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3 years ago

When you step inside a warm ski lodge on a cold day, you find your glasses fog. why does this occur?

1 answer:

5 0

Answer: Condensation of moisture in the air onto the glasses.

Explanation:
When you come from a cold environment into a warm ski lodge, the glasses are at a much lower temperature than that of the lodge.
Because the air in the room contains moisture (water vapor), it condenses on a surface whose temperature is lower than the dew point of the room.

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A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

$P=\rho \times g \times h$

Here

P is the pressure which is given as 68 kPa.
ρ is the density of the oil whose SG is 0.92. It is calculated as

$\rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\$

g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

So the height of column is 7.54m

a-3

By the relation of volume and density

$M=\rho \times V$

Here

ρ is the density of the oil which is 920 kg/m^3
V is the volume of cylinder with diameter 16m calculated as follows

$V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3$

Mass is given as

$M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg$

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

$\Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\$

Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

The original height of column is 5.98m

4 0

3 years ago

A 3,600 kg sphere is located at a distance of 5.2 m from a 1,200 kg sphere.what is the gravitational force between the two spher

storchak [24]

Gravitational push or pull.

5 0

2 years ago

A car rounds a flat curve and experiences a centripetal force directed toward the center of the curve and perpendicular to the d

Usimov [2.4K]

<h2>Answer:</h2>

If a car is rounding a flat curve, it experiences a centripetal force that pulls it towards the center of the circle it is rotating in.

Now,

The centripetal force can be balanced by the centrifugal force caused due to the acceleration of the body at the high speed which counters the centripetal force and in turn prevents the car from slipping down the curve.

So,

If the car doesn't hit the gas then the car will fall down from the curve as the Centripetal force will exceed the Centrifugal force of the car.

However, if the car doesn't hit the brake then the car will maintain it's position on the flat curve track as the centrifugal force will counter the effect of centripetal force directed towards the center.

3 0

2 years ago

Read 2 more answers

A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s

ICE Princess25 [194]

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

5 0

2 years ago

What condition is necessary for an object to make a good reference point?

777dan777 [17]

The object is fixed relative to the motion you are trying to describe.

8 0

3 years ago