All categories

3 years ago

Compare the freezing point of water in the aquanaut’s apartment to its value at the surface. Is it higher, lower, or the same?

Physics

1 answer:

Lelu [443]3 years ago

5 0

Answer:

Freezing Point - Lower

Boiling Point - Higher

Solid- liquid transition line in the phase diagram has a negative slope, but the liquid-gas transition line has a positive slope. Since there is more air pressure at 100m it will take less to freeze the water but more to boil it since it requires a larger temperature under larger pressures

You might be interested in

Calculate the inhomogeneity of a 1.5 t magnet

Ludmilka [50]

Need more answer choices

6 0

3 years ago

you happen to visit the moon when some people on earth see a total solar eclipse. who has a better experience of this event, you

Vinil7 [7]

The friends left on earth because they can see the total eclipse, where as you are on the moon witnessing sections get dark rather than the whole picture

5 0

3 years ago

Chapter 16, Problem 63. A person is standing in a room at 18 ◦C. The exposed surface area and skin temperature of the person are

Juliette [100K]

Answer: $Q=248.011 W$

Explanation:

Given

Temperature of Room $T_{\infty }=18^{\circ}\approx 291 K$

Area of Person $A=1.7 m^2$

Temperature of skin $T=32^{\circ}\approx 305 K$

Heat transfer coefficient $h=5 W/m^2.k$

Emissivity of the skin and clothes $\epsilon =0.9$

$\Delta T=32-18=14^{\circ}$

Total rate of heat transfer=heat Transfer due to Radiation +heat transfer through convection

Heat transfer due radiation $Q_1=\epsilon \sigma A(T^4-T_{\infty }^4 )$

where $\sigma =stefan-boltzman\ constant$

$Q_1=0.9\times 5.687\times 10^{-8}\times 1.7\times (305^4-291^4)$

$Q_1=129.01 W$

Heat Transfer due to convection is given by

$Q_2=hA(\Delta T)$

$Q_2=5\times 1.7\times 14=119 W$

$Q=Q_1+Q_2$

$Q=129.01+119=248.011 W$

7 0

3 years ago

A force can affect the _____ of an object.

BARSIC [14]

Flow. such as running into something or friction

Flow Final Answer

7 0

3 years ago

Read 2 more answers

A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° no

Rzqust [24]

Answer:

a) R = 2.5 mi b) To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

Cos 20 = Aₓ / A

sin 20 = $A_{y}$ / A

Aₓ = A cos 160 = 2 cos 160

$A_{y}$ = A sin160 = 2 sin160

Aₓ = -1,879 mi

$A_{y}$ = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

cos 2600 = Bₓ / B

sin 260 = $B_{y}$ / B

Bₓ = B cos 260

$B_{y}$ = B sin 260

Bₓ = 4 cos 260

$B_{y}$ = 4 sin 260

Bₓ = -0.6946mi

$B_{y}$ = - 3,939 mi

Third vector C = 3 mi to 15 north east

cos 15 = Cₓ / C

sin15 = $C_{y}$ / C

Cₓ = C cos 15

$C_{y}$ = C sin15

Cₓ = 3 cos 15

$C_{y}$ = 3 sin 15

Cₓ = 2,898 mi

$C_{y}$ = 0.7765 mi

Now we can find the final position of the person

X = Aₓ + Bₓ + Cₓ

X = -1.879 -0.6949 + 2.898

X = 0.3241 mi

Y = $A_{y}$ + $B_{y}$ + $C_{y}$

Y = 0.684 - 3.939 +0.7765

Y = -2.4785 mi

a) We use Pythagoras' theorem

R = √ (x2 + y2)

R = √ (0.3241 2 + (-2.4785) 2)

R = 2.4996 mi

R = 2.5 mi

b) let's use trigonometry

Tan θ = y / x

Tanθ = -2.4785 / 0.3241

θ = tan⁻¹ (-7,647)

θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

3 0

3 years ago