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kari74 [83]
3 years ago
13

Magnetic field strength decrease when what is increases

Physics
1 answer:
Elis [28]3 years ago
6 0
Ans: One way to increase or decrease the strength of the magnetic field is to change the number of loops in the coil. The more loops you add, the stronger the field will become. The more loops you remove, the weaker the field will become.
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Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the intensity of electric field E mid
algol13

Answer:

The intensity of the net electric field will:

E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C

Explanation:

Here we need first find the electric field due to the first charge at the midway point.

The electric field equation is given by:

|E_{1}|=k\frac{q_{1}}{d^{2}}

Where:

  • k is Coulomb's constant
  • q(1) is 20.00 μC or 20*10⁻⁶ C
  • d is the distance from q1 to the midpoint (d=10.0 cm)

So, we will have:

|E_{1}|=(9*10^{9})\frac{20*10^{-6}}{0.1^{2}}

|E_{1}|=1.8*10^{7}\: N/C

The direction of E1 is to the right of the midpoint.

Now, the second electric field is:

|E_{2}|=k\frac{q_{2}}{d^{2}}

|E_{2}|=(9*10^{9})\frac{8*10^{-6}}{0.1^{2}}

|E_{2}|=7.2*10^{6}\: N/C

The direction of E2 is to the right of the midpoint because the second charge is negative.

Finally, the intensity of the net electric field will:

E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C

I hope it helps you!

5 0
3 years ago
Automobile engines AND steam engines are examples of ______. a. internal combustion engines b. heat engines c. external combusti
uranmaximum [27]

Answer: B. Heat engines

Explanation:

8 0
3 years ago
Read 2 more answers
Explain how Law 1 applies to the image to the left.
alisha [4.7K]

Answer:

12

Explanation:

3 0
3 years ago
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How much would a man on Mars weigh if he had a mass of 75 kilograms and the acceleration due to gravity on the surface of Mars i
blsea [12.9K]

Answer:

F = 75(3.7) = 277.5 ≈ 280 N

Explanation:

6 0
3 years ago
Two resistors with values of 6.0 W and 12 W are connected in parallel. This combination is connected in series with a 2.0 W resi
Murljashka [212]
Well, first of all, you really shouldn't use ' W ' for the unit when you
talk about resistors.

You may have seen the resistors written as 6ω, 12ω, and 2ω in your
book or on the homework sheet. But that little symbol ' ω ' is not a ' w '.
It's the small Greek letter 'omega'. The CAPITAL omega is ' Ω '. It's used
to label resistors because it's short for "ohms". So the resistors in this
problem have resistances of 6Ω, 12Ω, and 2Ω, and we have to do some
manipulating of the individual resistors to find out what resistance the
battery actually sees.

The parallel combination of the first two resistors looks like a single
resistor, whose value is

1 / (1/6 + 1/12)

= 1 / (2/12 + 1/12)

= 1 / (3/12)

= 12/3 = 4Ω .

Now, that parallel combination is connected in series with 2Ω .
All three resistors together look like a single resistor of

4Ω + 2Ω = 6Ω .

So the battery thinks there's a single resistor connected to it,
with 6Ω of resistance. The current out of the battery is

I = V / R = (24v) / (6Ω) = 4 Amperes.

That 4 Amperes of current will split between the parallel resistors,
but it will ALL flow through the series 2Ω resistor because there's
no other path through that part of the circuit.

So the current through the 2Ω resistor is 4 Amperes. (B).

Note:
The POWER dissipated by the 2Ω resistor is

P = I² R = (4A)² · (2Ω) = 32 watts .

This is a fair amount of heat, so you'll need to provide some way
to remove the heat from the resistor, otherwise it'll burn or crack.
8 0
3 years ago
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