:

Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude spends 3.79 s in the air.The initial vertical velocity is the vertical component of the initial velocity: v 0 y = v 0 sin θ 0 = ( 30.0 m / s ) sin 45 ° = 21.2 m / s .

3 0

2 years ago

Differentiate the following functions with respect to x <br>xsin x

gtnhenbr [62]

Answer:

Here is your answer

Hope it helps

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3 years ago

Read 2 more answers

How can you find the reading of main scale and vernier scale

anygoal [31]

Answer:

this pdf should help you out

Explanation:

Download pdf

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2 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago

3) A dock worker pushes a 72 kg crate up a 2.0 m high,

Vlad [161]

Work done on the crate is 1411.2 J

Explanation:

Work done is defined as the product of force and the distance moved by the object. The unit of work done is in joules and denoted by the symbol J.

Work done = F * d

where F represents the force and d represents the distance moved by the object.

mass = 72 kg , distance moved by the object is given by 2.0 m

Force F = mass * gravity = 72 * 9.8

= 705.6 N =706 N.

Work done = 706 * 2.0 = 1412 J.

7 0

3 years ago