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2 years ago

10. If one body is positively charged and another body is negatively charged, free electrons tend to

Physics

1 answer:

marta [7]2 years ago

7 0

Answer: B.move from the negatively charged body to the positively charged body.

Explanation: If one body is positively charged and another body is negatively charged, free electrons tend to move from the negatively charged body to the positively charged body. This happens because they have different charges and therefore electrons are more attracted to the positive or its opposite charge.

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Which of the following describes a chemical change

marshall27 [118]

The answer is A .

<span>Rusting of iron. combustion (burning) of wood. metabolism of food in the body. mixing an acid and a base, such as hydrochloric acid (HCl) and sodium hydroxide (NaOH) .</span>

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3 years ago

Read 2 more answers

A glass optical fiber is used to transport a light ray across a long distance. The fiber has an index of refraction of 1.550 and

devlian [24]

To solve this exercise it is necessary to apply the concepts related to the Snells law.

The law defines that,

$n_1 sin\theta_1 = n_2 sin\theta_2$

$n_1 =$ Incident index

$n_2 =$ Refracted index

$\theta_1$ = Incident angle

$\theta_2 =$ Refracted angle

Our values are given by

$n_1 = 1.550$

$n_2 = 1.361$

$\theta_2 =90\° \rightarrow$ Refractory angle generated when light passes through the fiber.

Replacing we have,

$(1.55)sin \theta_1 = (1.361) sin90$

$sin \theta_1 = \frac{(1.361) sin90}{(1.55)}$

$\theta_1 =sin^{-1} \frac{(1.361) sin90}{(1.55)}$

$\theta_1 =61.4\°$

Now for the calculation of the maximum angle we will subtract the minimum value previously found at the angle of 90 degrees which is the maximum. Then,

$\theta_{max} = 90-\theta \\\theta_{max} =90-61.4\\\theta_{max}=28.6\°$

Therefore the critical angle for the light ray to remain insider the fiber is 28.6°

6 0

3 years ago

Adjacent antinodes of a standing wave on a string are 15.0 cm apart. A particle at an antinode oscillates in simple harmonic mot

timama [110]

Answer:

Explanation:

A ) Distance between two adjacent anti-node will be equal to distance between two adjacent nodes . So the required distance is 15 cm .

B ) wave-length, amplitude, and speed of the two traveling waves that form this pattern are as follows

wave length = same as wave length of wave pattern formed. so it is 30 cm

amplitude = 1/2 the amplitude of wave pattern formed so it is .850 / 2 = .425 cm

Speed = frequency x wavelength ( frequency = 1 / time period )

= 1 / .075) x 30 cm

400 cm / m

C ) maximum speed

= ω A

= (2π / T) x A

= 2 X 3.14 x .85 / .075 cm / s

= 71.17 cm / s

minimum speed is zero.

D ) The shortest distance along the string between a node and an antinode

= Wavelength / 4

= 30 / 4

= 7.5 cm

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3 years ago

If you are going 60 mph what is your speed in m per second

LenKa [72]

Answer:

1/60 mps

Explanation:

We would first have to divide 60 by 60 because there is 60mins per hour to get 1mpm. After that we would have to divide 1 by 60 because there are 60 secs in a min. So our final answer after doing 1/60 would be a fraction.

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3 years ago

Determine the minimum applied force p required to move wedge a to the right. the spring is compressed a distance of 175 mm. negl

BARSIC [14]

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3 years ago