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Aleks [24]
3 years ago
7

The universal force that is effective over the longest distance is called?

Physics
1 answer:
gulaghasi [49]3 years ago
3 0
The answer is D. gravity
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if a diffraction grating produces a third-order bright spot for red light (of wavelength 650 nm ) at 68.0 ∘ from the central max
Mashutka [201]

Answer:

The angle is 25.34°.

Explanation:

Given that,

Wave length = 650 nm

Angle = 68.0°

We need to calculate the distance

For a diffraction grating

d\sin\theta=m\lambda

d=\dfrac{2\times650\times10^{-9}}{\sin68.0}

d=2.10\times10^{-6}\ m

We need to calculate the angle

Using formula for angle

d\sin\theta=m\lambda

\sin\theta=\dfrac{m\times\lambda}{d}

\sin\theta=\dfrac{2\times450\times10^{-9}}{2.10\times10^{-6}}

\sin\theta=25.34^{\circ}

Hence, The angle is 25.34°.

8 0
3 years ago
(a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0
Sergio039 [100]
<h2>Answer:</h2>

(a) 6.95 x 10⁻⁸ C

(b) 6.25N/C

<h2>Explanation:</h2>

The electric field (E) on a point charge, Q, is given by;

E = k x Q / r²              ---------------(i)

Where;

k = constant = 8.99 x 10⁹ N m²/C²

r = distance of the charge from a reference point.

Given from the question;

E = 10000N/C

r = 0.250m

Substitute these values into equation(i) as follows;

10000 = 8.99 x 10⁹ x Q / (0.25)²

10000 = 8.99 x 10⁹ x Q / (0.0625)

10000 = 143.84 x 10⁹ x Q

Solve for Q;

Q = 10000/(143.84 x 10⁹)

Q = 0.00695 x 10⁻⁵C

Q = 6.95 x 10⁻⁸ C

The magnitude of the charge is 6.95 x 10⁻⁸ C

(b) To get how large the field (E) will be at r = 10.0m, substitute these values including Q = 6.95 x 10⁻⁸ C into equation (i) as follows;

E = k x Q / r²

E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 10²

E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 100

E = 6.25N/C

Therefore, at 10.0m, the electric field will be just 6.25N/C

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Answer:

Food, clothes, language, and belief

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Which of the following describes the bending of light due to a change in its speed?
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The Answer is This sounds like refraction.
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