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3 years ago

Which statement is true about white dwarf?

Physics

2 answers:

kramer3 years ago

8 0

C is the correct answer

galben [10]3 years ago

6 0

White dwarf are very dense and small in size is a true statement.

Answer: Option C

<u>Explanation:</u>

A white dwarf also known as degenerate dwarf. The white dwarf "mass is comparable to the Sun" and its "volume is comparable to Earth".

The white dwarf is a stellar whose size is near to the size of the earth. It had a gravitational collapse. The diameter of the white dwarf is just one percent of the Sun. It is extremely dense where there is no free space between the atoms.

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Write down formula of power

r-ruslan [8.4K]

There u go -> P=f/a

8 0

3 years ago

Read 2 more answers

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

Determine the binding energy of an F-19 nucleus. The F-19 nucleus has a mass of 18.99840325 amu. A proton has a mass of 1.00728

Anvisha [2.4K]

Answer:

Energy = 1.38*10^13 J/mol

Explanation:

Total number of proton in F-19 = 9

Total number of neutron in F-19 = 10

Expected Mass of F-19

= 9*1.007 + 10*1.008 = 19.152 u

Actual mass of F-19 = 18.998 u

Energy of one particle of F-19 = 931.5*Δm = 931.5*(19.152-18.998)

= 143.234 MeV

Energy of one mole of F-19 = 143.234*10^6*1.6*10^-19*6.022*10^23

= 1.38*10^13 J/mol

8 0

3 years ago

Please help on this one!!

inysia [295]

The energy transformations that occur as you coast down long hill on a bicycle, including the brakes to make the bike stop at the bottom, is that at the top of the hill you have high GPE AND LOW KE, on your way down you have HIGH KE AND LOW GPE, and at the bottom you have thermal energy due to the stop of the brakes.

the law of conversation of energy and describe the energy transformations that occur as you coast down a long hill on a bicycle and then apply the brakes to make the bike stop at the bottom.

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2 years ago

Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. Th

zlopas [31]

When they meet the 40-kg boy would have moved a distance of 6 m.

<h3>Distance moved by the 40 kg boy</h3>

Apply the principle of center mass;

Take the 40 kg mass as the reference point;

X(40 kg) = (40kg x 0 + 60kg x 10 m)/(40 kg + 60 kg)

X(40 kg) = (600)/(100)

X(40 kg) = 6 m

Thus, when they meet the 40-kg boy would have moved a distance of 6 m.

Learn more about distance here: brainly.com/question/2854969

#SPJ1

4 0

2 years ago