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2 years ago

Why do areas with low altitude have warmer air than areas with high altitude

Physics

1 answer:

barxatty [35]2 years ago

6 0

Answer:

Air at higher altitude is under less pressure than air at lower altitude because there is less weight of air above it, so it expands (and cools), while air at lower altitude is under more pressure so it contracts (and heats up).

Explanation:

Hope that helped

You might be interested in

Which symbol represents a type of electromagnetic radiation released during radioactive decay?

Lena [83]

The correct option is (D) Gamma ( $_0^0\gamma$ )

Explanation:
Now this question is a tricky one because all of these options are somehow involved in radioactive decay; however, in this case the SYMBOL is required NOT the elements. There are three symbols involved in radioactive decay, which are:
1. α for alpha decay
2. β for beta decay
3. γ for gamma decay

In the options only one symbol is present which is gamma. Hence option (D) Gamma ( $_0^0\gamma$ ) is the correct answer.

-i

4 0

3 years ago

You are a member of an alpine rescue team and must get a box of supplies, with mass 3.00 kg , up an incline of constant slope an

Sever21 [200]

Answer:

v = 8.45 m/s

Explanation:

given,

mass = 3 kg

angle = 30.0°

vertical distance = 3.3 m

μ = 0.06

according to conservation of energy

KE(loss) = PE(gain) + Work done (against\ friction)..............(1)

frictional Force

$F_f = \mu N$

$F_f = \mu m g cos \theta$

work against friction

W = F d

$W = \mu m g cos \theta \times h l sin\theta$

$W = \dfrac{\mu m g \times h}{tan\theta}$

Potential energy

PE = mgh

$\dfrac{1}{2}mv^2= \dfrac{\mu mgh}{tan \theta}+ mgh$

$\dfrac{1}{2}v^2= \dfrac{0.06 \times 9.81 \times 3.3}{tan 30^0}+ 9.8\times 3.3$

v = 8.45 m/s

the minimum speed is equal to 8.45 m/s

6 0

3 years ago

What is the stopping distance if the car is initially traveling at speed 6.0v? assume that the acceleration due to the braking i

Sunny_sXe [5.5K]

The distance for any rectilinear motion at constant acceleration is:

d = v₀t + 0.5at²
where
v₀ is the initial velocity

So, if v₀ = 6v, and it stopped to 0 m/s, then the acceleration is equal to:
a = (0 - 6v)/t = -6v/t
Thus,
d = (6v)(t) + (0.5)(-6v/t)(t²)
d = 6vt - 3t
d = 3t(2v - 1)

4 0

3 years ago

Three points A, B, and C of unknown charges are at the corners of an equilateral triangle.

nasty-shy [4]

Answer: option d.

$Q_A= \frac{1}{2} Q_B=- \frac{1}{3} Q_C$

Explanation:

1) The direction of the field lines inform about the sign of the charges.

The field lines extend from the positive charges to the negative charges, so you can conclude that the charge C is positve and both charge A and charge B are negative:


Charge C: positive

Charge A: negative

Charge B: netative


2) The density of the lines (number of lines in a region) inform about the magnitude of the electric field.

Since the charges are at the same distance, the magnitude of the electric field informs directly about the magnitude of the force and that about the magnitude of the charges.

Since, there are the double of lines between C and B than between C and A, the magnitude of charge B is the double than the magnitud of charge A.

From the five options given (a throug e) the only that is consistent with that charges A and B have the same sign, that charge C has different sign, and that charge B is the double of charge A is:

$Q_A= \frac{1}{2} Q_B=- \frac{1}{3} Q_C$

3 0

3 years ago

List of most launched orbital launch vehicles

weqwewe [10]

Falcon 9PegasusProtonSatellite Launch VehicleAntaresZenitAtlas VSoyuzAriane 5Delta IVPolar Satellite Launch VehicleGeosynchronous Satellite Launch VehicleVegaRokotAngaraSoyuz-2ShavitH-IIBDneprSafirFalcon 1Long March 4BLong March 2DDelta IIH-IILong March 3CSoyuz-ULong March 4CBlack ArrowLong March 3ANaro-1Zenit-3SLSputnikLong March 2FVLS-1VulcanLong March 2CAthenaLong March 3BJuno IN1Ariane 6KuaizhouZenit-2SimorghLong March 11StrelaMinotaur-CAriane 1Augmented Satellite Launch VehicleEuropa

4 0

3 years ago