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Minchanka [31]
2 years ago
12

In how much time will the voyagar I and II comelete the rotation of milky way galaxy?​

Physics
2 answers:
Art [367]2 years ago
8 0

Answer:

How long till Voyager 1 leaves the Milky Way?

Image result for In how much time will the voyager I and II complete the rotation of milky way galaxy?​

40,000 years

Voyager 1 becomes the first manmade object to leave the Solar System, and in 40,000 years it will come within 1.7 light years of star AC+793888, before continuing on its millions-of-years journey to the core of the Milky Way.

help me by marking as brainliest....

ss7ja [257]2 years ago
5 0

Answer:

By 500 million years from now, the solar system and the Voyagers alike will complete a full orbit through the Milky Way.

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Given an electron beam whose electrons have kinetic energy of 10.0 kev , what is the minimum wavelength λmin of light radiated b
kolbaska11 [484]
To answer the problem we would be using this formula which isE = hc/L where E is the energy, h is Planck's constant, c is the speed of light and L is the wavelength 
L = hc/E = 4.136×10−15 eV·s (2.998x10^8 m/s)/10^4 eV 

= 1.240x10^-10 m 

= 1.240x10^-1 nm
3 0
3 years ago
Read 2 more answers
A power plant produces 1000 MW to supply a city 40 km away. Current flows from the power plant on a single wire of resistance 0.
nikitadnepr [17]

Answer:

Current = 8696 A

Fraction of power lost = \dfrac{80}{529} = 0.151

Explanation:

Electric power is given by

P=IV

where I is the current and V is the voltage.

I=\dfrac{P}{V}

Using values from the question,

I=\dfrac{1000\times10^6 \text{ W}}{115\times10^3\text{ V}} = 8696 \text{ A}

The power loss is given by

P_\text{loss} = I^2R

where R is the resistance of the wire. From the question, the wire has a resistance of 0.050\Omega per km. Since resistance is proportional to length, the resistance of the wire is

R = 0.050\times40 = 2\Omega

Hence,

P_\text{loss} = \left(\dfrac{200000}{23}\right)^2\times2

The fraction lost = \dfrac{P_\text{loss}}{P}=\left(\dfrac{200000}{23}\right)^2\times2\div (1000\times10^6)=\dfrac{80}{529}=0.151

3 0
3 years ago
Um ladrão tenta fugir sozinho carregando em suas mãos uma mala cheia de barras de ouro. A densidade do ouro é igual a 20 g/cm³,
Roman55 [17]
I believe the answer to your question is A
4 0
3 years ago
A 3.00 kg block moving 2.09 m/s
Talja [164]

Answer:11.64kgm/s

Explanation:

4 0
3 years ago
I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?
Ad libitum [116K]

Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

\frac{1}{2} m {v}^{2}  = mgh

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

\sin(15)  =  \frac{h}{d}  \\ or \: h = d \sin(15)

Plugging this into the energy conservation equation and cancelling m, we get

{v}^{2}  = 2gd \sin(15)

Solving for d,

d =  \frac{ {v}^{2} }{2g \sin(15) }  =  \frac{ {(7.5 \:  \frac{m}{s}) }^{2} }{2(9.8 \:  \frac{m}{ {s}^{2} })(0.259)}   \\ = 11.1 \: m

3 0
3 years ago
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