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puteri [66]
3 years ago
8

Which is the larger change in temperature 1 Kelvin or 1 degree Celsius

Physics
1 answer:
Pachacha [2.7K]3 years ago
5 0
Those changes in temperature are equal. The absolute (Kelvin) scale and the Celsius scale use the same size degrees, (but their zeros are set at different temperatures).

Their 'degree' is 0.01 of the difference between the boiling and freezing temperatures of water.
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When did agriculture first began?​
timofeeve [1]

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  • Humans invented agriculture between 7,000 and 10,000 years ago, during the Neolithic era, or the New Stone Age. There were eight Neolithic crops: emmer wheat, einkorn wheat, peas, lentils, bitter vetch, hulled barley, chickpeas, and flax.

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Which of the following statements are true about the international system of measurement?
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8 0
3 years ago
Calculate the heat, in kilocalories, that is absorbed if 183 g of ice at 0.0 ∘C is placed in an ice bag, melts, and warms to bod
boyakko [2]

Answer:

The total amount of heat needed will be Q_T=21.411kcal.

Explanation:

We will divide the calculation in two: First, the heat needed to melt the ice, and then the heat needed to warm the resulting liquid from 0°C to 37°C.

m=183g

l_f=80\frac{cal}{g} =334\frac{J}{g}

l_w=1\frac{cal}{g} =4.184\frac{J}{g}

<em>i) </em>The fusion heat will be:

Q_f=l_fm=14640cal=14.640kcal

<em>ii)</em> The heat needed to warm the water from T_i=0^{\circ}C to T_i=37^{\circ}C will be:

Q_w=l_wm(T_f-T_i)=6771cal=6.771kcal

So, the total amount needed will be the sum of these two results:

Q_T=Q_f+Q_w=14.640kcal+6.771kcal=21.411kcal.

8 0
3 years ago
When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di
AleksandrR [38]

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

F=kx

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

k= \frac{F}{x}

k= \frac{mg}{x}

k= \frac{(0.35)(9.8)}{12*10^{-2}}

k = 28.58N/m

Perioricity in an elastic body is defined by

T = 2\pi \sqrt{\frac{m}{k}}

Where,

m = Mass

k = Spring constant

T = 2\pi \sqrt{\frac{0.35}{28.58}}

T = 0.685s

Therefore the period of the oscillations is 0.685s

4 0
3 years ago
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