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Gemiola [76]
2 years ago
15

Which of the following is one of the three main fuels that a star uses for fusion

Physics
2 answers:
Colt1911 [192]2 years ago
8 0
Hydrogen, helium, and carbon.
Elanso [62]2 years ago
8 0

Answer:

OXYGEN

Explanation:

from odyssey

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8. List three temperature scales. Then write the boiling and freezing
Tomtit [17]

Explanation:

For most temperature scales, the boiling point of water and the freezing point is used to calibrate it.

Three known temperature scales;

  • Kelvin scale
  • Celcius scale
  • Fahrenheit scale

                               Kelvin scale              Celcius scale          Fahrenheit scale

Freezing point             273K                            0°C                        32°F

Melting point                373K                          100°C                     212°F

5 0
2 years ago
A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then
Paha777 [63]

Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J

6 0
3 years ago
Use Hooke's Law to determine the work done by the variable force in the spring problem. Nine joules of work is required to stret
natima [27]

Answer:

29.16 J

Explanation:

From Hook's law,

W = 1/2(ke²)..................... Equation 1

Where W = work done, k = Spring constant, e = extension.

Given: W = 9 J, e = 0.5 m.

Substitute into equation 1

9 = 1/2(k×0.5²)

Solve for k

k = 18/0.5²

k = 72 N/m.

The work done required to stretch the spring by additional 0.4 m is

W = 1/2(72)(0.4+0.5)²

W = 36(0.9²)

W = 29.16 J.

6 0
2 years ago
(a) Find the frequency of revolution of an electron with an energy of 114 eV in a uniform magnetic field of magnitude 46.7 µT. (
stira [4]

Answer:

(a)  1.3 x 10^6 Hz

(b) 76.73 cm

Explanation:

(a)

the formula for the frequency is given by

f = B q / 2 π m

where, B be the strength of magnetic field, q be the charge on one electron, m is the mass of one electron.

B = 46.7 micro tesla = 46.7 x 10^-6 T

q = 1.6 x 10^-19 C

m = 9.1 x 10^-31 kg

f = (46.7 x 10^-6 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31) = 1.3 x 10^6 Hz

(b) K = 114 eV = 114 x 1.6 x 10^-19 J = 182.4 x 10^-19 J

K = 1/2 mv^2

182.4 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v = 6.3 x 10^6 m/s

r = m v / B q

Where, r be the radius of circular path

r = (9.1 x 10^-31 x 6.3 x 10^6) / (46.7 x 10^-6 x 1.6 x 10^-19)

r = 0.7673 m = 76.73 cm  

5 0
3 years ago
Find the equation of the line below.
Musya8 [376]

Answers all in picture below

:

7 0
2 years ago
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