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4 years ago

Pluto's orbit brings it inside the orbit of _____. Neptune Saturn Venus Uranus

Physics

2 answers:

sergejj [24]4 years ago

7 0

It will be neptune..

Margarita [4]4 years ago

6 0

Answer:

Pluto's orbit brings it inside the orbit of <u>Neptune</u>.

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According to peacemaking criminology, toward which option should a country redirect the money it currently spends on imprisoning

ratelena [41]

Answer:

Explanation:

It is C. Environmental campaigns

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3 years ago

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Calculate the power if the current in a circuit is 3A and the resistance of the circuit is 50hms

Tresset [83]

Answer:

450W

Explanation:

V= I*R

V=3A*50 Ohms

V=150V

P=I*V

P=3A*150V

P=450W

3 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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3 years ago

You want to examine the hairy details of your favorite pet caterpillar, using a lens of focal length 8.9 cm 8.9 cm that you just

Zepler [3.9K]

Answer:

The angular magnification is $M = 2.808$

Explanation:

From the question we are told

The focal length is $f = 8.9cm$

The near point is $n_p = 25.0cm$

The angular magnification is mathematically represented as

$M = \frac{n_p}{f}$

Substituting values

$M = \frac{25}{8.9}$

$= 2.808$

4 0

3 years ago

What is the force on an electron in a CRT when it’s moving at 2.5 × 105 meters/second perpendicular to a magnetic field of 1.5 t

Firdavs [7]

F = Magnetic Force
B = Magnetic Field
V = Velocity

*The vectors from the photo you get doing the left-hand rule.

The magnetic force is always perpendicular to the magnetic field.

And as told in the statement, the electron is moving perpendicular to a magnetic field, that is, the Velocity forms an 90 degree angle / Right angle with the magnetic field.

The formula to find the Magnetic Force is:

$f = |q| \times v \times b \times sin \: \theta$

Where "q" is the Charge and the sin theta is the angle formed by the Velocity and Magnetic Field, in this case it's 90°. Sin 90° = 1.

$f = |- 1.6 \times {10}^{ - 19} | \times 2.5 \times {10}^{5} \times 1.5 \times 1 \\ f = 6 \times {10}^{ - 19 + 5} \\ f = 6 \times {10}^{ - 14} \: newtons$
Newton (N) = C x m/s x T = (C x m x T)/s

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4 years ago