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babymother [125]
4 years ago
14

The difference in temperature between 100°C and 101°C is _____ the difference between 100K and 101K.

Physics
1 answer:
zloy xaker [14]4 years ago
3 0
The missing word is "equal".

In fact, The relationship between temperature in Celsius and temperature in Kelvin is
T(^{\circ}C) = T(K)-273.15
So, the absolute value of the two temperatures are different. However, there is a direct proportionality between the temperature in Celsius and the temperature in Kelvin, this means that temperature differences are equal in both units.

Let's see this with an example: let's call T_1 (C) and T_2 (C) two different temperatures in Celsius, and let's rewrite them in Kelvin using the previous relationship:
T_2 (C)-T_1 (C) = T_2(K)-273.15-(T_1(K)-273.15) = T_2(K)-T_1(K)
as we can see, both the 273.15 cancel out and so the temperature difference in Celsius is equal to the temperature difference in Kelvin.
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A rocket travels 1.3 km in 62 ms. What is its average speed in m⋅s−1? Do not give your answer in scientific notation. The answer
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Answer:

Average speed = 0.35 m/s

Explanation:

Given the following data;

Distance = 1.3 Km

Time = 62 minutes

To find the average speed in m/s;

First of all, we would convert the quantities to their standard unit (S.I) of measurement;

Conversion:

1.3 kilometres to meters = 1.3 * 1000 = 1300 meters

For time;

1 minute = 60 seconds

62 minutes = X

Cross-multiplying, we have;

X = 62 * 60

X = 3720 seconds

Now, we can calculate the average speed in m/s using the formula;

Speed = \frac {distance}{time}

Speed = \frac {1300}{3720}

Average speed = 0.35 m/s

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3 years ago
Which of the following is accurate in describing converging and diverging lenses?
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3 years ago
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Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a
allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2

Acceleration of the frog is 2874.33 m/s²

6 0
4 years ago
(a) Are the radiating electric field lines around a charged particle straight lines when the particle is stationary?
s2008m [1.1K]

Answer:

Yes

Explanation:

(a)

Yes, the radiating electric field lines around a stationary charged particle are straight lines when the particle is stationary.

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4 years ago
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