How would you change the distance between two charged particles to increase the electric force between them by a factor of 16?

What are two reasons a common measuring system is <br> important

sergey [27]

One common reason why it is important is so you could properly measure the length or width of a wooden log another reason is how you could measure how tall somebody is in centimeters or feet. Hope this helped give me brainliest please

6 0

3 years ago

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camcorder has a power rating of 18 watts. If the output voltage from its battery is 6 volts, what current does it use?

DerKrebs [107]

Power = (voltage) x (current)

IF it's using the full 18 watts of power right now, then . . .

18 W = (6 V) x (current)

Current = (18 W) / (6 V)

<em>Current = 3 Amperes</em>

8 0

3 years ago

On Earth, 1 kg = 9.8 N = 2.2 lbs. On the Moon, 1 kg = 1.6 N = 0.37 lbs. Use these relationships to answer the following question

romanna [79]

Answer:

(a) 490 N on earth

(b) 80 N on earth

(c) 45.4545 kg on earth

(d) 270.27 kg on moon

Explanation:

We have given 1 kg = 9.8 N = 2.2 lbs on earth

And 1 kg = 1.6 N = 0.37 lbs on moon

(a) We have given mass of the person m = 50 kg

As it is given that 1 kg = 9.8 N

So 50 kg = 50×9.8 =490 N

(b) Mass of the person on moon = 50 kg

As it is given that on moon 1 kg = 1.6 N

So 50 kg = 50×1.6 = 80 N

(c) We have given that weight of the person on the earth = 100 lbs

As it is given that 1 kg = 2.2 lbs on earth

So 100 lbs = 45.4545 kg

(d) We have given weight of the person on moon = 100 lbs

As it is given that 1 kg = 0.37 lbs

So 100 lbs $\frac{100}{0.37}=270.27kg$

8 0

4 years ago

Problem 21-40a:

Greeley [361]

Answer:

Part a)

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

$q = 3.37 \times 10^{-4} C$

Explanation:

As we know that electric force on electric charge is given as

$F = qE$

here we have

$q = 1.6 \times 10^{-19}C$

E = 153 N/C

now force is given as

$F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N$

Gravitational force on electric charge near surface of earth is given as

$F_g = mg$

$F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N$

now the ratio of two forces is given as

$\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}$

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

$F_g = qE$

$mg = qE$

$(5.25 \times 10^{-3})(9.81) = q(153)$

here we have

$q = 3.37 \times 10^{-4} C$

8 0

4 years ago

An electron is traveling horizontally toward the north in a uniform magnetic field that is directed vertically downward. In what

Arte-miy333 [17]

Answer:

e) upwrad z axis

Explanation:

To know the direction of the force we use the right hand rule.

The thumb points in the direction of the velocity, the fingers extended in the direction of the magnetic field and the palm of the hand in the direction of the force, this is for a positive charge if the charge is negative the force is in the opposite direction of The palm of the hand.

Let's apply to our case.

The thumb is in the x direction, the fingers in the vertical direction and since the electron has a negative charge, the force is on the z axis (perpendicular to the blade, coming out)

In general, in the nomenclature of the cardinal points the positive x-axis is the East, the positive y-axis is the North. Therefore the answer must be up on the z axis

8 0

3 years ago