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3 years ago

3 light-years =________.

1 answer:

6 0

   ( 3 yr) · (186,282.397 mile/s) · (86,400 s/day) · (365 day/yr)

   = (3 · 186,282.397 · 86,400 · 365) mile

   = 1.762380502 x 10¹³ miles

   = 1.8 x 10¹³ miles (rounded to the nearest trillion miles)

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Two point charge

timama [110]

Answer:

Showing results for Two point charge q, separated by 1.5cm have change value of +2.0 and -4.0AND/C respectively what is the magnitude of the Electric force midway between them?

Search instead for Two point charge q, seperated by 1.5cm have change value of +2.0 and -4.0N/C respectively what is the magnitude of the Electric force midway between them?

4 0

1 year ago

What is an neutral atom?

BlackZzzverrR [31]

Atoms are neutral; they contain the same number of protons as electrons. By definition, an ion is an electrically charged particle produced by either removing electrons from a neutral atom to give a positive ion or adding electrons to a neutral atom to give a negative ion.

7 0

3 years ago

Read 2 more answers

Where do people get their energy to turn a wheel??

Vesna [10]

Answer:

from food eaten (i.e contain carnlbohyadrates)

5 0

2 years ago

A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

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

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

3 0

2 years ago

4- by eating food only will you get energy?

Serga [27]

Answer:

yes

Explanation:

All parts of the body (muscles, brain, heart, and liver) need energy to work. This energy comes from the food we eat. Our bodies digest the food we eat by mixing it with fluids (acids and enzymes) in the stomach.

4 0

3 years ago