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3 years ago

Suppose that you're facing a straight current-carrying conductor, and the current is flowing toward you.

2 answers:

Ivan3 years ago

5 0

<span>Right-hand rule reveals a connection between the current and the magnetic field lines in the magnetic field. It is also called Ampere`s right hand screw rule. It determines the direction of the magnetic field surrounding a current carrying wire. The thunb points direction of current and fingers point the direction of magnetic lines of force. Answer: C ) a counterclockwise rotation.</span>

Nostrana [21]3 years ago

4 0

B. A clockwise direction.

Good luck!

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Form concentric circles around the wire

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3 years ago

What two units of measure are necessary for describing speed

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Distance and time, distance because that's how far and time because that's how long

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3 years ago

Biologists use optical tweezers to manipulate micron-sized objects using a beam of light. In this technique, a laser beam is foc

vekshin1

Answer:

Explanation:

Part A) Using

light intensity I= P/A

A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2

Radius= Diameter/2

P= power= 10*10^-3=0.01 W

light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2

Part B) Using

I=c*ε*E^2/2

rearrange to solve for E= $\sqrt{$ ((I*2)/(c*ε))

c is the speed of light which is 3*10^8 m/s^2

ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1

I= the already solved light intensity= 8.85*10^10 W/m^2

amplitude of the electric field E= $\sqrt{$ (9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)

---> E= $\sqrt{$ (1.8*10^11) / (2.66*10^-3) = $\sqrt{$ (6.8*10^13) = 8.25*10^6 V/m

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2 years ago

What causes earthquake s to occur?

Sliva [168]

Answer:

tilte in the earth

Explanation:

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2 years ago

Read 2 more answers

A car has a kinetic energy of 1.9 × 10^3 joules. If the velocity of the car is decreased by half, what is its kinetic energy?

VLD [36.1K]

The initial kinetic energy of the car is
$E_1 = \frac{1}{2}mv_1^2 = 1.9 \cdot 10^3 J$

Then, the velocity of the car is decreased by half: $v_2 = \frac{v_1}{2}$
so, the new kinetic energy is
$E_2 = \frac{1}{2}mv_2 ^2 = \frac{1}{2} m ( \frac{v_1}{2} )^2= \frac{1}{2}m \frac{v_1^2}{4}= \frac{E_1}{4}$
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
$E_2 = \frac{1.9 \cdot 10^3 J}{4}=475 J$

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3 years ago