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Check all that apply. The magnetic force on the current-carrying wire is strongest when the current is parallel to the magnetic

dedylja [7]

Answer:

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.

Explanation:

The magnitude of the magnetic force exerted on a current-carrying wire due to a magnetic field is given by

$F=ILB sin \theta$ (1)

where I is the current, L the length of the wire, B the strength of the magnetic field, $\theta$ the angle between the direction of the field and the direction of the current.

Also, B, I and F in the formula are all perpendicular to each other. (2)

According to eq.(1), we see that the statement:

"The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines."

is correct, because when the current is perpendicular to the magnetic field, $\theta=90^{\circ}, sin \theta = 1$ and the force is maximum.

Moreover, according to (2), we also see that the statements

"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field. "

"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current. "

because F (the force) is perpendicular to both the magnetic field and the current.

5 0

3 years ago

A 4.0kg bowling ball sliding to the right at 8.0 m/s has an elastic head-on collision with another 4.0 kg bowling ball initially

Kisachek [45]

a)

We use the formula :

m1v1i + m2v2i = m1v1f + m2v2f

Substituting the values in:

4.0kg*8.0m/s + 4.0kg*0m/s = 4.0kg*0m/s +4.0kg*v2f

Calculating this we get:

32.0kg*m/s + 0kg*m/s = 0kg*m/s + 4.0kg*v2f

Rearrange for v2f:

v2f = $\frac{32.0kg*m/s}{4.0kg}$

This gives us 8.0 m/s as the final velocity of the second ball.

b)

Since the collision is assumed to be elastic it means that the kinetic energy must be equal before and after the collision.

This means we use the formula:

Ek = $\frac{1}{2} *m*v^{2}$ + $\frac{1}{2} *m*v^{2}$ = $\frac{1}{2} *m*v^{2}$ + $\frac{1}{2}*m*v^{2}$

Substituting in values:

Ek = 0.5*4.0kg*(8.0m/s)^2 + 0.5*4.0kg*(0m/s)^2 = 0.5*4.0kg*(0m/s)^2 + 0.5*4.0kg*(8.0m/s)^2

This simplifies to:

Ek= 128J + 0J = 0J + 128J

This shows us that the kinetic energy is equal on each side therefore the collision is Elastic and no energy has been lost.

6 0

2 years ago

A 2.0-μF capacitor and a 4.0-μF capacitor are connected in series across a 1.0-kV potential. The charged capacitors are then dis

vekshin1

Answer:

Explanation:

Given that,

We have two capacitors connected in series

C1=2.0-μF

C2=4.0-μF

Then the equivalent of their series connection

1/Ceq = ½ + ¼

1/Ceq= (2+1)/4

1/Ceq=¾

Taking the reciprocal

Ceq= 4/3 μF

The capacitors are connected to a battery of 1kv

V=1000Volts

We know that,

Q=CV

Where Q is charge

C is capacitance and

V is voltage

Then, Q=4/3 ×1000

Q=4000/3 -μC

Since the capacitors are in series, then the charge pass through them, so each charge on the capacitors are 4000/3 μF

After the capacitor has been charge, the capacitor are disconnect and reconnected in parallel to each other,

For parallel connection, they have the same voltage but different charges.

When connected in parallel, there is a charge redistribution,

And the total charge will be 2•4000/3=8000/3 -μF

Then, Q1 +Q2= 8000/3 μF

Now the charge on each capacitor will be, let them have a common voltage V

Q=CV

Then, Q1=C1V

Q1= 2×V=2V

Q2= 4×V=4V

Then, Q1+Q2=8000/3

4V+2V=8000/3

6V=8000/3

V=8000/(3×6)

V=4000/9

V=444.44Volts

Now, Q1=2V

Q1=2×4000/9

Q1=8000/9 μF

Also, Q2=4V

Q2=4×4000/9

Q2=16000/9 μF

4 0

2 years ago

What is the name given to the syestem of units used in the united states

kozerog [31]

Answer:

Inches. (in)

Explanation:

7 0

3 years ago

Read 2 more answers

How can you tell which substances are creating the light of the sun or a lightbulb?

Westkost [7]

Through spectroscopy. Each atom in a gas absorbs and emits light at very specific and unique frequencies. Heating up a gas causes it to glow at these frequencies. If you put the light from a mercury lamp, for example, through a prism there will not be a rainbow. There will only be specific bands of light at certain colors.
On the other hand, white light comes from the sun. The inner part of the sun creates white light as it isn't just a gas state (this specific choosing of frequencies is a gaseous phenomenon) but the atmosphere of the sun is a gas. So when the white light passes through it, it absorbs specific frequencies specific to the elements in the gas. These get scattered (released at random directions) and so many of them don't reach our telescopes. So in a rainbow from stars (including the sun) have dark bands at specific frequencies. You need equipment to focus and see the spectrum closely to notice this. But the missing frequencies are the EXACT frequencies that the gas of the sun's atmosphere would release if heated in a lamp.
So based on what light is emitted (in gas bulbs) or missing from a spectrum (from stars) we can tell what elements are present there.

4 0

3 years ago