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3 years ago

What is the upper temperature range for stars?

2 answers:

6 0

The upper temperature range for stars is : c. 40,000 K

Every object has their own upper and lower temperature
This indicate the maximum possible energy that a star could release to its surrounding if its somehow exploded.

azamat3 years ago

3 0

The answer is C 40,000 Kelvin.

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PLZ HELP! I WILL GIVE BRAINLIEST OR WHATEVER U PPL WANT JUST HELP!

Tpy6a [65]

Newton's first law is sometimes known as the law of inertia. It is the law that states that an object at rest will stay at rest and an object in motion will stay in motion unless a force acts upon it. For example, if I was working with a wrench in space an it slipped, it would keep on going in one direction with a constant speed unless it hits something. Hope this helps!

5 0

3 years ago

What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a

Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

$M=-\frac{f_o}{f_e}$

Here, M is the magnification of an astronomical telescope, $f_e$ is the focal length of the eyepiece lens and $f_o$ is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put $f_o=74 cm$ and $f_e=2.4 cm$ in the above expression.

$M=-\frac{74}{2.4}$

M=-30.83

Therefore, the magnification of an astronomical telescope is -30.83.

5 0

4 years ago

If a cart is accelerating downhill under a net force of 25 N, what additional force would cause the cart to have a constant velo

Vladimir [108]

No additional force is required because it's already going downhill

7 0

3 years ago

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

$T^2 = \frac{4\pi^2 d^3}{2GM}$

The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

$T = 120290789.7s$

$T = 120290789.7s(\frac{1year}{31536000s})$

$T \approx 3.8143 years$

Therefore the period of this star is 3.8years

7 0

3 years ago

A technician at a semiconductor facility is using an oscilloscope to measure the AC voltage across a resistor in a circuit. The

d1i1m1o1n [39]

Answer:

The value to be reported is 5.48V

Explanation:

The RMS (root mean square) is defined as the value of voltage that will produce the same heating effect, or power dissipation, in circuit, as this AC voltage.

The RMS voltage is also called effective voltage because it is just as effective as DC voltage in providing power to an element.

It is expressed as $V_{rms}$ = $\frac{V_{m} }{\sqrt{2} }$

where Vm is the maximum or peak value of the voltage

In calculating the RMS of the voltage , we simply divide the peak voltage by square root of 2 (√2)

$V_{rms}$ = $\frac{7.75}{\sqrt{2} }$

= $\frac{7.75}{1.414}$

= 5.48 V

6 0

3 years ago