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grin007 [14]
3 years ago
6

What temperature has the same value in both the fahrenheit and kelvin scales?

Physics
1 answer:
jeka57 [31]3 years ago
6 0

Answer:

574.25 °

Explanation:

To know this, we need to get the expressions to calculate °F and °K.

In the case of Kelvin:

°K = °C + 273

in fahrenheit:

°F = 9/5(°C + 32)

Now, in order to know the temperature which both Kelvin and Fahrenheit are the same, we need to use both above equations, and solve for °C.

°C = °K - 273 (1)

°C = 5/9(°F - 32) (2)

Using 1 and 2 into a same expression:

°K - 273 = 5/9(°F - 32)

With this, we need to know now the moment which K = F, so, all we need to do is replace the F for the K in the above expression. Doing this, we have:

°K - 273 = 5/9(°K - 32)

°K = 0.555(°K - 32) + 273

°K = 5/9K - 17.78 + 273

°K - 5/9K = 255.22

4/9°K = 255.22

°K = 255.22 * 9 / 4

°K = 574.25 °

And this is the temperature in which both Kelvin and Fahrenheit are the same.

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Answer:

4 Joules (if you mean 2000 grams)

4000 Joules (if you mean 2000 kilograms)

Explanation:

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3 years ago
The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the averag
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The speed of tsunami is a.0.32 km. 

Steps involved  :

The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?

Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32

As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.

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2 years ago
Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The mo
ddd [48]

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j +  5k

Therefore,

M₀ = (i + j + k) x (1i + 0j +  5k)

M₀ = \left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]

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M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

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An object that completes 20 vibrations in 10 seconds has a frequency of
nika2105 [10]

Answer:

<em> The object has frequency of 2 Hz and time period of 0.5 s.</em>

Explanation:

<em>Frequency</em> is defined as number of oscillation per second ie back and forth swings done in single second.

Here it is given that the object oscillates 20 times in 10 seconds.

So f = \frac{20}{10} = 2Hz

The <em>time period</em> is defined as time taken by the object to complete one full oscillation.

T = \frac{1}{f}

T= \frac{1}{2} =0.5 s

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How many protons would the element with the atomic number 10 contain?
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