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2 years ago

Viewing moving objects from pond water under his microscope and named then animalcules

Physics

1 answer:

pogonyaev2 years ago

3 0

I think it was either Robert Hook or Anton Van Leevwen Hoek

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A 2-kg toy car accelerates from 0 to 5 m/s2. It

yan [13]

10 joules of work is done by the object

8 0

2 years ago

When gravity increases what happens to weight?

Fynjy0 [20]

Weight increases but mass stays the same

7 0

2 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

One gallon of gasoline in an automobiles engine produces on average 9.50 kg of carbon dioxide, which is a greenhouse gas; that i

Musya8 [376]

The annual production of carbon dioxide is 124121.49×10^{6}[/tex] kg.

First we calculate the fuel consumed by each car in a year

Fuel consumed=6990/21.4=326.63 gallon

Now we calculate the amount of fuel consumed by 40 million cars in a year

Fuel consumed=326.63*40*10^6=13065.42 million gallon,

Now we can calculate the annual production of carbon dioxide in the USA

CO2 production rate=9.50*13065.42=124121.49*10^6 kg

Therefore the annual production of carbon dioxide in USA is 124121.49×10^{6}[/tex] kg

4 0

3 years ago

Hope you all doing okay

Vsevolod [243]

Answer:Thank you

Explanation:

8 0

1 year ago