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3 years ago

When the temperature of air rises, the amount of water needed for saturation

Physics

2 answers:

Oliga [24]3 years ago

7 0

It Increases. I just took a quiz with the same question.

Lemur [1.5K]3 years ago

6 0

<span>When the temperature of air rises, the amount of water needed for saturation increases. </span>

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reviews the approach taken in problems such as this one. A bird watcher meanders through the woods, walking 0.916 km due east, 0

Verizon [17]

Answer:

Displacement: 2.230 km Average velocity: 1.274 $\frac{km}{h}$

Explanation:

Let's represent displacement by the letter S and the displacement in direction 49.7° as A. Displaement is a vector, so we need to decompose all the bird's displacement into their X-Y compoments. Let's go one by one:

0.916 km due east is an horizontal direction and cane be seen as direction towards the negative side of X-axis.

0.928 km due south is a vertical direction and can be seen as a direction towards the negative side of Y-axis.
3.52 km in a direction of 49.7° has components on X and Y axes. It is necessary to break it down using trigonometry,

First of all. We need to sum all the X components and all the Y componets.

∑ $Sx = Ax -0.916$ ⇒ ∑ $Sx = [tex]3.52cos(49.7) - 0.916$

∑ $Sx = 1.361 km$

∑ $Sy = Ay - 0.918$ ⇒ ∑ $Sy = 3.52sin(49.7) - 0.918$

∑ $Sy = 1.767$

The total displacement is calculated using Pythagoeran therorem:

$S_{total} =\sqrt{Sx^{2}+ Sy^{2} }$ ⇒

$S_{total} = 2.230 km$

With displacement calculated, we can find the average speed as follows:

$V = S/t$ ⇒ $V = \frac{2.230}{1.750}$

$V = 1.274\frac{km}{h}$

7 0

3 years ago

g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn

krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

T² = ( $\frac{4\pi }{G M_s}$ ) r³ (1)

in this case the period of the season is

T₁ = 93 min (60 s / 1 min) = 5580 s

r₁ = 410 + 6370 = 6780 km

r₁ = 6.780 10⁶ m

for the satellite

T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

T² = K r³

K = T₁²/r₁³

K = $\frac{ 5580^2}{ (6.780 10^6)^2}$

K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

r³ = T² / K

r³ = 86400² / 9.99 10⁻¹⁴

r = ∛(7.4724 10²²)

r = 4.21 10⁷ m

this distance is from the center of the earth

7 0

3 years ago

A land breeze forms when:

victus00 [196]

Explanation:

Its D. The warm air from the land moves towards the water

7 0

3 years ago

1. Mr. Ure has a mass of 65 kg, due to the fact that he is WAY too skinny! What is the force of Earth's gravity on him?

shepuryov [24]

m = mass of Mr. Ure = 65 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Ure is given as

F = mg

F = 65 x 9.8

F = 637 N

F = force of gravity on car = 3050 N

m = mass of the car = ?

g = acceleration due to gravity = 9.8 m/s²

force of gravity on car is given as

F = mg

3050 = m (9.8)

m = 3050/9.8

m = 311.22 kg

m = mass of Mr. Rees = 90 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Rees is given as

F = mg

F = 90 x 9.8

F = 882 N

7 0

3 years ago

On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the

IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = $\frac{1}{2}\varepsilon _{0}E^{2}$

$=\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119$

= 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = $\frac{B^{2}}{2\mu _{0}}$

$=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}$

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

8 0

3 years ago