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AlekseyPX
3 years ago
7

How do giant stars differ from sequence stars ?

Physics
2 answers:
Norma-Jean [14]3 years ago
8 0
<span>A star is born as a main sequence star when the temperature of the star's core becomes high enough to fuse hydrogen into helium. ... The story is different in evolved stars like giant stars. Once hydrogen in the core of the star is exhausted, the energy generation becomes small and the core starts to collapse. Hope this helps.</span>
soldi70 [24.7K]3 years ago
8 0
When a star is born it is a main sequence  when the temperature of the star's core becomes high enough to fuse hydrogen into helium. The giant stars are the adult ones, it has a larger radius and luminosity than a main-sequence star. The giant star is the aging process of the star. 
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NASA developed a reflective foil barrier to help shield spacecraft from heat transfer in space. These reflective barriers are no
morpeh [17]

Answer: Radiation

Explanation: Radiation is the energy that comes from a source in form of electromagnetic waves, subatomic particles, light, or heat which travels through space.

Examples of radiation include the light, heat, and particles emitted from the Sun.

Using a foil barrier to prevent heat transfer is possible because foil has a silver color, and silver reflects light and heat instead of absorbing them. This is the opposite of black surfaces that absorb heat.

So in homes where these foil reflective barriers are used, the transfer of heat through Radiation is highly reduced.

3 0
3 years ago
WILL NAME THE BRAINLIEST! An airplane undergoes the following displacements: It first flies 72 km in a direction of 30° East of
nekit [7.7K]

Answer:

82.1 km

Explanation:

We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:

A_x = (72) sin 30^{\circ} =36.0 km\\A_y = (72) cos 30^{\circ} = 62.4 km

Vector B is 48 km south, so:

B_x = 0\\B_y = -48

Finally, vector C:

C_x = -(100) cos 30^{\circ} =-86.6 km\\C_y = (100) sin 30^{\circ} = 50.0 km

Now we add the components along each direction:

R_x = A_x + B_x + C_x = 36.0 + 0 +(-86.6)=-50.6 km\\R_y = A_y+B_y+C_y = 62.4+(-48)+50.0=64.6 km

So, the resultant (which is the distance in a straight line between the starting point and the final point of the motion) is

R=\sqrt{R_x^2+R_y^2}=\sqrt{(-50.6)^2+(64.6)^2}=82.1 km

4 0
3 years ago
The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ????0 . Find the minimu
Reptile [31]

Answer:

Explanation:

Threshold frequency = 4.17 x 10¹⁴ Hz .

minimum energy required = hν where h is plank's constant and ν is frequency .

E = 6.6 x 10⁻³⁴ x 4.17 x 10¹⁴

= 27.52 x 10⁻²⁰ J .

wavelength of radiation falling = 245 x 10⁻⁹ m

Energy of this radiation = hc / λ

c is velocity of light and  λ  is wavelength of radiation .

= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 245 x 10⁻⁹

= .08081 x 10⁻¹⁷ J

= 80.81 x 10⁻²⁰ J

kinetic energy of electrons ejected = energy of falling radiation - threshold energy

= 80.81 x 10⁻²⁰  - 27.52 x 10⁻²⁰

= 53.29 x 10⁻²⁰ J .

4 0
3 years ago
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?
Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight =   60 .3 (9.81) = 591.54 N

Accel of lift changes this to    60.3 ( 9.81 - L)     where L - accel of lift

                                           60.3 ( 9.81 - L ) = 564

                                               solve for L = .457 m/s^2  DOWNWARD

                                                        so L = - .457 m/s^2

4 0
2 years ago
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