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poizon [28]
3 years ago
6

If the only force exerted on a star far from the center of the Galaxy (r = 7.40 ✕ 1020 m) is the gravitational force exerted by

the ordinary matter (Mord = 1.90 ✕ 1041 kg), find the speed of the star. Assume a circular orbit and assume all the Galaxy's matter is concentrated at the center.
Physics
1 answer:
lina2011 [118]3 years ago
8 0

Answer:

The value is  v = 1.309*10^{5}\ m/s

Explanation:

The radius is r = 7.40 *10^{20} \  m

The mass of the ordinary matter is M_{rod} =  1.90 *10^{41}\  kg

Generally the speed of the star is mathematically represented as

         v = \sqrt{\frac{G * M}{r} }

Here G is the gravitational constant with a value

        G = 6.67384 * 10^{-11}

So

      v = \sqrt{\frac{6.67384 * 10^{-11} * 1.90 *10^{41}}{7.40 *10^{20}} }

=>    v = 1.309*10^{5}\ m/s

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una onda longitudinal tiene una frecuencia de 200 hz y una longitud de onda de 4.2m ¿cual es la rapidez de la onda?​
swat32

Answer:

v = 8.4 m/s

Explanation:

The question ays, "A longitudinal wave has a frequency of 200 Hz and a wavelength of 4.2m. What is the speed of the wave?".

Frequency of a wave, f = 200 Hz

Wavelength = 4.2 cm = 0.042 m

We need to find the speed of the wave. The formula for the speed of a wave is given by :

v=f\lambda\\\\v=200\times 0.042\\\\=8.4\ m/s

So, the speed of the wave is equal to 8.4 m/s.

4 0
3 years ago
A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a potential differ
ra1l [238]

Answer:

The correct question is:

"Find the energy each gains"

The energy gained by a charged particle accelerated through a potential difference is given by

\Delta U = q\Delta V

where

q is the charge of the particle

\Delta V is the potential difference

For a proton,

q=+e=1.6\cdot 10^{-19}C

And since \Delta V=100 V

The energy gained by the proton is

\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J

For an alpha particle,

q=+2e=3.2\cdot 10^{-19}C

Therefore, the energy gained is

\Delta U=(3.2\cdot 10^{-19})(100)=3.2\cdot 10^{-17}J

Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)

q=+e=1.6\cdot 10^{-19}C

So the energy gained is the same as the proton:

\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J

6 0
3 years ago
What is the average velocity of atoms in 1.00 mol of neon (a monatomic gas) at 465 K? For m, use 0.0202 kg.
TiliK225 [7]

Answer: 757m/s

Explanation:

Given the following :

Mole of neon gas = 1.00 mol

Temperature = 465k

Mass = 0.0202kg

Using the ideal gas equation. For calculating the average kinetic energy molecule :

0.5(mv^2) = 3/2 nRt

Where ;

M = mass, V = volume. R = gas constant(8.31 jK-1 mol-1, t = temperature in Kelvin, n = number of moles

Plugging our values

0.5(0.0202 × v^2) = 3/2 (1 × 8.31 × 465)

0.0101 v^2 = 5796.225

v^2 = 5796.225 / 0.0101

v^2 = 573883.66

v = √573883.66

v = 757.55109m/s

v = 757m/s

5 0
3 years ago
At constant volume, the heat of combustion of a particular compound is − 3550.0 kJ / mol. When 1.075 g of this compound ( molar
swat32

Answer:

C=1,25\cdot 10^{5} kJ/^{\circ}C

Explanation:

First of all let's define the specific molar heat capacity.

C = \frac{-Q}{n\cdot \Delta T} (1)

Where:

Q is the released heat by the system

n is the number of moles

ΔT is the difference of temperature of the system  

Now, we can find n with the molar mass (M) the mass of the compound (m).

n=\frac{m}{M}=6.95\cdot 10^{-3} moles      

Using (1) we have:

C=\frac{-3550}{6.95\cdot 10^{-3} 4.073}

C=1,25\cdot 10^{5} kJ/^{\circ}C

I hope it helps!

6 0
3 years ago
The smallest division value of electronic balance​
lara [203]

Answer:

0.1g to 0.0000001g hope it helps uu

4 0
3 years ago
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