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jeyben [28]
3 years ago
15

What principle of light’s behavior can be used to make a far away object appear closer and also make a small object appear large

r?
Physics
2 answers:
Alex_Xolod [135]3 years ago
8 0

Explanation:

Refraction is the change in the direction of the wave passing from one medium to another.

The change in the direction of the propagation of any wave is due to the different speed at different points.

A far away object appears closer and a small object appears larger.

Take an example. Suppose, an object lies in water.

A far away object appears closer. The angle from which the rays of light reach the observer is larger than the angle that it makes in air.

This makes the angular size larger to the observer's eye which makes the object look larger relative what it looks in air. Similar but vice versa condition occurs in the case when a small object appears larger.

Therefore, refraction can be used to make a far away object appear closer and also make a small object appear larger.

4vir4ik [10]3 years ago
3 0
It would be refraction
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Answer:

I am pretty sure it is B.

Explanation:

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Two small, identical conducting spheres repel each other with a force of 0.045 N when they are 0.15 m apart. After a conducting
Levart [38]

Answer:

q_1 = \pm 1.68 \times 10^{-7} C

q_2 = \pm 6.68 \times 10^{-7} C

Explanation:

As we know that the force between two small spheres is given as

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 , q_2 = charges on two small spheres

r = distance between two spheres = 0.15 m

now the force between them is given as

0.045 = \frac{(9\times 10^9)(q_1q_2)}{0.15^2}

q_1q_2 = 1.125 \times 10^{-13}

now when two spheres are connected together then the charge on them is equally divided

q = \frac{q_1+q_2}{2}

now the force between them is given as

F = \frac{k(\frac{q_1+q_2}{2})^2}{0.15^2}

0.070 = \frac{(9\times 10^9)(\frac{q_1+q_2}{2})^2}{0.15^2}

q_1 + q_2 = 8.37\times 10^{-7}

so here we have

q_1 = \pm 1.68 \times 10^{-7} C

q_2 = \pm 6.68 \times 10^{-7} C

5 0
3 years ago
A charge of 8.0 pc is distributed uniformly on a spherical surface (radius = 2.0 cm), and a second charge of â3.0 pc is distribu
loris [4]
According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.

Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k  · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²

We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10<span>⁻¹² C
</span><span>r = 5.0 cm = 0.05 m

Now, we can apply the formula:
</span><span>E(r) = k · (q₁ + q₂) / r²
      = </span>9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
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Hence, <span>the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C</span>
4 0
4 years ago
Given the information below, estimate the total distance travelled during these 6 seconds using a left endpoint approximation. t
Nutka1998 [239]

Answer:

184 feets

Explanation:

Given the data:

time (sec) __ velocity (ft/sec)

0 __________30

1 __________ 54

2 __________56

3 __________34

4 __________ 8

5 __________ 2

6 __________22

Using left end approximation:

(0,1) ___ f(0) = 30

(1,2) ___ f(1) = 54

(2,3) ___f(2) = 56

(3,4) ___f(3) = 34

(4,5) ___f(4) = 8

(5,6) __ f(5) = 2

Hence, the Total distance traveled during the 6 second interval is:

Change ; dT = 1

1 * (30 + 54 + 56 + 34 + 8 + 2) = 184

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sweet-ann [11.9K]

Answer:

B

Explanation:

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