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3 years ago

8

A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is ma

de with the battery, the voltage across the resistor is

1 answer:

Vanyuwa [196]3 years ago

8 0

Answer:

At the moment contact is made with the battery, the voltage across the resistor is equal to the batteries terminal voltage

Explanation;

Because at series connection the battery and resistor have equal voltage

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Why skies are blue but not other colours

wolverine [178]

Answer:

The Short Answer: Sunlight reaches Earth's atmosphere and is scattered in all directions by all the gases and particles in the air. Blue light is scattered more than the other colors because it travels as shorter, smaller waves. This is why we see a blue sky most of the time.

Explanation:

4 0

3 years ago

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a

IRISSAK [1]

A. $4.64\cdot 10^{11}m$

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

$v=\frac{2\pi r}{T}$

where we have:

$v=30,000 km/s = 3\cdot 10^7 m/s$ is the orbital speed

r is the orbital radius

$T=27 h \cdot 3600 =97,200 s$ is the orbital period

Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

$r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m$

B. $6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s$

The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

$m\frac{v^2}{r}=\frac{GMm}{r^2}$

where

m is the mass of the clumps of matter

G is the gravitational constant

M is the mass of the black hole

Solving the formula for M, we find the mass of the black hole:

$M=\frac{v^2 r}{G}=\frac{(3\cdot 10^7 m/s)^2(4.64\cdot 10^{11} m)}{6.67\cdot 10^{-11}}=6.26\cdot 10^{36}kg$

and considering the value of the solar mass

$M_s = 2\cdot 10^{30}kg$

the mass of the black hole as a multiple of our sun's mass is

$M=\frac{6.26\cdot 10^{36} kg}{2\cdot 10^{30} kg}=3.13\cdot 10^6 M_s$

C. $9.28\cdot 10^9 m$

The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by

$R=\frac{2MG}{c^2}$

where M is the mass of the black hole and c is the speed of light.

Substituting numbers into the formula, we find

$R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m$

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2 years ago

Haileys comet is a comet that passes close to earth every 75 years. based on this passage, which gravitational attraction has th

yawa3891 [41]

Answer:

gravity lol

Explanation:

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3 years ago

Read 2 more answers

Du een Shem.<br>when we shake mango free, the mangoes fall<br>down, why?

Vanyuwa [196]

Answer:

Because it is a stone fruit

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3 years ago

Light from a laser with wavelength 400 nm passed through two small openings and produces an interference pattern on a screen 1 m

alexira [117]

To solve this problem we will apply the concepts related to destructive interference from double-slit experiments. For this purpose we will define the path difference as,

$\text{Path difference}= dsin\theta = (2n-1)\frac{\lambda}{2}$

Here,

$\lambda$ = Wavelength

$\theta$ = Angle when occurs the interference point of destructive interference

Our values are given as,

$\text{Wavelength} = \lambda = 400nm = 4*10^{-7}m$

$\text{Distance of Screen} = D = 1m$

Using the previous expression we have,

$d \times \theta = \frac{\lambda}{2}$

$d \times (0.1) = \frac{4*10^{-7}}{2}$

$d = 2*10^{-6} m$

$d = 2\mu m$ $d = 2\mu m$

Therefore the distance between the two openings is $2\mu m$

8 0

3 years ago