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3 years ago

In what layer does the temperature rise the fastest?

Physics

1 answer:

mihalych1998 [28]3 years ago

4 0

<h2><u>STRATOS</u><u>PHERE</u></h2>

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Because warm air rises and cool air sinks, the troposphere is unstable. In the stratosphere, temperature increases with altitude. The stratosphere contains the ozone layer, which protects the planet from the Sun's harmful UV radiation.

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The colors of the stars in the sky range from red to blue. Assuming that the color indicates the frequency at which the star rad

Crazy boy [7]

Answer:

a) 4458K b) 5048K, c) 6166K, d) 5573K

Explanation:

The temperature of the stars and many very hot objects can be estimated using the Wien displacement law

$\lambda_{max}$ T = 2,898 10⁻³ [m K]

T = 2,898 10⁻³ / $\lambda_{max}$

a) indicate that the wavelength is

Lam = 650 nm (1 m / 109 nm) = 650 10⁻⁹ m

Lam = 6.50 10⁻⁷ m

T = 2,898 10⁻³ / 6.50 10⁻⁷

T = 4,458 10³ K

T = 4458K

b) lam = 570 nm = 5.70 10⁻⁷ m

T = 2,898 10⁻³ / 5.70 10⁻⁷

T = 5084K

c) lam = 470 nm = 4.70 10⁻⁷ m

T = 2,898 10⁻³ / 4.7 10⁻⁷

T = 6166K

d) lam = 520 nm = 5.20 10⁻⁷ m

T = 2,898 10⁻³ / 5.20 10⁻⁷

T = 5573K

5 0

4 years ago

Read 2 more answers

How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s

poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

$W = \Delta KE$

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

Here,

m = mass

$v_{f,i}$ = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2$

$W_1 = 150(m) J$

Second case) When the particle goes from 20m/s to 30m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2$

$W_1 = 250(m) J$

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0

4 years ago

Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject

Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

$t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}$

where

$\mu_{sun}$ is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2. $t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}$

t = 12,105.96 sec

6 0

3 years ago

An 800-kHz radio signal is detected at a point 2.7 km distant from a transmitter tower. The electric field amplitude of the sign

dimaraw [331]

Answer:

Option D is correct: 170 µW/m²

Explanation:

Given that,

Frequency f = 800kHz

Distance d = 2.7km = 2700m

Electric field Eo = 0.36V/m

Intensity of radio signal

The intensity of radial signal is given as

I = c•εo•Eo²/2

Where c is speed of light

c = 3×10^8m/s

εo = 8.85 × 10^-12 C²/Nm²

I = 3×10^8 × 8.85×10^-12 × 0.36²/2

I = 1.72 × 10^-4W/m²

I = 172 × 10^-6 W/m²

I = 172 µW/m²

Then, the intensity of the radio wave at that point is approximately 170 µW/m²

7 0

3 years ago

How did the social structure of ancient Rome affect the lives of its people?

Neporo4naja [7]

Answer:

Explanation:

5 0

3 years ago