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AURORKA [14]
3 years ago
8

This diagram of the first right-hand rule relates which two quantities?

Physics
2 answers:
Reptile [31]3 years ago
8 0

PLATO ANSWER!!!!!!

current (thumb) to magnetic field (fingers)

AURORKA [14]3 years ago
7 0

Option A, current (thumb) to magnetic field (fingers)

As per the First right-hand rule,

Using right hand, if we suppose that thumb points towards the electric current

fingers curl towards the magnetic field

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A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. An identical charge particle enters
emmasim [6.3K]

Answer:

<em>The second particle will move through the field with a radius greater that the radius of the first particle</em>

Explanation:

For a charged particle, the force on the particle is given as

F = \frac{mv^{2} }{r}

also recall that work is force times the distance traveled

work = F x d

so, the work on the particle = F x d,

where the distance traveled by the particle in one revolution = 2\pi r

Work on a particle = 2πrF = 2\pi mv^{2}

This work is proportional to the energy of the particle.

And the work is also proportional to the radius of travel of the particles.

Since the second particle has a bigger speed v, when compared to the speed of the first particle, then, the the second particle has more energy, and thus will move through the field with a radius greater that the radius of the first particle.

3 0
2 years ago
Ah electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following stat
Advocard [28]

Answer:

The correct statement is "The electric field is directed toward the electron and has a magnitude of \rm \dfrac{ke}{r^2} ".

Explanation:

According to Coulomb's law, the magnitude of the electric field due to a static point charge q at a point r distance away from it is given by

\rm E = \dfrac{k|q|}{r^2}.

  • k is the Coulmob's constant.

The direction of the electric field along the line joining the charge and the point where electric field is to be found and it is directed from positive charge to negative charge.

Conventionally, we assume a positive test charge placed at the point where electric field is to be found, the test charge has very small charge such that its charge does not affect the electric field due to the given charge.

The charge on the electron = -e.

The electric field due to an electron is given by

\rm E = \dfrac{k|-e|}{r^2}=\dfrac{ke}{r^2}.

The direction of this electric field is from positive test charge, placed at the point where electric field is to be found, towards the electron along the line joining the two.

Thus, the correct statement is "The electric field is directed toward the electron and has a magnitude of \rm \dfrac{ke}{r^2} ".

5 0
2 years ago
If a star is moving away from you at a constant speed, how do the wavelengths of the absorption lines change as the star gets fa
Minchanka [31]

Answer:

they stay shifted the same amount to the red

Explanation:

Redshift is given by

z=\dfrac{\lambda_o-\lambda_e}{\lambda_e}

Where,

\lambda_o = Wavelength observed

\lambda_e = Wavelength emitted

Also

Transverse redshift is given by

1+z=\dfrac{1}{\sqrt{1-v^2/c^2}}

v = Velocity of object

c = Speed of light = 3\times 10^8\ m/s

So, if the velocity is constant the redshift remains the same

8 0
2 years ago
A 1500 W electric heater is plugged into the outlet of a 120 V circuit that has a 20 A circuit breaker. You plug an electric hai
Mila [183]

Answer:

Explanation:

Current drawn by electric heater = power/volt =1500/120 = 12.5 A.

current drawn by hair drier at 600 watt = 600/120 =5 A

current drawn by hair drier at 900 watt = 900/120 = 7.5 A.

Total current drawn by heater and hair drier used at 900 watt

= 12.5 + 7.5 = 20 A

Breaking current =20 A

So fuse will trip at this point .

3 0
2 years ago
A student uses 60 Newtons of force to climb 18 meters in 6 seconds. Calculate her Power.
puteri [66]

Answer:

15

Explanation:

P=W/T

T=6sec

W=?

F=60N

S=18m

W=F X S. .s indicate displacement

W=60x18

W=108

So p=108 j/6sec

P=15watt

5 0
2 years ago
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