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natita [175]
3 years ago
6

A negatively charged balloon has 4 μC of charge. How many excess electrons are on this bal- loon? The elemental charge is 1.6 ×

10−19 C.
Answer in units of electrons.
Physics
1 answer:
bagirrra123 [75]3 years ago
7 0
Data:

The charge of a body depends on the amount of electrons it gains or loses. Q = n * e, where "Q" is charge, "n" is the number of plus or minus electrons, and "e" is the fundamental charge of an electron 1,6 * 10 ^{-19}C<span>. To know if the body has gained or lost, we look at the signal of its charge, remembering that the electron is negative. The charge of the body is 4 μC (positive), so there is a lack of electrons! 

Q = 4 </span>μC → Q = 4*10^{-6}
e = 1,6 * 10 ^{-19}C
n = ?<span>

We have:
</span>Q = n*e
n =  \frac{Q}{e}
n =  \frac{4*10^{-6}}{1,6 * 10 ^{-19}}
n = 2,5*10^{-6-(-19)}
n = 2,5*10^{-6+19}
\boxed{n = 2,5*10^{13}electrons}
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Answer:

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making m the subject of the formula in the equation above to calculate the mass of the black hole:

M = \frac{v^{2} r}{G}.............(1)

<u>For M84:</u>

r = 8 pc = 8 * 3.08 * 10¹⁶

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G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

M = \frac{( 4*10^{5}) ^{2} *24.64* 10^{16} }{6.674 * 10^{-11} }

M = 590.7 * 10³⁶ kg

<u>For M87:</u>

r = 20 pc = 20 * 3.08 * 10¹⁶

r = 61.6* 10¹⁶ m

v = 500 km/s = 5 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

M = \frac{( 5*10^{5}) ^{2} *61.6* 10^{16} }{6.674 * 10^{-11} }

M = 2307.46 * 10³⁶ kg

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when object goes under acceleration

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Two lasers, one red (with wavelength 633.0 nmnm) and the other green (with wavelength 532.0 nmnm), are mounted behind a 0.150-mm
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(a) The screen  is 3.20m from the split.

(b) The closest minima for green, distance Δy = 0.45 cm.

When a wave hits a barrier or opening, numerous events are referred to as diffraction. It is described as the interference or bending of waves via an aperture or around the corners of an obstruction into the area that forms the geometric shadow of the obstruction or aperture.

(a)Equation of minima = sinθ  = mλ/α

Given, m = 3, λ = 6.33X10⁻⁷, α = 0.00015

Putting the values in formula to get θ.

  θ = sin⁻¹ ( \frac{3 X 6.33X10^{-7} }{0.00015} ) = 0.01266 rad

triangle need to be drawn to find relationship between θ, y$ and L

tan(θ) = y/L  where; y = 4.05 cm

L = y/tan(θ) = 3.20

Hence, the screen is 3.20m from the split.

(b) Find the closest minima for green

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sinθ  = 4λ/α

θ = sin⁻¹ (\frac{4X6.33X10^{-7} }{0.00015}) = 0.01688 rad

Calculate L using

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From equation subtract y₃ from y:

                 4.50 cm - 4.05 cm = 0.45 cm

Hence, distance Δy = 0.45 cm.

Learn more about the Diffraction with the help of the given link:

brainly.com/question/12290582

#SPJ4

I understand that the question you are looking for is "Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the ot

Question

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the other side of the slit is a white screen. When the red laser is turned on, it creates a diffraction pattern on the screen.

a. The distance y3,red from the center of the pattern to the location of the third diffraction minimum of the red laser is 4.05 cm. How far L is the screen from the slit? Express this distance L in meters to three significant figures.

b. With both lasers turned on, the screen shows two overlapping diffraction patterns. The central maxima of the two patterns are at the same position. What is the distance Δy between the third minimum in the diffraction pattern of the red laser (from Part A) and the nearest minimum in the diffraction pattern of the green laser?

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2 years ago
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