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3 years ago

Convert the temperature 288 K to degrees Celsius

2 answers:

6 0

Answer: 14.85°C

Justification:

1) To convert Kelvin temperature tu Celsius temperature you subtract 273.15

2) This is the formula.

°C = K - 273.15

3) The conversion of 288 K to degrees Celsius is:

°C = 288 - 273.15 = 14.85 K

Kelvin temperature is called absolute temperature. The lower bound of the absolute temperature is 0 K.<span>, that means absolute zero and it is physically impossible to reach.

At 0 K or absolute temperature the atoms would stop vibrating meining that the entropy of the substance reaches it's minimum value, i.e. 0

</span>Zero K is equivalent to - 273.15°C

Bumek [7]3 years ago

6 0

i dont know how to convert it but you can use goolge to transfer it

288 kelvin = 14.85 celsius

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What does it mean for water to have a high heat capacity

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Its heat capacity is higher than that of any other liquid or solid, its specific heat being 1 cal / g, this means that to raise the temperature of 1 g of water by 1 ° C it is necessary to provide an amount of heat equal to a calorie . Therefore, the heat capacity of 1 g of water is equal to 1 cal / K.

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The water has a very high heat capacity, a large amount of heat is necessary to raise its temperature 1.0 ° K. For biological systems this is very important because the cellular temperature is modified very little in response to metabolism. In the same way, aquatic organisms, if water did not possess that quality, would be very affected or would not exist.

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3 years ago

A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of

Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

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