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Naddika [18.5K]
3 years ago
14

PLEASE HELP WILL MEDAL !!!

Physics
2 answers:
vekshin13 years ago
7 0
Gas form is the answer
Tema [17]3 years ago
5 0
Water is vapor in gas. Solid - snow liquid - warer
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Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0
4 years ago
Most paper does not reflect light very well because its surface is somewhat rough.<br><br> T or F
Sloan [31]
I think the correct answer is true. Most paper does not reflect light very well because its surface is somewhat rough. Light only reflects to surfaces which has a smooth texture or have a uniform texture on the surface. Hope this answers the question. Have a nice day.
3 0
3 years ago
Read 2 more answers
what are the 3 properties of components of the universe that can be determined using electromagnetic radiation?
stepan [7]
Your answer is electricity, light and magnetism.  They can be determined usinf elecromagnetic radioation. 
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Even the energy  can't be detected by our eyes, there are a lot of measurement  instruments that can measure infrared (IR), gamma rays, radio or X-rays or ultraviolet (UV)</span>
4 0
3 years ago
S To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volum
erastovalidia [21]

To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be \frac{4.83598}{c^{\frac{1}{3} } }.

We know that the surface area and volume of the sphere is given by:

A=4 \pi r^{2}\\V=\frac{4}{3} \pi r^{3}

Therefore, the ratio between the surface area and the volume for the sphere will be:

\frac{A}{V}=\frac{4 \pi r^{2}\\}{\frac{4}{3} \pi r^{3}}=\frac{3}{r}

Equating the volume to the constant c, we will find the value of r.

V=c=\frac{4}{3} \pi r^{3}\\r= (\frac{3c}{4\pi} )^{\frac{1}{3} }

Substituting the value of r in the ration between surface area and volume, we get:

\frac{A}{V}=\frac{3}{ (\frac{3c}{4\pi} )^{\frac{1}{3} }}

Calculating the constants, we get:

\frac{4.83598}{c^{\frac{1}{3} } }

Hence, the ration between surface area and volume is \frac{4.83598}{c^{\frac{1}{3} } }

To learn more about surface area and volume of sphere, refer to:

brainly.com/question/4387241

#SPJ4

3 0
1 year ago
How much work does the electric field do in moving a proton from a point with a potential of +125 v to a point where it is -55 v
777dan777 [17]
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
W= qV_i - qV_f
where q is the charge of the proton, q=1 e = 1.6\cdot 10^{-19}C, with e being the elementary charge, and V_i = +125 V and V_f = -55 V are the initial and final voltage.

Substituting, we get (in electronvolts):
W=e(125 V-(-55 V))=180 eV
and in Joule:
W=(1.6 \cdot 10^{-19})(125 V-(-55V))=2.88 \cdot 10^{-17}J

5 0
3 years ago
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