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Elza [17]
4 years ago
8

Please help!! Will mark brainliest!!

Physics
1 answer:
Effectus [21]4 years ago
5 0

I believe the correct answer is the bottom one.  Hope this helped!

-TTL

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A wire with a resistance of 10 ohm is stretched so that its new length is four times its original length. find the resistance of
amm1812
Let original length be L. The new length is therefore 4L.

Let original cross sectional surface area of the wire be equal to πr^2.

This means original volume was L x πr^2 = Lπr^2

The volume is the same but the length is different so 4L x new surface area must be equal to Lπr^2. Let new surface area be equal to Y.

4L x Y = Lπr^2
=> Y = (πr^2 )/ 4


Using the resistivity formula,
R = pL/A. p which is resistivity is a constant so it stays the same

But this time, instead of L we have 4L and instead of πr^2 we have (πr^2)/4.

so the new resistance
= (4Lp)/ {(πr^2)/4}
= 16 (pL)/(πr^2)
= 16 (pL)/A. because πr^2 is A
since pL/A is equal to R from the formula, this is equal to
16 R.

R was 10 ohms
therefore new resistance is 16 x 10 = 160 ohms




6 0
4 years ago
How do u find net force?
Damm [24]
1Draw a quick sketch of the object.
2Draw an arrow showing every force acting on the object.
3<span>To calculate the net force, add any vectors acting on the same axis (x and y), making sure to pay attention to the directions.
Hope this helps :)
</span>
7 0
3 years ago
Which of the following does notaccurately describe the forces that exist with an atom? 
natima [27]
The believe the answer to this is B
8 0
3 years ago
An example of when total internal reflection occurs is when all the light passing from a region of higher index of refraction to
Amiraneli [1.4K]

Answer:

is reflected back into the region of higher index

Explanation:

Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.

According to Snell's law, refraction of ligth is described by the equation

n_1 sin \theta_1 = n_2 sin \theta_2

where

n1 is the refractive index of the first medium

n2 is the refractive index of the second medium

\theta_1 is the angle of incidence (in the first medium)

\theta_2 is the angle of refraction (in the second medium)

Let's now consider a situation in which

n_1 > n_2

so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

Where \frac{n_1}{n_2} is a number greater than 1. This means that above a certain value of the angle of incidence \theta_1, the term on the right can become greater than 1. So this would mean

sin \theta_2 > 1

But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by

\theta_c =sin^{-1}(\frac{n_2}{n_1})

8 0
4 years ago
? is the change in position relative to an initial location
umka21 [38]

Answer: c. displacement

Explanation:

Let's begin by stating clear that motion is the change of position of a body at a certain time. So, during this motion, the body will have a trajectory and a displacement.

In the specific case of the displacement, it is defined as the distance in a straight line between the initial and final position (is a vector magnitude).

Therefore:

Displacement s the the change in position relative to an initial location.

7 0
4 years ago
Read 2 more answers
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