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3 years ago

This diagram shows a ball rolling from left to right along a frictionless

Physics

1 answer:

yanalaym [24]3 years ago

5 0

I’d say C: 1,3,4 because when you think about it section 2 and five would be balanced right?

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In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

GaryK [48]

We will determine the wavelength through the relationship given by the distance between slits, this relationship is given under the function

$y = \frac{m\lambda}{d}$

Here,

m = Number of order bright fringe

$\lambda$ = Wavelength

d = Distance between slits

Both distance are the same, then

$y_1 = y_2$

$\frac{m_1\lambda_1 r}{d} = \frac{m_2\lambda_2 r}{d}$

$\frac{m_1\lambda_1}{m_2\lambda_2} =1$

Rearranging to find the second wavelength

$m_1 \lambda_1 = m_2 \lambda _2$

$\lambda_2 = \frac{m_1\lambda_1}{m_2}$

$\lambda_2 = \frac{7(629)}{8}$

$\lambda_2 = 550.3nm$

Therefore the wavelength of the light coming from the second monochromatic light source is 550.3nm

7 0

3 years ago

Why is static electricity worse in winter?

Ratling [72]

Because static electricity build up in our homes because we have the heaters on which suck moisture out of the air.

8 0

3 years ago

A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the obje

kobusy [5.1K]

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

3 0

3 years ago

How to find wavelength

Andrews [41]

If the wavelength<span> is given, the energy can be determined by first using the wave equation (c = λ × ν) to </span>find<span> the frequency, then using Planck's equation to </span>calculate<span> energy. Use the equations above to answer the following questions. 1. Ultraviolet radiation has a frequency of 6.8 × 1015 1/s.</span>

4 0

3 years ago

A point charge q1=+5.00nC is at the fixed position x=0, y=0, z=0. You find that you must do 8.10×10−6J of work to bring a second

Maru [420]

The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

<h3>Magnitude of second charge</h3>

The magnitude of the second charge is determined by applying Coulomb's law.

$E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 = \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC$

Thus, the value of the second charge is 1.2 nC.

Learn more about electric potential here: brainly.com/question/14306881

7 0

2 years ago