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2 years ago

12

Write the total equation for the proton-proton chain reaction that takes place in stars?

1 answer:

Sav [38]2 years ago

6 0

Supernova?? lollll.

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According to Newton's 2nd Law of Motion, if the mass of an object is 10 kg and the force is 10 newtons, then the acceleration is

Mkey [24]

According to Newton's second Law of motion, if the mass of an object is 10 kg and the force is 10 newtons, then the acceleration is 1m/s².

<h3>How to calculate acceleration?</h3>

The acceleration of a moving body can be calculated by dividing the force of the body by its mass.

According to this question, the mass of an object is 10 kg and the force is 10 newtons, then the acceleration can be calculated as follows:

acceleration = 10N ÷ 10kg

acceleration = 1m/s²

Therefore, according to Newton's second Law of motion, if the mass of an object is 10 kg and the force is 10 newtons, then the acceleration is 1m/s².

Learn more about acceleration at: brainly.com/question/12550364

#SPJ1

8 0

1 year ago

A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar

denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of $\beta$

Using formula of $\beta$

$\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}$

Put the value into the formula

$\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}$

$\beta=8.86\times10^{9}$

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

$T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}$

Put the value into the formula

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}$

$T=5.45\times10^{-4}$

(b). We need to calculate the tunnel probability for width 1.00 nm

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}$

$T=7.74\times10^{-8}$

Hence, The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

6 0

3 years ago

How could you increase the total energy of a system consisting of a jar full of water?

grin007 [14]

Answer:

D.) Heat the jar in a microwave oven.

Explanation:

5 0

3 years ago

Read 2 more answers

A single-turn circular loop of wire of radius 5.0 cm lies in a plane perpendicular to a spatially uniform magnetic field. During

Artemon [7]

Answer:

Induced EMF, $\epsilon=0.0143\ volts$

Explanation:

Given that,

Radius of the circular loop, r = 5 cm = 0.05 m

Time, t = 0.0548 s

Initial magnetic field, $B_i=200\ mT=0.2\ T$

Final magnetic field, $B_f=300\ mT=0.3\ T$

The expression for the induced emf is given by :

$\epsilon=\dfrac{d\phi}{dt}$

$\phi$ = magnetic flux

$\epsilon=\dfrac{d(BA)}{dt}$

$\epsilon=A\dfrac{d(B)}{dt}$

$\epsilon=A\dfrac{B_f-B_i}{t}$

$\epsilon=\pi (0.05)^2\times \dfrac{0.3-0.2}{0.0548}$

$\epsilon=0.0143\ volts$

So, the induced emf in the loop is 0.0143 volts. Hence, this is the required solution.

5 0

2 years ago

HELP!!! 20 POINTS!! TRUE OR FALSE??

ki77a [65]

Answer:

It might be true

Explanation:

6 0

3 years ago

Read 2 more answers