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4 years ago

The current in a wire increases from 5 A to 10 A, what happens to its magnetic field?

Physics

1 answer:

sdas [7]4 years ago

6 0

Answer:

The first one is twice as strong

The second one is half as strong

The third one is the force is reversed

Explanation: just did it

You might be interested in

Which pair of factors affects the force of gravity between objects?

butalik [34]

Answer:

mass and distance

Explanation:

dont need to summary this

5 0

3 years ago

What is the volume of an object that has a density of 65g/cm3 and a mass of 130g.

lora16 [44]

Density ρ is mass m per unit volume v, or

ρ = m / v

Solving for v gives

v = m / ρ

So the given object has a volume of

v = (130 g) / (65 g/cm³) = 2 cm³

5 0

3 years ago

What are the periodic variations in Earth's rotation and orbit around the sun that alter the way solar radiation is distributed

Dahasolnce [82]

Answer:

1. The precession of the equinoxes.

2. Changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun.

3. Variations in the eccentricity

Explanation:

These variations listed above; the precession of the equinoxes (refers, changes in the timing of the seasons of summer and winter), this occurs on a roughly about 26,000-year interval; changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun, this occurs roughly in a 41,000-year interval; and changes in the eccentricity (that is a departure from a perfect circle) of Earth’s orbit around the Sun, occurring on a roughly 100,000-year timescale. which influences the mean annual solar radiation at the top of Earth’s atmosphere.

5 0

3 years ago

A car travels 1800 m south in 35 s, what is it's velocity?

kaheart [24]

Answer:

its velocity is 51, or 51.42 to be exact it would be: 51.4285714286

8 0

3 years ago

(a) You short-circuit a 20 volt battery by connecting a short wire from one end of the battery to the other end. If the current

erica [24]

(a) $1.11 \Omega$

When the battery is short-cut, the only resistance in the circuit is the internal resistance of the battery. Therefore, we can apply Ohm's law:

$r=\frac{V}{I}$

where

V = 20 V is the voltage across the internal resistance of the battery

I = 18 A is the current flowing through it

Solving the equation,

$r=\frac{20 V}{18 A}=1.11\Omega$

(b) 360 W

The power generated by the battery is given by the equation

$P=VI$

where

V = 20 V is the voltage of the battery

I = 18 A is the current

Substituting into the formula,

$P=(20 V)(18 A)=360 W$

(c) 360 J

The energy dissipated by the internal resistance is given by

$E=Pt$

where

P = 360 W is the power generated

t = 1 s is the time

Solving the equation, we find

$E=(360 W)(1 s)=360 J$

(d) 1.65 A

The battery is now connected to a $R=11 \Omega$ resistor. This means that the internal resistance of the battery is now connected in series with the other resistor R: so, the total resistance of the circuit is

$R_T = r+R=1.11 \Omega +11 \Omega = 12.11 \Omega$

And so, the current flowing through the circuit is

$I=\frac{V}{R_T}=\frac{20 V}{12.11\Omega}=1.65 A$

(e) 29.9 W

The power dissipated in the external resistor is given by

$P=I^2 R$

where

I = 1.65 A is the current

$R=11 \Omega$ is the resistance

Solving the equation, we find

$P=(1.65 A)^2(11 \Omega)=29.9 W$

(f) 18.17 V

The terminals of the voltmeter are placed at the two end of the battery. The battery provides an emf of 20 V, however due to the internal resistance, some of this voltage is dropped across the internal resistance. Therefore, the actual potential difference that will be read by the voltmeter will be:

$V=\epsilon - Ir =20 V -(1.65 A)(1.11 \Omega)=18.17 V$

4 0

3 years ago