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3 years ago

What Energy source that produces wind

2 answers:

6 0

Wind is usually caused by a difference in pressure that causes wind to be "Pushed" on the surface of earth. Air usually moves from high to low pressure and that is what we call wind. Please hit the thanks button if this helped :)

arsen [322]3 years ago

6 0

The energy source that produces wind is solar

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The dimensions of a cylinder are changing, but the height is always equal to the diameter of the base of the cylinder. If the he

VashaNatasha [74]

Answer:

dV/dt = 9 cubic inches per second

Explanation:

Let the height of the cylinder is h

Diameter of cylinder = height of the cylinder = h

Radius of cylinder, r = h/2

dh/dt = 3 inches /s

Volume of cylinder is given by

$V = \pi r^{2}h$

put r = h/2 so,

$V = \pi \frac{h^{3}}{4}$

Differentiate both sides with respect to t.

$\frac{dV}{dt}=\frac{3h^{2}}{4}\times \frac{dh}{dt}$

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

$\frac{dV}{dt}=\frac{3\times 2\times 2}{4}\times 3$

dV/dt = 9 cubic inches per second

Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.

6 0

3 years ago

How much energy (in kilojoules) is required to convert 200 mL of diethyl ether at its boiling point from liquid to vapor if its

LiRa [457]

Answer:

55.96kJ

Explanation:

Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether

Volume (v) = 200mL, density (d) = 0.7138g/mL

Mass = d × v = 0.7138 × 200 = 142.76g

Enthalpy of vaporization of diethyl ether = 29kJ/mol

MW of diethyl ether (C2H5)2O = 74g/mol

Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g

Energy = 142.76g × 0.392kJ/g = 55.96kJ

4 0

3 years ago

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road