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3 years ago

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If q1 is 2μC, q2 is 5μC and r1 = 0.2m, then the force between the charges is

1 answer:

N76 [4]3 years ago

4 0

F= kq1q2/r²
F=9×10^9×2×10^-6×5×10^-6/(0.2)²
F=0.09/0.04
F=2.25N

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List and define all of Newton’s laws of motion.

Luba_88 [7]

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

Explanation:

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2 years ago

If the mass of a material is 41 grams and the volume of the material is 5 cm^3, what would the density of the material be? answe

Mila [183]

Density can be calculated using the following rule:
Density = mass/volume

We are given that:
mass = 41 grams
volume = 5 cm^3

Substitute with the givens in the above equation to get the density as follows:
Density = 41/5 = 8.2 g/cm^3

Based on the above calculations:
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2 years ago

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What is the amplitude and wavelength of the wave shown below?

masha68 [24]

Answer:

Option B

Explanation:

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2 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

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3 years ago

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aleksandr82 [10.1K]

Explanation:

<h2><u>Steps </u><u>:</u></h2>

<u>Move </u><u>decimal</u><u> </u><u>from</u><u> </u><u>left </u><u>to </u><u>right</u><u> </u><u>=</u><u>0</u><u> </u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u>
<u>Then </u><u>count </u><u>the</u><u> </u><u>numbers</u><u> </u><u>before</u><u> </u><u>decimal </u><u>and </u><u>w</u><u>rite </u><u>it </u><u>like</u><u> </u><u>this </u><u>=</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u><u>x</u><u>1</u><u>0</u><u> </u><u>power-</u><u>9</u><u> </u>
<u>That's</u><u> </u><u>all </u>

<u>hope</u><u> it</u><u> </u><u>help</u>

<h2><u>#</u><u>H</u><u>o</u><u>p</u><u>e</u></h2>

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2 years ago