All categories

3 years ago

How are red dwarf stars and yellow dwarf stars alike?

1 answer:

8 0

Yellow and red dwarf stars are normal stars - they burn hydrogen in their cores and live on the main sequence of stellar lifetimes.

You might be interested in

Answers to electrical?

Orlov [11]

Can you get a better pic so i can read it

3 0

3 years ago

What could be the possible answer to the question ?<br><br>thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$

F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0

2 years ago

Read 2 more answers

The internal energy of material is determined by

Mnenie [13.5K]

The combined amount of kinetic and potential energy of its molecules

6 0

3 years ago

A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm

iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

$m = m_{o}* m_{e}$

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

$m_{o} = \frac{v_{o}}{-u _{0}}$

and me as

$m_{e} = 1+\frac{D}{f_{e}}$

d is distant of distinct vision = 25.0 cm for normal eye

fe = focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}$

$\frac{1}{v_{o}} = 2.1 cm$

$m_{o} = \frac{2.1}{0.150} = 14$

$m_{e} = 1+\frac{25}{1}$

$m_{e} =26$

$m = m_{o}* m_{e}$

m = 14*26 = 364

4 0

3 years ago

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 3 months. The line

Sladkaya [172]

Answer:

Explanation:

given

T = 3months = 7.9 × 10⁶s

orbital speed = 88 × 10³m/s

V= 2πr÷T

∴ r = (V×T) ÷ 2π

r = (88km × 7.9 × 10⁶s) ÷ 2π

r = 1.10 × 10⁸km

using kepler's 3rd law

mass of both stars = (seperation diatance)³/(orbital speed)²

M₁ + M₂ = (2r)³/( $\frac{1}{4}$ year)²

= (1.06 × 10²⁵)/(6.2×10¹³)

1.71×10¹²kg

since M₁ = M₂ =1.71×10¹²kg ÷ 2

M₁ = M₂ = 8.55×10¹¹kg

6 0

3 years ago