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3 years ago

Can someone please explain to me what in the hell this asking?

1 answer:

4 0

As air heats up, air molecules occupy a larger area comparitively to cool air. To summerize, the ball will appear to be "more inflated."

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Areas near baton rouge louisiana recently received 36.2 cm of rain in a single day. how many meters of rain was this?

sergey [27]

100 cm is 1 meter. So your answer would be 0.362 meters.

5 0

3 years ago

Read 2 more answers

How much does 1 liter of water weigh

Zinaida [17]

Weight = (mass) x (gravity)

On Earth ...

Weight = (1 kg) x (9.8 m/s^2)

Weight = 9.8 Newtons

8 0

3 years ago

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

andrew11 [14]

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}$

For second order

$\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

8 0

3 years ago

In regions where few species existed before or where species were wiped out what occurs?

marysya [2.9K]

I cant really say but i believe it is called extinction

5 0

4 years ago

Two protons are 9.261 fm apart. (1 fm= 1 femtometer = 1 x 10^-15 m.) What is the ratio of the electric force to the gravitationa

Strike441 [17]

Answer:

$1.24\times 10^{36}$

Explanation:

$q$ = magnitude of charge on each proton = 1.6 x 10⁻¹⁹ C

$m$ = mass of each proton = 1.67 x 10⁻²⁷ kg

r = distance between the two protons = 1 x 10⁻¹⁵ m

Electric force between the two protons is given as

$F_{e} = \frac{kq^{2}}{r^{2}}$

$F_{e} = \frac{(9\times 10^{9})(1.6\times 10^{-19})^{2}}{(1\times 10^{-15})^{2}}$

$F_{e} = 230.4$ N

Gravitational force between the two protons is given as

$F_{g} = \frac{Gm^{2}}{r^{2}}$

$F_{g} = \frac{(6.67\times 10^{-11})(1.67\times 10^{-27})^{2}}{(1\times 10^{-15})^{2}}$

$F_{g} = 1.86\times 10^{-34}$ N

Ratio is given as

$Ratio =\frac{F_{e}}{F_{g}}$

$Ratio =\frac{230.4}{1.86\times 10^{-34}}$

$Ratio = 1.24\times 10^{36}$

3 0

3 years ago

Read 2 more answers