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3 years ago

Free electrons may be transferred between bodies by:

2 answers:

8 0

Free electrons may be transferred between bodies by "Contact Action"

When positively charged body comes close to a negatively charged body, then electron flows from positively to negatively charged body

In short, Your Answer would be Option B

Hope this helps!

elixir [45]3 years ago

6 0

Answer:

B) contact action

Explanation:

When two conductors are connected with each other then in that case the free charge of one conductor may be transferred to other if the potential of one conductor is not same as the potential of other conductor.

As we know that positive charge always move from higher potential to lower potential.

So here when two bodies are connected together then in that case electron from one body is transferred to other body if the the potential of two bodies are not same.

So here correct answer will be

B) contact action

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The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m2. What is the maximum va

Hitman42 [59]

Answer:

7.78x10^-8T

Explanation:

The Pointing Vector S is

S = (1/μ0) E × B

at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,

S = (1/μ0) E B

where S, E and B are magnitudes. The average value of the Pointing Vector is

<S> = [1/(2 μ0)] E0 B0

where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)

Also at any instant,

E = c B

where E and B are magnitudes, so it must also be true at the instant of peak values

E0 = c B0

Substituting for E0,

<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²

Solve for B0.

Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)

= 7.79 x10 ^-8 T

5 0

3 years ago

A ray of yellow light (f = 5.09 × 1014hz) travels at a speed of 2.04 × 108meters per second in

denis23 [38]

Velocity = fλ

where f is frequency in Hz, and λ is wavelength in meters.

2.04 * 10⁸ m/s = 5.09 * 10¹⁴ Hz * λ

(2.04 * 10⁸ m/s) / (5.09 * 10¹⁴ Hz ) = λ

4.007*10⁻⁷ m = λ

The wavelength of the yellow light = 4.007*10⁻⁷ m

8 0

3 years ago

Two black holes (the remains of exploded stars), separated by a distance of

jolli1 [7]

The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

distance between the two black holes, r = 10 AU = 1.5 x 10¹² m
gravitational force between the two black holes, F = 6.9 x 10²⁵ N.
combined mass of the two black holes = 5.20 x 10³⁰ kg

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

$F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2$

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁ + 2.328 x 10⁶⁰ = 0

solve the quadratic equation using formula method;

a = 1, b =- 5.2 x 10³⁰, c = 2.328 x 10⁶⁰

$m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20}) \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg$

The second mass is calculated as follows;

m₂ = 5.2 x 10³⁰ kg - m₁

m₂ = 5.2 x 10³⁰ kg - 4.7 x 10³⁰ kg

m₂ = 5 x 10²⁹ kg

m₂ = 5.2 x 10³⁰ kg - 4.9 x 10²⁹ kg

m₂ = 4.7 x 10³⁰ kg

Thus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

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3 0

2 years ago

Water waves that you might see at the beach move in circular or elliptical paths. What type of mechanical waves are water waves?

vazorg [7]

To find:

Which type of mechanical wave is a water wave?

Explanation:

The water wave is a combination of longitudinal and transverse waves. They are a type of wave called surface waves. The surface waves are the waves that transmit energy in the interface between two mediums.

Final answer:

Thus the correct answer is option C.

5 0

1 year ago

A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity

Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2 = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3