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3 years ago

15

How does the direction of the electric current, moving across a battery powered device, differ from the direction of travel in t

he AC current sent to homes and businesses?

2 answers:

svetoff [14.1K]3 years ago

8 0

They both have different wave traction's

amm18123 years ago

6 0

Different wave tractions

Hope this helps :D

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1. An object with a mass of m is thrown straight up near the surface of the earth. While the object is going up, the net force o

Vaselesa [24]

Answer:

C: equal to mg

Explanation:

in free-fall, gravity is always the net force on an object

5 0

3 years ago

Solve for x. 4(x+1)≤4x+3

alexdok [17]

4x + 4 < 4x + 3 (expand it)
4 < 3 (cancel 4x on both sides)
Since 4 < 3 is not true there is no solution.

Answer: NO SOLUTION.

3 0

3 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:
$E_2 = \frac{1.9 \cdot 10^3 J}{4}=475 J$

5 0

4 years ago

Is a 10 year old supposed to be doing algebra

forsale [732]

If he feels like, is interested in it, and is able to grasp it, then why not ? Why not indeed ?

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3 years ago

Read 2 more answers

There is a naturally occurring vertical electric field near the Earth’s surface that points toward the ground. In fair weather c

trasher [3.6K]

Answer:

$\frac{F}{W} = 9.37 \times 10^{-4}$

Explanation:

Radius of the pollen is given as

$r = 12.0 \mu m$

Volume of the pollen is given as

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi (12\mu m)^3$

$V = 7.24 \times 10^{-15} m^3$

mass of the pollen is given as

$m = \rho V$

$m = 7.24 \times 10^{-12}$

so weight of the pollen is given as

$W = mg$

$W = (7.24 \times 10^{-12})(9.81)$

$W = 7.1 \times 10^{-11}$

Now electric force on the pollen is given

$F = qE$

$F = (-0.700\times 10^{-15})(95)$

$F = 6.65 \times 10^{-14} N$

now ratio of electric force and weight is given as

$\frac{F}{W} = \frac{6.65 \times 10^{-14}}{7.1 \times 10^{-11}}$

$\frac{F}{W} = 9.37 \times 10^{-4}$

7 0

3 years ago