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3 years ago

Two forces (4n and 3n) pull to the left while a 12n force pulls to the right. What is the net force?

2 answers:

5 0

Call the positive direction the one that points to the right.

negative force <=== ===> positive force

Then the forces are +12n, -4n, and -3n.

The net force is their sum: (+12n - 4n - 3n) = +5 n .

(5 newtons, pointing to the right) .

Ahat [919]3 years ago

3 0

6 i believe because it makes sense too me

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A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative r

sergij07 [2.7K]

Answer:

The workdone is $W = 1.712 *10^{-20 } \ J$

Explanation:

From the question we are told that

The potential difference is $V = 107 mV = 107 *10^{-3} \ V$

Generally the charge on $Na^{+}$ is $Q_{Na^{+}} = 1.60 *10^{-19 } \ C$

Generally the workdone is mathematically represented as

$W = Q_{Na^{+}}V$

=> $W = 1.60 *10^{-19 } * 107 *10^{-3}$

=> $W = 1.712 *10^{-20 } \ J$

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2 years ago

_________ = voltage / resistance <br> a. power <br> b. energy <br> c. charge <br> d. current

Fittoniya [83]

c.charge due to the reaction process between the two

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3 years ago

How many mm are in a micro-m?

Dahasolnce [82]

For this case we have that by definition, a micrometer is equivalent to a thousandth of a millimeter, that is, 0.001 millimeters.

Then, in other words, we have 0.001 millimeters in a micrometer.

Answer:

In a micrometer there are 0.001 millimeters.

4 0

3 years ago

Please help!!!!!

adoni [48]

Gpe is basses on the force equation...
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8 0

2 years ago

Read 2 more answers

Find the ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance betw

Snezhnost [94]

Answer:

The ratio of the new force over the original force is 16

Explanation:

Recall the formula for the gravitational force between two masses M1 and M2 separated a distance D:

$F_G=G\,\frac{M_1\,\,M_2}{D^2}$

So now, if the masses M1 and M2 are quadrupled and the distance stays the same, the new force becomes:

$F'_G=G\,\frac{4M_1\,\,4M_2}{D^2}=G\,\frac{16\,\,M_1\,\,M_2}{D^2}=16\,\,G\,\frac{M_1\,\,M_2}{D^2}= 16\,\,F_G$

which is 16 times the original force.

So the ratio of the new force over the original force is 16

5 0

2 years ago