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4 years ago

A temperature of 50.0 degrees Celsius is the same as a Kelvin temperature of approximately

Physics

1 answer:

Mila [183]4 years ago

4 0

The answer is 323 K.

<u>Explanation:</u>

The Celsius temperature scale is equal to - 273.15.

One kelvin is equal to the change of thermodynamic temperature that results in a change of thermal energy kT by 1.380649 * 10^−23 J.

Convert degrees Celsius to kelvins with this simple formula:

kelvins = [degree C] + 273.15

The formula to convert Celsius to kelvins is degree C + 273.15.

50 degree celsius = 50 + 273.15

= 323.15 = 323 K.

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Calculate the change in entropy as 0.3071 kg of ice at 273.15 K melts. (The latent heat of fusion of water is 333000 J / kg)

Ostrovityanka [42]

Answer:

374.39 J/K

Explanation:

Entropy: This can be defined as the degree of disorder or randomness of a substance.

The S.I unit of entropy is J/K

ΔS = ΔH/T ..................................... Equation 1

Where ΔS = entropy change, ΔH = Heat change, T = temperature.

ΔH = cm................................... Equation 2

Where,

c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.

Substitute into equation 2

ΔH = 333000×0.3071

ΔH = 102264.3 J.

Also, T = 273.15 K

Substitute into equation 1

ΔS = 102264.3/273.15

ΔS = 374.39 J/K

Thus, The change in entropy = 374.39 J/K

3 0

3 years ago

Which of the following is example of convection check all that apply

sladkih [1.3K]

Answer air circulating in a hot air balloon

Explanation:

6 0

3 years ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

$F_D = c_D A \frac{\rho V^2}{2}$

Where,

F_D = Drag Force

$c_D$ = Drag coefficient

A = Area

$\rho$ = Density

V = Velocity

Our values are given by,

$c_D = 0.5$ (That is proper of a cone-shape)

$A = 9m^2$

$\rho = 1.2Kg/m^3$

$V = 6.5m/s$

Part A ) Replacing our values,

$F_D = 0.5*9*\frac{1.2*6.5^2}{2}$

$F_D = 114.075N$

Part B ) To find the torque we apply the equation as follow,

$\tau = F*d$

$\tau = (114.075N)(7)$

$\tau = 798.525N.m$

3 0

4 years ago

What will be the weight of a man of mass 90kg when he is on a planet with acceleration due to gravity of 15ms-2?

Feliz [49]

Answer:

1350N

Explanation:

Weight = Mass x Acceleration Due to Gravity

W=mg

W=90*15=1350N

7 0

4 years ago

When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of t

choli [55]

Answer:

Part a)

$\tau = 23.1 Nm$

Part b)

$\tau = 17.05 Foot pound force$

Explanation:

As we know that torque is defined as the product of force and its perpendicular distance from reference point

so here we have

$\tau = \vec r \times \vec F$

now we have

$\tau = (0.140)(165)$

$\tau = 23.1 Nm$

Part b)

Now we know the conversion as

1 meter = 3.28 foot

1 N = 0.225 Lb force

now we have

$\tau = 23.1 Nm$

$\tau = 23.1 (0.225 Lb)(3.28 foot)$

$\tau = 17.05 Foot pound force$

3 0

3 years ago