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Anarel [89]
3 years ago
6

Given: Saturated air changes temperature by 0.5°C/100 m. The air is completely saturated at the dew point. The dew point has bee

n reached, and the temperature is 2°C. The air continues to rise another 1,400 m. What is the final temperature?
Physics
1 answer:
erma4kov [3.2K]3 years ago
3 0

Answer:

Explanation:

Given

saturated air temperature by 0.5^{\circ}C/100 m

Dew point temperature is given by t=2^{\circ}C

Dew point is defined as the temperature after which air no longer to uphold the water vapor fuse with it and some water vapor may condense to a liquid.

air continues to rise for 1400 m

i.e. change in temperature would be \Delta t =\frac{0.5}{100}\times 1400=7^{\circ}C

Final temperature t_f

t_f+\Delta t=t

t_f=2-7=-5^{\circ}C

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Proablyly a or c idk 
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Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra
Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

s^2 = (x_A - x_B)^2+(y_A - y_B)^2  (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

s^2 = x_B^2+y_A^2 (2)

Taking the differential with respect to time:

\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}  (3)

where \displaystyle{\frac{dx_B}{dt}}=v_B and \displaystyle{\frac{dx_A}{dt}}=v_A are the respective given velocities of the boats. To find s and x_B we make use of the given position for A, y_A=30, the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h

with this time, we know can now calculate the distance at which B is:

\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi

and applying Pythagoras:

\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}

Now substituting all the values in (3) and solving for  \displaystyle{\frac{ds}{dt} } we get:

\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}

4 0
3 years ago
A scientist measures the growth of a bamboo plant over time. The table above shows the results. What is the best interference fo
lisabon 2012 [21]

(B) 2.25cm

<u>Explanation:</u>

Given:

At 40 hours, the height of the bamboo plant is 2.1cm

At 50 hours, the height of the bamboo plant is 2.4cm

Height of the bamboo plant after 45 hours = ?

The difference in length from 40 to 50 hours = 2.4 - 2.1cm

                                                                      = 0.3 cm

Mean of 40 and 50 is 45.

Thus,

At 45 hours, the height will increase by 0.3/2

                                                         = 0.15 cm

Height at 45 hour = 2.1 + 0.15cm

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7 0
3 years ago
Mathphys Help help help
Keith_Richards [23]

Answer:

1030 mph

Explanation:

The new velocity equals the initial velocity plus the wind velocity.

First, in the x (east) direction:

vₓ = 335 mph + 711 cos 19° mph

vₓ = 1007 mph

And in the y (north) direction:

vᵧ = 0 mph + 711 sin 19° mph

vᵧ = 231 mph

The net speed can be found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1007 mph)² + (231 mph)²

v ≈ 1030 mph

6 0
3 years ago
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mojhsa [17]

Answer:

6.38 x 10^4 J

Explanation:

d = 0.33 cm = 0.33 x 10^-2 m, Area = 87 x 36 cm^2 = 0.87 x 0.36 m^2

ΔT = 14 degree C, t = 1 min = 60 second

K = 0.8 W / m K

Heat  = K A ΔT t / d

H = 0.8 x 0.87 x 0.36 x 14 x 60 / (0.33 x 10^-2)

H = 6.38 x 10^4 J

7 0
3 years ago
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