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3 years ago

11

Estimate how much solar energy reaches the earth per year (in Joule).

1 answer:

Alexxandr [17]3 years ago

4 0

Each hour 430 quintillion Joules of energy from the sun hits the Earth.

In a year it is very hard to determine because of the night and different light levels.

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Can someone help me asap this is due at 2 50

Rom4ik [11]

The data is inadequate

8 0

3 years ago

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago

What metric unit would you use to measure the weight of an elephant

Tems11 [23]

Answer:

kilograms

Explanation:

5 0

3 years ago

Under certain circumstances, potassium ions (K+) in a cell will move across the cell membrane from the inside to the outside. Th

choli [55]

Answer:

$1.368\times 10^{-20}\ J$

Explanation:

q = Charge in the potassium ion = $19e-18e$

e = Charge of electron = $1.6\times 10^{-19}\ C$

$V_2-V_1$ = Change in potential = $0-(-85.5\times 10^{-3})$

Change in electric potential is given by

$E=q(V_2-V_1)\\\Rightarrow E=(19e-18e)(0-(-85.5\times 10^{-3})\\\Rightarrow E=1.6\times 10^{-19}\times 85.5\times 10^{-3}\\\Rightarrow E=1.368\times 10^{-20}\ J$

The energy is $1.368\times 10^{-20}\ J$

3 0

3 years ago

When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to

Serjik [45]

Answer:

E) d/sqrt2

Explanation:

The initial electric force between the two charge is given by:

$F=k\frac{q_1 q_2}{d^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

d is the separation between the two charges

We can also rewrite it as

$d=\sqrt{k\frac{q_1 q_2}{F}}$

So if we want to make the force F twice as strong,

F' = 2F

the new distance between the charges would be

$d'=\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{1}{\sqrt{2}}\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{d}{\sqrt{2}}$

so the correct option is E.

8 0

2 years ago