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3 years ago

What is the energy equivalent of an object with a mass of 1.83 kg? 1.65 × 1017 J 2.92 × 1017 J 3.10 × 1017 J 5.49 × 1017 J

Physics

2 answers:

Charra [1.4K]3 years ago

6 0

Answer:

1.65 x 1017j

Explanation:

Kryger [21]3 years ago

5 0

Answer:

Explanation:

We can use the Einsten's relation which states the equivalence between mass and energy:

where

E is the energy

m is the rest mass of the object

c is the speed of light

By using this formula and m=2.5 kg, we can find the equivalent energy of the object:

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Which explains upwelling in the oceans?

Leokris [45]

Winds blowing across the ocean surface push water away. Water then rises up from beneath the surface to replace the water that was pushed away. This process is known as “upwelling.”

Upwelling occurs in the open ocean and along coastlines. The reverse process, called “downwelling,” also occurs when wind causes surface water to build up along a coastline and the surface water eventually sinks toward the bottom.

Water that rises to the surface as a result of upwelling is typically colder and is rich in nutrients. These nutrients “fertilize” surface waters, meaning that these surface waters often have high biological productivity. Therefore, good fishing grounds typically are found where upwelling is common.

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3 years ago

By how much does the volume of an aluminum cube 4.00 cm on an edge increase when the cube is heated from 19.0°C to 67.0°C? The l

Genrish500 [490]

Answer:

The volume of an aluminum cube is 0.212 cm³.

Explanation:

Given that,

Edge of cube = 4.00 cm

Initial temperature = 19.0°C

Final temperature = 67.0°C

linear expansion coefficient $\alpha=23.0\times10^{-6}/C^{\circ}$

We need to calculate the volume expansion coefficient

Using formula of volume expansion coefficient

$\beta=3\alpha$

Put the value into the formula

$\beta=3\times23.0\times10^{-6}$

$\beta=0.000069=69\times10^{-6}/C^{\circ}$

We need to calculate the volume

$V= a^3$

$V=4^3$

$V=64\ cm^3$

The change temperature of the cube is

$\Delta T=T_{f}-T_{i}$

Put the value into the formula

$\Delta T=67-19 = 48^{\circ}C$

We need to calculate the increases volume

Using formula of increases volume

$\Delta V=V\beta\Delta T$

Put the value into the formula

$\Delta V=64\times69\times10^{-6}\times48$

$\Delta V=0.212\ cm^3$

Hence, The volume of an aluminum cube is 0.212 cm³.

5 0

3 years ago

For a Physics course containing 10 students, the maximum point total for the quarter was 200. The point totals for the 10 studen

Talja [164]

Answer:

130.5

Explanation:

According to the stemplot attached (Which I think it is, and if not, then you only need to replace the procedure with your data and you should be fine), you need to calculate first the points of all ten students. In that plot, we can easily calculate the points.

The first number in the colum represents the centen of the point, while the numbers of the second column, represents the units of that centen, for example if you see:

16 | 8 5 6

This means that the point for the students are 168, 165 and 166. Three students, three points.

If you watch the stemplot, the points for the students are:

116, 118, 121, 124, 128, 133, 137, 142, 146 and 179.

The median can be calculated using the mean between the two values in the middle of the sequence.

In this case, half of ten is 5, so, the numbers from the middle in this sequence are 128 and 133, therefore:

Median = 128 + 133 / 2 = 130.5

5 0

3 years ago

What is one fun fact about both rubies and sapphires?

Rufina [12.5K]

Answer:

A ruby is measures a 9 on the Mohs Scale of Hardness, second only to diamond and matched with sapphire. This makes a ruby an extremely hard and durable gemstone.

A sapphire is the birthstone for September. It is also a traditional gift for those celebrating 5th or 45th anniversaries.

8 0

3 years ago

Read 2 more answers

Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt

Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

$= \frac{4}{3}\pi(6400+ 50)^3 - \frac{4}{3}\pi (6400)
^3\\\\= \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3$

Hence the volume of atmosphere is $2.6\times 10^{19} m^3$

(b)

Write the ideal gas equation as foll ows:

$PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles$

$no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules
$

Hence the required molecules is $13.3\times10^{43} molecules
$

(c)

Write the ideal gas equation as follows:

$PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles$

$no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules$

Hence the required molecules in Caesar breath is $1.2\times 10^{23} molecules$

(d)

Volume fraction in Caesar last breath is as follows:

$Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}$

(e)

Since the volume capacity of the human body is 500 mL.

$Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath$

5 0

3 years ago