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ELEN [110]
3 years ago
7

The Sun orbits the center of the Galaxy in 225 million years at a distance of 26,000 lightyears. Given that a^3 = (M1 +M2) x P^2

, where a is the semimajor axis in AU and P is the orbital period in years, what is the mass of the Galaxy within the Sun’s orbit?
Physics
1 answer:
ss7ja [257]3 years ago
5 0

According to Kepler third law the relation between orbital period and radio of matter in the Galaxy is given by

a^3 = (M_1+M_2)p^2

Where,

a = Radius of start orbit

p = Orbital Period

M = Total mass in a sphere of radius centered on galactic center

Our values are given as

a = 26000ly (\frac{9.461*10^{15}m}{1LY}) = 2.45*10^{20}m

p = 225million year = 225*10^6 year

Replacing we have,

M = \frac{2.45*10^{20}}{225*10^6}

M = 1.088*10^{12}M_{sun}

M = 108Billion M_{sun}

Therefore the mass of the galaxy within the sun's orbit is 108Billion the mass of the sun.

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Indicate the result of each of the following unit vector cross products (unit vector hat-symbols not shown): . kxi = . jxi= -jxk
notka56 [123]

Answer:

Explanation:

The cross product of two vectors is given by

\overrightarrow{A}\times \overrightarrow{B}=\left | \overrightarrow{A} \right |\left | \overrightarrow{B} \right |Sin\theta \widehat{n}

Where, θ be the angle between the two vectors and \widehat{n} be the unit vector along the direction of cross product of two vectors.

Here, K x i = - j

As K is the unit vector along Z axis, i is the unit vector along X axis and j be the unit vector along  axis.

The direction of cross product of two vectors is given by the right hand palm rule.

So, k x i = j

j x i = - k

- j x k = - i

i x i = 0

4 0
3 years ago
A hot-air balloon is rising upward with a constant speed of 2.03 m/s. When the balloon is 8.13 m above the ground, the balloonis
HACTEHA [7]
The answer is 4.2s...
7 0
3 years ago
The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N
guajiro [1.7K]

Answer:

\frac{r_1}{r_2}=6.9

Explanation:

According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:

P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}

where,

F₁ = Load lifted by output plunger = 2100 N

F₂ = Force applied on input piston = 44 N

r₁ = radius of output plunger

r₂ = radius of input piston

Therefore,

\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9

4 0
3 years ago
How many times does a human heart beat during a person’s lifetime? How many gallons of blood does it pump? (Estimate that the he
Nostrana [21]

Answer: The heart pumps 124.2 billion cm³ of blood in a lifetime

Explanation:

as an adult the pulse rate average must be around 72 beats per minute.

The heart beats about 103,680 times in a day.

There are 365 days in a year

number of heart beat in a year = 365 days x 103,680 = 37,843,200 beats in a year

For every the heart pumps 50cm³ of blood,

Hence,

Amount of blood pump in a year = 50 x 37,843,200 = 1,892,160,000cm³ of blood pumped in a year.

Using the estimated lifespan average an individual is 69 years

So in a life time,

The human heart pumps = 1,892,160,000 x 69 years = 124,200,000,000

If the heart pumps 50cm³ of blood per beat, the heart pumps a total of 130,559,040,000 cm³ (130.6 billion cm³) of blood in a LIFETIME.

3 0
3 years ago
A child (approximately 4 years old) takes her metal “slinky Toy” (a flexible coiled metal spring) and does various tests to dete
anzhelika [568]

Answer:

248

Explanation:

L = Inductance of the slinky = 130 μH = 130 x 10⁻⁶ H

l = length of the slinky = 3 m

N = number of turns in the slinky

r = radius of slinky = 4 cm = 0.04 m

Area of slinky is given as

A = πr²

A = (3.14) (0.04)²

A = 0.005024 m²

Inductance is given as

L = \frac{\mu _{o}N^{2}A}{l}

130\times 10^{-6} = \frac{(12.56\times 10^{-7})N^{2}(0.005024)}{3}

N = 248

8 0
3 years ago
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