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kotegsom [21]
3 years ago
12

A light wave passes through an aperture (that is, a narrow slit). When it does so, the degree to which the wave spreads out will

be...
Physics
1 answer:
crimeas [40]3 years ago
7 0

Explanation:

Single slit diffraction

Diffraction is the phenomenon of spreading out of waves as they pass through an aperture or around objects. Diffraction occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave. For very small aperture sizes, the vast majority of the wave is blocked. in case of  large apertures the wave passes by or through the obstacle without any significant diffraction.

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Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after
adoni [48]

Answer: v = 1.19 * 10^{6} m/s

Explanation: q = magnitude of electronic charge = 1.609 * 10^{-19} c

mass of an electronic charge = 9.10 * 10^{-31} kg

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = \frac{mv^{2} }{2},  potential energy = qV

hence, \frac{mv^{2} }{2} = qV

\frac{9.10 *10^{-31} * v^{2}  }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31}  * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s

7 0
3 years ago
A cube of plastic 1.2x10^-5 km on a side has a mas of 1.1 g. what is its density in g/cm^3?
Mamont248 [21]
Density = mass / volume

mass = 1.1 g

volume = length of side ^ 3 = [1.2 * 10^-5 km * 100000 cm/km]^3 = [1.2 cm]^3 = 1.728 cm^3

density = 1.1 g / 1.728 cm^3 = 0.64 g / cm^3

 
5 0
3 years ago
1. A limiting factor for using nuclear energy is the
Goryan [66]

Answer:

waste

Explanation:

took da test. nhgvgg

8 0
3 years ago
Read 2 more answers
A body moves in a straight line. At time, it's acceleration is given by a = 12t + 1. When t =0, the velocity of the body v is 2
madreJ [45]

Answer:

v = 6t² + t + 2, s = 2t³ + ½ t² + 2t

59 m/s, 64.5 m

Explanation:

a = 12t + 1

v = ∫ a dt

v = 6t² + t + C

At t = 0, v = 2.

2 = 6(0)² + (0) + C

2 = C

Therefore, v = 6t² + t + 2.

s = ∫ v dt

s = 2t³ + ½ t² + 2t + C

At t = 0, s = 0.

0 = 2(0)³ + ½ (0)² + 2(0) + C

0 = C

Therefore, s = 2t³ + ½ t² + 2t.

At t = 3:

v = 6(3)² + (3) + 2 = 59

s = 2(3)³ + ½ (3)² + 2(3) = 64.5

5 0
2 years ago
Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.
mr_godi [17]

Answer:

The magintude of the acceleration for both objects is 3.11m/s^2

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

T_A-m_A*g=m_A*a

and for mB (descending):

T_B-m_B*g=-m_B*a

T_A=T_B

because the two boxes has the same acceleration because they are attached together:

m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2

So the magintude of the acceleration for both objects is 3.11m/s^2

5 0
3 years ago
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