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Rudik [331]
3 years ago
10

Which property is closely related to its temperature

Physics
2 answers:
Lilit [14]3 years ago
8 0

Answer:

Mass – The single most important property that determines other properties of the star. Luminosity – The total amount of energy (light) that a star emits into space. Temperature – surface temperature, closely related to the luminosity and color of the star

Explanation:

Ber [7]3 years ago
6 0
Mass, bc it’s just science
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The chart shows data for a moving object.
Ipatiy [6.2K]

Answer:

number 2

Explanation:

8 0
3 years ago
คลื่นกลเกิดขึ้นได้อย่างไร
Feliz [49]

Mechanical waves are oscillation of matter, they are important because they all transfer energy from one place to another. There are 2 types of mechanical waves. A transverse wave where the particles vibrate perpendicular to the direction of energy travel and a longitudinal wave where particle vibrations are parallel to the direction of the energy transfer.


I hope it helps, Regards.

4 0
3 years ago
Before designing the filter, one must understand the relationship between the output voltage of the circuit and the frequency. F
Kamila [148]

Answer:

Option b. is correct

Explanation:

An RLC electrical circuit consists of constituent components: a resistor (R), an inductor (L), and a capacitor (C). A resistor, an inductor, and a capacitor are connected in series or parallel.

The impedances of the circuit elements depend on the frequency.

Both impedance magnitudes decrease when the frequency increases

5 0
3 years ago
a. When the electric field between the plates is 75% of the dielectric strength and energy density of the stored energy is 2800
Olegator [25]

Answer: The value of the dielectric constant k = 1.8

Explanation:

If C= ε A/d and

Electrostatic energy W = 1/2CV^2

Substitutes C in the first formula into the energy formula.

W = 1/2 ε A/d × V^2

Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed

Substitute V = Ed into the energy W.

W = 1/2 × ε A/d ×( Ed )^2

W = 1/2 × ε A/d × E^2 × d^2

d will cancel one of the ds

W = 1/2 × ε Ad × E^2

W/Ad = 1/2 × ε × E^2

W/V = 1/2 × ε E^2

Where Ad = volume V

E = dielectric strength

εo = permittivity of free space = 8.84 x 10^-12 F/m

W/V = 2800 J/m^3

Let first calculate the dielectric strength

2800 = 1/2 × 8.84×10^-12 × E^2

5600 = 8.84×10^-12E^2

E^2 = 5600/8.84×10^-12

E = sqrt( 6.3 × 10^14)

E = 25 × 10^7

75% of E = 18.9 × 10^6Jm

The permittivity of the material will be achieved by using the same formula

2800 = 1/2 × ε E^2

2800 = 0.5 × ε × (18.9×10^6)^2

2800 = ε × 1.78 × 10^14

ε = 2800/1.78×10^14

ε = 1.57 × 10^-11

Dielectric constant k = relative permittivity

Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is

k = 1.57×10^-11/8.84×10^-12

k = 1.776

k = 1.8 approximately

Therefore, the value of the dielectric constant k is 1.8

3 0
3 years ago
Imagine that you are going to visit your friend. Before you get there, you decide to stop at the variety store. If you walk 200
SashulF [63]

Answer:

400m

Explanation:

Brainliest? :))

Let your initial displacement from your home to the store be

Dd

>

1 and your displacement from the store to your friend’s house

be Dd

>

2.

Given: Dd

>

1 = 200 m [N]; Dd

>

2 = 600 m [S]

Required: Dd

>

T

Analysis: Dd

>

T 5 Dd

>

1 1 Dd

>

2

Solution: Figure 6 shows the given vectors, with the tip of Dd

>

1

joined to the tail of Dd

>

2. The resultant vector Dd

>

T is drawn in red,

from the tail of Dd

>

1 to the tip of Dd

>

2. The direction of Dd

>

T is [S].

Dd

>

T measures 4 cm in length in Figure 6, so using the scale of

1 cm : 100 m, the actual magnitude of Dd

>

T is 400 m.

Statement: Relative to your starting point at your home, your

total displacement is 400 m [S].

6 0
2 years ago
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