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Amanda [17]
1 year ago
9

Why does temperature decrease with higher altitude?multiple choice of:A- air is held strongly together by gravity, so there is l

ess heat transfer B- Air density is lower, so there is less hear transferC- Air density is higher, so there is less hear transferD- Air pressure is higher, so there is less heat transfer.
Physics
1 answer:
stiv31 [10]1 year ago
6 0

ANSWER

B - Air density is lower, so there is less heat transfer

EXPLANATION

We want to identify why temperature decreases with higher altitude.

When radiation is absorbed by the earth's surface, it heats the surroundings and the air through convection and conduction.

However, as we ascend higher into the atmosphere, the amount of solar radiation absorbed by the troposphere's gases decreases significantly.

Due to less heat transfer, air molecules start to move away from each other causing a decrease in air density

Hence, we see that temperature decreases with increasing altitude because there is less heat transfer due to the lower air density.

The answer is option B.

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A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and
cricket20 [7]

Answer:

The frequency f = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = \frac{d}{2} =\frac{44}{2} = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = \frac{d}{2} =\frac{3.0}{2} = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = \frac{d}{2} = \frac{2.0}{2} = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that  the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2

pgy_1=\frac{1}{2}pv^2_2 +pgy_2

v_2=\sqrt{2g(y_1-y_2)}

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = \sqrt{2*9.8*(0.35-0)}

v₂ = \sqrt {6.86}

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

v_3 = \frac{v_2r_2^2}{v_3^2}

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = (2.62  m/s)* (\frac{1.5mm^2}{1.0mm^2} )

v₃ = 0.0589 m/s

∴ velocity  of the fluid in the cylinder =  0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

f=\frac{v_s}{4(h-v_3t)}

where;

v_s = velocity of sound

h = height of the fluid

v₃ = velocity  of the fluid in the cylinder

f=\frac{343}{4(0.40-(0.0589)(0.4)}

f= \frac{343}{0.6576}

f = 521.59 Hz

∴ The frequency f = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})

\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)

\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}

where;

v_s (velocity of sound) = 343 m/s

v₃ (velocity  of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s  when the cylinder has been filling for 4.0 s.

8 0
3 years ago
A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes
OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

y(t) = h +u_y t - \frac{1}{2}gt^2

where

h = 2.5 km = 2500 m is the initial height

u_y = 0 is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

y(t) = 0

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s

So the horizontal distance travelled is

d=v_x t

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

d=(333.3)(22.6)=7533 m = 7.5 km

7 0
3 years ago
You adjust the temperature so that a sound wave travels more quickly through the air. You increase the temperature from 30°C to
earnstyle [38]

The correct answer to the question is : D) 352.6 m/s.

CALCULATION :

As per the question, the temperature is increased from 30 degree celsius to 36 degree celsius.

We are asked to calculate the velocity of sound at 36 degree celsius.

Velocity of sound is dependent on temperature. More is the temperature, more is velocity of sound.

The velocity at this temperature is calculated as -

                            V = 331  + 0.6T m/s

                               = 331 + 0.6 × 36 m/s

                               = 331 + 21.6 m/s

                               = 352.6 m/s.

Here, T denotes the temperature of the surrounding.

Hence, velocity of the sound will be 352.6 m/s.

5 0
3 years ago
Read 2 more answers
A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a
AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

q = -Kb\frac{dh}{dl}

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25  m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

        = 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours

4 0
3 years ago
How do human break N2 into a useable form
Fiesta28 [93]
A process known as fixation<span>. the majority of nitrogen is fixed by </span>bacteria<span>, most of which are </span>symbiotic<span> with plants</span>


<span />
6 0
3 years ago
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