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2 years ago

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The surface pressures at the bases of warm and cold columns of air are equal. air pressure in the warm column of air will ______

with increasing height __________ than in the cold column.

1 answer:

EastWind [94]2 years ago

3 0

In atmospheric science, surface pressure<span> is the atmospheric </span>pressure<span> at a location on Earth's </span>surface<span>. It is directly proportional to the mass of air over that location. For numerical reasons, atmospheric models such as general circulation models (GCMs) usually predict the nondimensional logarithm of </span>surface pressure<span>.

The answer is decrease more slowly

</span>

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A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate

liq [111]

Answer:

3 cm

Explanation:

According to the question,

$D=1.5 m$ .

$d=9 cm$ .

$\lambda =0.9 mm$ .

Now the approximate slit's image width is equal to width of central maxima.

And width of central maxima is twice the width from center to first maxima

So,

$y=2\frac{\lambda (D)}{d}$ .

Substitute all the variable in above equation.

$y=\frac{(2)0.9\times 10^{-2} m(1.5 m) }{0.09 m}$ .

$y=3 cm$ .

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2 years ago

he first excited state of the helium atom lies at an energy 19.82 eV above the ground state. If this excited state is three-fold

bekas [8.4K]

Answer:

Relative population is 2.94 x 10⁻¹⁰.

Explanation:

Let N₁ and N₂ be the number of atoms at ground and first excited state of helium respectively and E₁ and E₂ be the ground and first excited state energy of helium respectively.

The ratio of population of atoms as a function of energy and temperature is known as Boltzmann Equation. The equation is:

$\frac{N_{1} }{N_{2} }$ = $\frac{g_{1}e^{\frac{-E_{1} }{KT} } }{g_{2}e^{\frac{-E_{2} }{KT} }}$

$\frac{N_{1} }{N_{2} }$ = $\frac{g_{1}e^{\frac{-(E_{1}-E_{2}) }{KT} } }{g_{2}}$

Here g₁ and g₂ be the degeneracy at two levels, K is Boltzmann constant and T is equilibrium temperature.

Put 1 for g₁, 3 for g₂, -19.82 ev for (E₁ - E₂) and 8.6x10⁵ ev/K for K and 10000 k for T in the above equation.

$\frac{N_{1} }{N_{2} }$ = $\frac{1\times e^{\frac{-(-19.82)}{8.6\times 10^{-5}\times 10000} } }{3}$

$\frac{N_{1} }{N_{2} }$ = 3.4 x 10⁹

$\frac{N_{2} }{N_{1} }$ = 2.94 x 10⁻¹⁰

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2 years ago

A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 10 m/s. It is hit straight back a

Maru [420]

Answer:

-5.1 kg m/s

Explanation:

Impulse is the change in momentum.

Change in momentum= final momentum - initial momentum=m $v_{2}$ +m $v_{1}$

Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)

Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)

4 0

2 years ago

A 2600-m-high mountain is located on the equator. how much faster does a climber on top of the mountain move than a surfer at a

puteri [66]

The climber move 0.19 m/s faster than surfer on the nearby beach.

Since both the person are on the earth, and moves with the constant angular velocity of earth, however there linear velocity is different.

Number of seconds in a day, t=24*60*60=86400 sec

The linear speed on the beach is calculated as

V1= $\frac{2πr}{t}$

Here, t is the time

Plugging the values in the above equation

V1= $\frac{2π*6.4*10^6}{86400}$ =465.421 m/s

Velocity on the mountain is calculated as

V2= $\frac{2π(r+h)}{t}$

Plugging the values in the above equation

V2= $\frac{2π(6.4*10^6+2600}{86400}$ =465.61 m/s

Therefore person on the mountain moves faster than the person on the beach by 465.61-465.421=0.19 m/s

5 0

2 years ago

Marcus can drive his boat 24 miles down the river in 2 hours but takes 3 hours to return upstream. Find the rate of the boat in

notsponge [240]

Answer:

speed of boat as

$v_b = 10 mph$

river speed is given as

$v_r = 2 mph$

Explanation:

When boat is moving down stream then in that case net resultant speed of the boat is given as

since the boat and river is in same direction so we will have

$v_1 = v_r + v_b$

Now when boat moves upstream then in that case the net speed of the boat is opposite to the speed of the river

so here we have

$v_2 = v_b - v_r$

as we know when boat is in downstream then in that case it covers 24 miles in 2 hours

$v_1 = \frac{24}{2} = 12 mph$

also when it moves in upstream then it covers same distance in 3 hours of time

$v_2 = \frac{24}{3} = 8 mph$

$v_b + v_r = 12 mph$

$v_b - v_r = 8 mph$

so we have speed of boat as

$v_b = 10 mph$

river speed is given as

$v_r = 2 mph$

8 0

2 years ago

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