All categories

3 years ago

Explain how climbing a mountain is similar to hiking from the equator to one of the poles

Physics

2 answers:

4vir4ik [10]3 years ago

7 0

Answer:

Explanation:

SRRY God bless

spin [16.1K]3 years ago

3 0

Climbing an mountain and hiking are similar because both includes walking and climbing and they both are like the same thing

You might be interested in

According to Einsteins explanation, what causes gravitation?

olga2289 [7]

<span>Answer: Gravity is most accurately described by the general theory of relativity (proposed by Albert Einstein in 1915) which describes gravity not as a force, but as a consequence of the curvature of spacetime caused by the uneven distribution of mass.</span>

4 0

4 years ago

Describe how the boiling point of water on top of a mountain would be different from its boiling point at sea level.

Sonbull [250]

Answer:

At elevated altitudes, any cooking that involves boiling or steaming generally requires compensation for lower temperatures because the boiling point of water is lower at higher altitudes due to the decreased atmospheric pressure.

Explanation:

7 0

3 years ago

The phase change in which a substance changes from a gas directly to a solid is

vlabodo [156]

<span>Condensation is the change of the substance from liquid to solid phase. Example of this is the formation of ice. Vaporization is the change of substance from liquid to gas phase. Example of this is the boiling of water. Deposition is the change of a substance from gas to solid phase. Example of this is the formation of ice on a winter day. Sublimation is the change of a substance from solid to gas phase. Example of this is dry ice. The answer is letter C.</span>

7 0

3 years ago

Read 2 more answers

An electric furnace runs 13 hours a day to heat a house during January (31 days). The heating element has a resistance of 7.2 an

liq [111]

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = $\frac{I^2}{R}$ ........1

put here value we get

Power consumed = $\frac{16.7^2}{7.2}$

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

Power consumed by furnace in one hour is = Power consumed × 3600

Power consumed by furnace in one hour is = 38.7347 × 3600

Power consumed by furnace in one hour is 139.445kWh

and

Power consumed by furnace in the month of January is

Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

Power consumed by furnace in the month of January = 56196.335 kWh

cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

4 0

3 years ago

Read 2 more answers

Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What p

Temka [501]

Answer:

The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

Explanation:

From General gas equation.

PV = nRT............................... Equation 1

Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.

n = mass/molar mass .................. Equation 2

substituting equation 2 into equation 1.

PV = (mass/molar mass)RT

⇒ Mass/molar mass = PV/RT..................... Equation 3

But mass = Density × Volume

⇒ M = D × V.................... Equation 4

Where D = density, M = mass

Substituting equation 4 into equation 3

DV/molar mass = PV/RT............ Equation 5

Dividing both side of the equation by Volume (V) in Equation 5

D/molar mass = P/RT .............. Equation 6

Cross multiplying equation 6

D × RT = P × molar mass

∴ Molar mass = (D × RT)/P.................. Equation 7

Where D = 0.518 g/L , R = 0.0821 atm dm³/K.mol,

T = 25°C = 25 + 273 = 298 K,

P =721 mmHg = (721/760) atm= 0.949 atm

Substituting these values into equation 7

Molar mass = (0.518 × 0.0821 × 298)/0.949

Molar mass = 13.35 g/mole

The molar mass of the mixture is =13.35 g/mole

Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.

∴ 13.35 = 4(y) + 32(1-y)

13.35 = 4y + 32 - 32y

Collecting like terms in the equation,

32y - 4y = 32 - 13.35

28y = 18.65

y = 18.65/28

y =0.666

y = 0.666 × 100 = 66.6%

∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

6 0

3 years ago