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2 years ago

How do spectroscopes help with studying of distance stars?

Physics

1 answer:

Svetradugi [14.3K]2 years ago

8 0

Answer:

You take the light from a star, planet or galaxy and pass it through a spectroscope, which is a bit like a prism letting you split the light into its component colours. "It lets you see the chemicals being absorbed or emitted by the light source. From this you can work out all sorts of things," says Watson

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We know that the magnitudes of the negative charge on the electron and the positive charge on the proton are equal. Suppose, how

TEA [102]

Answer:3,45 x 10^9 N

Explanation: We have considered the total charge for each coin , this is the total atoms x 29 electrons for cooper and multiplier by electron charge, the total charge for each coin is 0,464 C

Finally we use the Coulomb law,

F=k Q/ (r)^2

6 0

2 years ago

what times are the moon phases visible? My science teacher said that the phases of the moon are only visible at sertant times of

klio [65]

Your teacher is right. The moon can be seen early in the morning sometimes and late at night. Different phases are only visible on certain days as one day might be full quarter, the next full moon, the next first quarter, etc.

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2 years ago

Taylor mixes two liquids together in a beaker. A solid forms at the bottom of the beaker, and the liquid changes from pink to bl

Dimas [21]

It is a chemical change and a physical change

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2 years ago

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

The orbital speed is $v= 2.6*10^{3} m/s$

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

3 years ago

A 1500 kg tractor pulls a 750 kg trailer north and applies a 2250 N force on it. What is the force on the tractor?

Masja [62]

Answer:

law of Action and Reaction F = 2250 N

Explanation:

The tractor and the trailer are two bodies that interact, therefore, by the law of Action and Reaction, the force that one applies on the other is equal to the force that the second body (trailer) applies on the first (tractor), but with opposite direction

F = 2250 N

directed from trailer to tractor

3 0

2 years ago