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3 years ago

Demonstrate the path light takes when traveling through the convex lens. Include an explanation of the drawing.

1 answer:

4 0

The Convex lens is the oval shaped one mainly used to magnify things so therefore we use the Convex lens for glasses and magnifying glasses

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topjm [15]

When light passes through one transparent medium to another transperent medium it bends and that beding of light is know as refraction of light !

8 0

3 years ago

Read 2 more answers

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.

myrzilka [38]

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane. You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

5 0

3 years ago

Which two options would INCREASE the electric force between two charged particles?

atroni [7]

The electric force between two charged particles can be increased by decreasing the distance between the two particles.

<h3>How to increase electric force between two charged particles.</h3>

The technique of decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. while

increasing the separation distance between objects decreases the force of attraction or repulsion between the objects.

Read more on Electric Force:

brainly.com/question/17692887

#SPJ1

7 0

2 years ago

Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 80 mm2 when carryin

Vedmedyk [2.9K]

Answer:

The correct answer is $2.8*10^{-5}ms^{-1}$

Explanation:

The formula for the electron drift speed is given as follows,

$u=I/nAq$

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.

Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is $8.93/63.5 molcm^{-3}$ . Converting this number to m³ using very elementary unit conversion we get $140384molm^{-3}$ . If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

$n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}$

if we convert the area from mm³ to m³ we get $A=80*10^{-6}m^{2}$ .So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,