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Anastaziya [24]
3 years ago
12

What is the difference between a bar magnet and an electromagnet? (check all that are

Physics
1 answer:
lukranit [14]3 years ago
5 0

Answer:

D and A

Explanation:

Hint The basic difference between them is that a bar magnet is a permanent magnet whereas an electromagnet is a temporary magnet. An electromagnet is formed when an electric current is passed through wires wound around soft metalcore. An electromagnet loses its magnetism once the current flow is stopped.

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Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
4 years ago
A 80 kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 25 degrees hill.
zloy xaker [14]

Answer:

P=28.085\,hp

Explanation:

Given that:

  • mass of 1 skier, m=80kg
  • inclination of hill, \theta=25^{\circ}
  • length of inclined slope, l=220m
  • time taken to reach the top of hill, t=2.3 min= 138 s
  • coefficient of friction, \mu=0.15

<em>Now, force normal to the inclined plane:</em>

F_N=m.g.cos\theta

F_N=80\times 9.8\times cos25^{\circ}

F_N=710.54\,N

<em>Frictional force:</em>

f=\mu.F_N

f=0.15\times 710.54

f=106.58\,N

<em>The component of weight along the inclined plane:</em>

W_l=m.g.sin\theta

W_l=80\times 9.8\times sin25^{\circ}

W_l=331.33\,N

<em>Now the total force required along the inclination to move at the top of hill:</em>

F=f+W_l

F=106.58+331.33

F=437.91\,N

<em>Hence the work done:</em>

W=F.l

W=437.91\times 220

W=96340.80\,J

<em>Now power:</em>

P=\frac{W}{t}

P=\frac{96340.80}{138}

P=698.12\,W

<u>So, power required for 30 such bodies:</u>

P=30\times 698.12

P=20943.65\,W

P=\frac{20943.65}{745.7}

P=28.085\,hp

8 0
3 years ago
The subject of the second part of messiah is:
Zielflug [23.3K]
The second part is about <span>his suffering and the spread of his doctrine.</span>
5 0
4 years ago
An electric current in a conductor varies with time according to the expression I(t) = 100 sin (120πt) , where I is in amperes a
Mice21 [21]

The  total charge passing a given point in the conductor from t = 0 to t = 1/240 s is 12000π Coulombs

<h3>Total charge through a conductor</h3>

A conductor is a substance that allow current and electricity to pass through it.

The expression that will be used to calculate the current is expressed s:

I(t) = 100 sin (120πt)

where

I is in amperes and t is in seconds.

Since dQ = Idt =  I(t) = 100 sin (120πt)

On integrating

Q = 100*120πcos(120πt) |¹/¹²⁴⁰₀

Substitute the limits

Q = 100 * 120π

Q = 12000π Coulombs

Hence the  total charge passing a given point in the conductor from t = 0 to t = 1/240 s is 12000π Coulombs

Learn more on total charge through a conductor here: brainly.com/question/15083960

#SPJ4

7 0
2 years ago
a washing machine uses 750 watts of power and costs $0.15 per kilowatt-hour. How much does it cost to run a washing machine for
Alja [10]

Answer:$0.1125

Explanation:

power=750 watts

Power=750/1000=0.75 kilowatts

$0.15 for 1 kilowatts-hour

$h for 0.75 kilowatts-hour

$h=(0.75 x 0.15)

$h=$0.1125

7 0
3 years ago
Read 2 more answers
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