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ch4aika [34]
4 years ago
5

Where did the element heavier than hydrogen and helium come from?

Physics
2 answers:
k0ka [10]4 years ago
7 0

Answer:

After the hydrogen in the star's core is exhausted, the star can fuse helium to form progressively heavier elements, carbon and oxygen and so on, until iron and nickel are formed. Elements heavier than that are formed in supernovas.

Explanation:

Snezhnost [94]4 years ago
6 0

They were produced inside stars.

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A net force of 25N causes an object to accelerate at 4m/s^2. what is the mass of the object?
Doss [256]

Answer:

6.25 kg

Explanation:

Fnet=ma

m=Fnet/a

m=25/4

m=6.25kg

3 0
3 years ago
Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init
Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ=(m+\frac{1}{2})\frac{lamda}{n}  ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= \frac{3*lamda}{2}

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= \sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }

δ= \sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}

δ= \sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }

δ= \sqrt{4900+43^2}-\sqrt{4900+23^2}

δ= \sqrt{4900+1849}-\sqrt{4900+529}

δ= \sqrt{6749}-\sqrt{5429}

δ=  82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= \frac{3*lamda}{2}

δ= 8.47

8.47 = \frac{3*lamda}{2}

λ = \frac{8.47*2}{3}

λ = 5.65m

3 0
4 years ago
A box having a mass of 0.2 kg is dragged across a horizontal floor by
Lyrx [107]

you can check attachment for answer.

kind regards

7 0
3 years ago
A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se
Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

v = \frac{Bqr}{m}

v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}

v = 661098.9 = 661099 m/s

(b)

\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}

e / m = 1.76 x 10^14 C / kg

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K = 0.5 x mv²

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K = 1.99 x 10^-19 J

K = 1.24 eV

So, the potential difference is

V = 1.24 V

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V = 2 x 1.24 = 2.48 V

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Let v is the speed

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3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

r=\frac{mv}{Bq}

r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}

r = 0.0113 m = 1.13 cm

7 0
3 years ago
What is the difference between a series circuit and a parallel circuit?
Angelina_Jolie [31]
Parallel has more than one circuit or form of energy

series has only one form of energy circuit
6 0
3 years ago
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