Why is earth's temperature just right for life?

It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T

tekilochka [14]

Answer:

ω = 0.05 rad/s

Explanation:

We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

$Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\$

where,

ω = angular velocity of cylinder = ?

g = required acceleration = 9.8 m/s²

r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m

Therefore,

$\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\$

7 0

3 years ago

When you boil potatoes, will your cooking time be reduced with vigorously boiling water instead of gently boiling water? (Direct

lara [203]

Answer:

The cooling time will not be reduced.

Explanation:

The time to cook is virtually the same in both types, vigorously and gently boiling water.

The reason cooking of spaghetti calls for vigorously boiling water is to keep the pasta agitated so that they do not stick to one another.

The temperature of boiling water is the same for both vigorously boiling water and gently boiling water, therefore there will be little time difference in when the potatoes will cook when it is done with vigorously boiling water than when it is cooked with gently boiling water.

However cooking potatoes in vigorously boiling water may cause the water to dry up on time and the potatoes get burnt.

8 0

3 years ago

What is the meaning of the reference point in electric potential?.

mrs_skeptik [129]

Answer:

<h3><u>ELECTRIC POTENTIAL</u></h3>

• the amount of work needed to move a unit charge from a reference point to a specific point against an electric field.

4 0

2 years ago

You are testing a new amusement park roller coaster with an empty car with a mass 120 kg. One part of the track is a vertical lo

Kay [80]

Answer:

- 5436 J

Explanation:

mass of car, m = 120 kg

radius of loop, r = 12 m

velocity at the bottom (A) = Va = 25 m/s

Velocity at the top(B) = Vb = 8 m/s

Vertical distance from A to B = diameter of loop, h = 2 x 12 = 24 m

by use of Work energy theorem

Work done by all the forces = change in kinetic energy of the body

Work done by the force + Work done by the friction = Kinetic energy at B - kinetic energy at A

- m x g x h + Work done by friction = 0.5 x 120 x (Vb^2 - Va^2)

- 120 x 9.8 x 24 + Work done by friction = 60 x (64 - 625)

- 28224 + Work done by friction = - 33660

Work done by friction = -33660 + 28224 = - 5436 J

4 0

3 years ago

A 75.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 10.79 de

jeyben [28]

Answer:

$c_{e1}$ = 0.331 J / g ° C

Explanation:

We have a calorimetry exercise where all the heat yielded by one of the components is absorbed by the other.

Heat ceded Qh = m1 ce1 ( $T_{h}$ - $T_{f}$ )

Heat absorbed Qc = m2 ce2 ( $T_{f}$ - T₀)

Body 1 is metal and body 2 is water . Where m are the masses of the two bodies, ce their specific heat and T the temperatures

Qh = Qc

m₁ $c_{e1}$ ( $T_{h}$ - $T_{f}$ ) = m₂ $c_{e2}$ ( $T_{f}$ - T₀)

we clear the specific heat of the metal

$c_{e1}$ = m₂ $c_{e2}$ ( $T_{f}$ - T₀) / (m₁ ( $T_{h}$ - $T_{f}$ ))

$c_{e1}$ = 50.00 4.184 (20.15 -10.79) / (75.00 (99.0-20.15))

$c_{e1}$ = 209.2 (9.36) / (75 78.85)

$c_{e1}$ = 1958.11 / 5913.75

$c_{e1}$ = 0.331 J / g ° C

5 0

4 years ago