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3 years ago

Simplifying fractions30 over 54

Physics

2 answers:

g100num [7]3 years ago

5 0

Answer:

5/9

Explanation:

Veronika [31]3 years ago

4 0

Answer:

5/9

Explanation:

First, find a common factor for 30 and 54.

30 and 54 can both be divided by 2.

30 divided by 2 = 15

54 divided by 2 = 27

So, your new fraction is 15/27.

However, don't stop there.

15 and 27 can both be divided by 3 as well.

15 divided by 3 = 5

27 divided by 3 = 9

So, your final answer is 5/9.

I hope this helps! :)

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An electron is accelerated within a particle accelerator using a 100 MV electric potential. The 100 MeV electron moves along an

Delicious77 [7]

Answer:

The length of the tube is 3.92 m.

Explanation:

Given that,

Electric potential = 100 MV

Length = 4 m

Energy = 100 MeV

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Using formula of relativistic energy

$E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)$

Put the value into the formula

$1.6\times10^{-15}= 9.1\times`10^{-31}\times9\times10^{16}(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)$

$(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)=\dfrac{1.6\times10^{-15}}{9.1\times10^{-31}\times9\times10^{16}}$

Here, $\gamma-1=(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)$

$\gamma-1=0.01953$

$\gamma=0.01953+1$

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We need to calculate the length

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3 years ago

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Now we know that

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Where v is the final velocity

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3 years ago

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 15.1 K.

Zarrin [17]

Answer: (a) The magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .

(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$ .

Explanation:

(a) Expression for change in temperature is as follows.

$|\Delta T| = |x - y|K$

= 15.1 K

= $|(x - 273.15) - (y - 273.15)|^{o}C$

= $|x - y|^{o}C$

= $15.1^{o}C$

Therefore, the magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .

(b) Change in temperature from Celsius to Fahrenheit is as follows.

F = 1.8C + 32

C = $\frac{F - 32}{1.8}$

Since, K = C + 273

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= $27.18^{o}F$

or, = $27.2^{o}F$

Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$ .

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Simplifying fractions30 over 54​

Simplifying fractions30 over 54