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3 years ago

Why does it take mars over 500 days to orbit the sun?

Physics

1 answer:

kakasveta [241]3 years ago

6 0

Answer:

Mar's orbital path is more than that of Earth, thus it takes more number of days to orbit around the sun.

Explanation:

Mars takes over 500 days to orbit all the way around the sun than Earth because its distance from the sun (228 million kilometers) is greater than that of Earth (150 million kilometers) which takes it 365 days.

Planets that orbit closer to the sun take shorter time to orbit around the sun because the cover a shorter orbital distance and orbit faster than those planets further from the sun.

<u>For example</u>

Using Earth's distance from the sun, 150 million kilometers and the number of days taken to orbits the sun ,365 days and the distance Mars is from the Earth, 228 million kilometers, you can approximate the time Mar takes to orbit the sun as:

Earth 150 million kilometers = 365 days

Mars 228 million kilometers= ?

Cross product ; (228 *365) /150 =555 -----(a value closer to that in the question)

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In proton beam therapy, a beam of high-energy protons is used to kill cancerous cells in a tumor. In one system, the beam, which

Thepotemich [5.8K]

The number of protons that strike the tumor each second is $1.75 \times 10^8$ protons.

<u>Given the following data:</u>

Energy of proton = $2.8 \times 10^{-11}$ Joules.

Current = 84 nA to A = $84 \times 10^{-9}$ Ampere.

Energy of total dose = $3.4 \times 10^{-3}$ Joules.

<u>Scientific data:</u>

Charge of proton = $1.602 \times 10^{-19}$ C.

To determine the number of protons that strike the tumor each second:

<h3>How to determine the number of protons.</h3>

Mathematically, the quantity of charge per unit time is given by this formula:

$Q =It=Ne\\\\N = \frac{Q}{e}$

<u>Where:</u>

N is the number of protons.

t is the time measured in seconds.

I is the current.

e is the charge of protons.

Substituting the parameters into the formula, we have;

$N=\frac{2.8 \times 10^{-11} }{1.602 \times 10^{-19}} \\\\N = 1.75 \times 10^{8}$ protons.

Read more on charges here: brainly.com/question/14372859

8 0

2 years ago

Set up a parallel circuit with three resistors. Set up the battery to be 9 V, one resistor at 10 Ohms, the other at 21 Ohms, and

valina [46]

In a parallel circuit, the equivalent resistance is the reciprocal of (the sum of the individual reciprocals).

1/R = 1/10 + 1/21 + 1/13

1/R = 0.225 mhos

R = 4.45 ohms

I = V / R

The total current out of the battery is

I = (9v)/(4.45ohms)

I = 2.02 Amperes

As the total current leaves the battery, it splits into 3 paths, and each resistor gets part of it. The 10ohm resistor gets the most current; the 21ohm resistor gets the least current. After flowing through the resistors, the 3 currents join and add up to 2.02 Amperes again, and the same current returns to the battery.

Each resistor has the same 9v of EMF across it.

4 0

3 years ago

The percentage of fe in an iron ii sample was determined by redox titration with cr2o7 calculate the molarity of the standard k2

Archy [21]

The molarity is 0.018M and the percentage is 8.46%.

Net ionic reaction is

$\mathrm{Cr} 2 \mathrm{O}^{2-}+6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+} \rightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H} 2 \mathrm{O}$

$1 \mathrm{~mol}$ of $\mathrm{Cr} 2 \mathrm{O} 7(-2)$ reacts with $6 \mathrm{~mol}$ of $\mathrm{Fe}(2+)$ to form $6 \mathrm{~mol}$ of $\mathrm{Fe}(3+)$ and $2 \mathrm{~mol}$ of $\mathrm{Cr}(3+)$

Find molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$

Molarity = moles of $\mathrm{K} 2 \mathrm{Cr} 207 /$ Volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7(\mathrm{~L}$ )

Moles of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$ = grams of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 /$ molar mass of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$

$=1.2275 \mathrm{~g} / 294.19 \mathrm{~mol}=0.0042 \mathrm{~g} / \mathrm{mol}$

Molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7=$ moles of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$ / Volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$ (L)

$=0.0042 \mathrm{~g} / \mathrm{mol} / 0.250 \mathrm{~L}=0.017 \mathrm{M}$

Find molarity of $\mathrm{Fe}$ (II)

Molarity of $\mathrm{Fe}(\mathrm{II})=6 \times$ molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 \times$ volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 /$ volume of $\mathrm{Fe}$ (II)

$=6 \times 0.017 \mathrm{M} \times 0.03598 \mathrm{~L} / 0.200 \mathrm{~L}$

$=0.018 \mathrm{M}$

Find moles of $\mathrm{Fe}(\mathrm{II})$

Moles of $\mathrm{Fe}($ II) =$ molarity of $\mathrm{Fe}$ (II) $\times$ volume of $\mathrm{Fe}$ (II)

$=0.018 \mathrm{M} \times 0.200 \mathrm{~L}=0.004 \mathrm{~mol}$

Find mass of Fe(II)

Mass of $\mathrm{Fe}( II )=$ moles of $\mathrm{Fe}( II$ ) $\times$ molar mass of $\mathrm{Fe}(\mathrm{II})$

$=0.004 \mathrm{~mol} \times 55.85 \mathrm{~g} / \mathrm{mol}$

$=0.205 \mathrm{~g}$

Find $\% \mathrm{Fe}(\mathrm{II})$ in unknown sample

$\begin{aligned}\% \mathrm{Fe} &=(\text { mass of } \mathrm{Fe} / \text { weight of unknown sample }) \times 100 \\&=(0.205 \mathrm{~g} / 2.4234 \mathrm{~g}) \times 100 \\&=8.46 \%\end{aligned}$