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3 years ago

Light travels at 3.0 × 108 m/s in a vacuum and slows to 2.0 × 108 m/s in glass. What is the index of refraction of glass?

2 answers:

8 0

We are asked to solve for the index of refraction and the formula is n = c/v where "n" represents the index of refraction, "c" represents the speed of light in the vacuum while "v" represents the speed of another medium.
In the problem, we have the given values below:
c = 3 x 10^8 m/s
v = 2 x 10^8 m/s
n =?

Solving for n, we have the solution below:
n = 3x10^8 / 2x10^8
n = 1.5

The answer is 1.5 for the index of refraction.

Dafna1 [17]3 years ago

8 0

it is 1.5 just passed the question on edge

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object carries a charge of -8.1 ÂµC, while another carries a charge of -2.0 ÂµC. How many electrons must be transferred from the

LenKa [72]

Number of electrons transferred: $1.91\cdot 10^{13}$

Explanation:

The charge on the first object is

$Q_1 = -8.1\mu C$

while the charge on the 2nd object is

$Q_2=-2.0 \mu C$

When they are in contact, the final charge on each object will be

$Q=\frac{Q_1+Q_2}{2}=\frac{-8.1+(-2.0)}{2}=-5.05 \mu C$

So, the amount of charge (electrons) transferred from the 1st object to the 2nd object is

$\Delta Q = Q_1 - Q = -8.1 -(5.05)=-3.05 \mu C = -3.05\cdot 10^{-6}C$

The charge of one electron is

$e=-1.6\cdot 10^{-19}C$

Therefore, the number of electrons transferred is

$N=\frac{Q}{e}=\frac{-3.05\cdot 10^{-6}}{-1.6\cdot 10^{-19}}=1.91\cdot 10^{13}$

Learn more about electrons:

brainly.com/question/2757829

#LearnwithBrainly

8 0

3 years ago

What was the hiker's total displacement? A.24 km B.4 km north C.14 km northeast D.0 km

kipiarov [429]

We can see it as two movements: one on the north-south axis and one on the west-east axis.

On the north-south Axis, the hiker went 8 km north and 4 km south, so in total the hiker went 4 km north (we can subtract one for the other and the direction remains that which was bigger)

On the west-easth Axis, the hiker went 6 km east and 6 km west, so in total the hiker didn't move on this axis! or, he/she moved but then "came back"

So in total, the hiker only moved4 km north - answer B.

7 0

3 years ago

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Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete

Sonbull [250]

Answer:

work done in pumping the entire fuel is 466587 J

Explanation:

weight of the gasoline per volume = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 2 m

height of the tractor tank above the top of the tank = 5 m

work done in pumping fuel to this height = ?

First, we find the volume of the fuel

since the tank is cylindrical, we assume that the fuel within also takes the cylindrical shape.

Also, we assume that the fuel completely fills the tank.

volume of a cylinder = $\pi r^{2}l$

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x $1.5^{2}$ x 2 = 14.139 m^3

we then find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 14.139 = 93317.4 N

work done = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 93317.4 x 5 = 466587 J

8 0

3 years ago

At one point in space, the electric potential energy of a 15 nC charge is 42 μJ . Part A) What is the electric potential at this

Anastasy [175]

Answer:

Part A:

$\rm 2.8\times 10^3\ Volts.$

Part B:

$\rm 5.6\times 10^{-5}\ J.$

Explanation:

Part A:

Potential energy of charge at the given point, $\rm U=42\ \mu J=42\times 10^{-6}\ J.$

Charge, $\rm q=15\ nC = 15\times 10^{-9}\ C.$

The potential energy at a point due to a charge is defined as

$\rm U=qV$ .

where,

V = electric potential at that point.

Therefore,

$\rm V=\dfrac{U}{q}=\dfrac{42\times 10^{-6}}{15\times 10^{-9}}=2.8\times 10^3\ Volts.$

Part B:

Now, if the charge at that point is replaced with $\rm q_1 = 20\ nC = 20\times 10^{-9}\ C.$ , then the electric potential energy at that point is given by

$\rm U=q_1V = 20\times 10^{-9}\times 2.8\times 10^3=5.6\times 10^{-5}\ J.$

5 0

3 years ago

As you run , your feet push on the ground with a force 500N . The ground pushes back on your feet with a force of 500N . (Please

Mademuasel [1]

Answer:

true

Explanation:

newtons third law says to every action there is always an equal and opposite reaction. so the answer is true

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3 years ago

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