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4 years ago

Evidence of wave theory light

1 answer:

7 0

All of these things were seen in formal experiments by the 19th century. But some of them are easy to see in your own home. It's obvious that light can reflect - you just have to look in a mirror. Light bounces off the mirror and goes into your eye so you can see yourself. It's also obvious that light can refract: All you have to do is put a spoon in a large glass of water and watch how the spoon appears to bend.

That happens because the light is bending as it moves between air and water. Both of these things can be seen even more clearly in a laboratory using beams of light or lasers.

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A positively charged particle initially at rest on the ground accelerates upward to 160 m/s in 2.10 s. The particle has a charge

mamaluj [8]

Answer:

The magnitude of the electric field is $8.6\times10^{2}\ N/C$

Explanation:

Given that,

Time t = 2.10 s

Speed = 160 m/s

Specific charge =Ratio of charge to mass = 0.100 C/kg

We need to calculate the acceleration

Using equation of motion

$a=\dfrac{v-u}{t}$

Put the value into the formula

$a=\dfrac{160-0}{2.10}$

$a=76.19\ m/s^2$

We need to calculate the magnitude of the electric field

Using formula of electric field

$E=\dfrac{F}{q}$

$E=\dfrac{ma}{q}$

$E=\dfrac{a+g}{\dfrac{q}{m}}$

Put the value into the formula

$E=\dfrac{76.19+9.8}{0.100}$

$E=8.6\times10^{2}\ N/C$

The direction is upward.

Hence, The magnitude of the electric field is $8.6\times10^{2}\ N/C$

4 0

4 years ago

Read 2 more answers

Problem 31:

blagie [28]

Answer:

Bi. Current in 15.4 Ω (R₁) is 7.14 A.

Bii. Current in 21.9 Ω (R₂) is 5.02 A.

Biii. Current in 11.7 Ω (R₃) is 9.40 A.

C. Total current in the circuit is 21.56 A.

Explanation:

Bi. Determination of the current in 15.4 Ω (R₁)

Voltage (V) = 110 V

Resistance (R₁) = 15.4 Ω

Current (I₁) =?

V = I₁R₁

110 = I₁ × 15.4

Divide both side by 15.4

I₁ = 110 / 15.4

I₁ = 7.14 A

Therefore, the current in 15.4 Ω (R₁) is 7.14 A.

Bii. Determination of the current in 21.9 Ω (R₂)

Voltage (V) = 110 V

Resistance (R₂) = 21.9 Ω

Current (I₂) =?

V = I₂R₂

110 = I₂ × 21.9

Divide both side by 21.9

I₂ = 110 / 21.9

I₂ = 5.02 A

Therefore, the current in 21.9 Ω (R₂) is 5.02 A

Biii. Determination of the current in 11.7 Ω (R₃)

Voltage (V) = 110 V

Resistance (R₃) = 11.7 Ω

Current (I₃) =?

V = I₃R₃

110 = I₃ × 11.7

Divide both side by 11.7

I₃ = 110 / 11.7

I₃ = 9.40 A

Therefore, the current in 11.7 Ω (R₃) is 9.40 A.

C. Determination of the total current.

Current 1 (I₁) = 7.14 A

Current 2 (I₂) = 5.02 A

Current 3 (I₃) = 9.40 A

Total current (Iₜ) =?

Iₜ = I₁ + I₂ + I₃

Iₜ = 7.14 + 5.02 + 9.40

Iₜ = 21.56 A

Therefore, the total current in the circuit is 21.56 A

7 0

3 years ago

How do we find work from a force-displacement graph when the force exerted is not constant?

Artemon [7]

Answer:

Explanation:

Read forces directly from the graph.

Read displacements directly from the graph.

Use the area under the graph to find the work done by the force. This is equal to the kinetic or potential energy the object gains due to the application of the force.

6 0

3 years ago

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

$V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

The distance from the center of the square to one of the corners is $\sqrt2 L/2 = 0.035m$

$V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0$

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) $V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

$r_1 = 0.05\sqrt2m\\r_2 = 0.05m$

$V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a$

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3$

$W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}$

4 0

3 years ago

Intelligence tests that are given to participants in groups

White raven [17]

a. are typically paper-and-pencil measures.

Similar with psychological tests, mostly structured personality tests.

Psychological tests comes two ways: 
The structure psychological tests or, objectives tests and unstructured psychological tests or, also called projective tests. By what you are referring the responder strongly asserts a projective tests which in definition comes with an unambiguous stimuli or no paper test just drawings and images. If what the responder’s suggesting is correct you are referring to the Rorschach projective tests, these tests are a figure symmetrically placed in an inkblot that lets you visualize or create a mental picture out of it, and makes you describe what you in see much detail as you can.

4 0

3 years ago

Evidence of wave theory light​

Evidence of wave theory light