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krok68 [10]
2 years ago
14

I need a diagram for how a scrap heap magnet works

Physics
1 answer:
cupoosta [38]2 years ago
5 0
Will this one work?...................

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An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after
geniusboy [140]

Answer:

E=\frac{K}{ed}

Explanation:

We are given that

Initial kinetic energy of an electron=K

Distance=d

Final velocity=v=0

Charge,q=-1e

We have to find the magnitude of electric field.

Work done=Force\times displacement

Using the formula

Work done=qE\times d=-eEd

Using work energy theorem

Work done=Final K.E-Initial K.E=0-K

Work done=-K

Substitute the values

-K=-eEd

K=eEd

E=\frac{K}{ed}

Hence, the magnitude of the electric field=E=\frac{K}{ed}

3 0
2 years ago
Explain using physics vocabulary WHY the volume increases as the temperature increases.
Sveta_85 [38]

Answer:

P V = n R T      ideal gas equation

V = k T     where k = a constant and equals k = n R / P

V is proportional to T when other factors are constant

4 0
2 years ago
Ishani and John now try a problem involving a charging capacitor. An uncharged capacitor with C = 6.81 μF and a resistor with R
DENIUS [597]

Answer:

Q=81.72\times10^{-6}C

I=2.1\times10^{-5}A

Explanation:

The maximum charge on the capacitor will be, at the end of the process, given by the formula (and for our values):

Q=CV=(6.81\times10^{-6}F)(12V)=81.72\times10^{-6}C

The maximum current on the resistor will be, at the beginning of the process, given by the formula (and for our values):

I=\frac{V}{R}=\frac{12V}{5.8\times10^{5}\Omega}=2.1\times10^{-5}A

5 0
2 years ago
(a) Calculate the self-inductance (in mH) of a 55.0 cm long, 10.0 cm diameter solenoid having 1000 loops.
DedPeter [7]

Explanation:

(a) We have,

Length of solenoid, l = 55 cm = 0.55 m

Diameter of the solenoid, d = 10 cm

Radius, r = 5 cm = 0.05 m

Number of loops in the solenoid is 1000.

(a) The self inductance in the solenoid is given by :

L=\dfrac{\mu_o N^2A}{l}

A is area

L=\dfrac{4\pi \times 10^{-7}\times (1000)^2\times \pi (0.05)^2}{0.55}\\\\L=0.0179\ H\\\\L=17.9\ mH

(b) The energy stored in the inductor is given by :

E=\dfrac{1}{2}LI^2\\\\E=\dfrac{1}{2}\times 0.0179\times (19.5)^2\\\\E=3.4\ J

Hence, this is the required solution.

4 0
2 years ago
Can somebody help me with this quiz please
mezya [45]
Sure what do u need help with
4 0
2 years ago
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