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2 years ago

Why are the layers of the atmospher in the order that they are in?

Physics

1 answer:

aliya0001 [1]2 years ago

4 0

Answer:

TROPOSPHERE the lowest layer of the atmosphere closest to the earth

STRATOSPHERE constant temperature

MESOSPHERE layer with the strongest wind and below the themosphere

Themosphere absorbs UV radiation

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An object's describes it's distance and direction from a reference point .

Ainat [17]

6: velocity i hpe this is what your looking for

4 0

2 years ago

Read 2 more answers

How do conservation tillage practices lead to agricultural sustainability?

Zielflug [23.3K]

Conservation tillage practices help reduce soil erosion and maintain soil nutrient levels.

Explanation:

The approach that helps in the reduction of doing tillage practices and also reducing its frequency. this is done for obtaining certain benefits for both environment and economic. This mainly focuses on providing sustainability by leaving some plants remaining in the soil.

It aims in decreasing the emission of gases of greenhouse effects like carbon dioxide. Using these practices helps in reducing the erosion and runoffs. This will promote health of the soil because the nutrients are not take off form the soil due to soil erosion and runoffs.

8 0

3 years ago

A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the

zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

$E= \frac{kqr}{R^3}$

The potential at a distance can be represented as:

V(r) - V(0) = $-\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

V(r) = $-[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

= 0.56 × 10⁻² m

R = 1.29 cm

= 1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

$V(r)$ $= -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}$

$V(r)$ = -2.15 × 10⁻⁴ V

The difference between the radial distance and center can be expressed as:

V(r) - V(0) = $-\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]^R$

V(r) = $-\frac{qR^2}{8 \pi E_0R^3 }$

V(r) = $-\frac{q}{8 \pi E_0R }$

V(r) $= -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}$

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0

2 years ago

A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5\times 10^{-

joja [24]

Answer:

$5.25\cdot 10^{40} kg m^2/s$

Explanation:

The angular momentum of the pulsar is given by:

$L=m\omega r^2$

where

$m=2.8\cdot 10^{30} kg$ is the mass of the pulsar

$r = 10.0 km = 1\cdot 10^4 m$ is the radius

$\omega$ is the angular speed

Given the period of the pulsar, $T=33.5\cdot 10^{-3} s$ , the angular speed is given by

$\omega=\frac{2\pi}{T}=\frac{2 \pi}{33.5\cdot 10^{-3}s}=187.5 rad/s$

And so, the angular momentum is

$L=m\omega r^2=(2.8\cdot 10^{30}kg)(187.5 rad/s)(1\cdot 10^4 m)^2=5.25\cdot 10^{40} kg m^2/s$

8 0

2 years ago

A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during

Katarina [22]

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:

From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.

We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

The speed increases from 0 m/s to 1341 m/s.
The time elpased is 2.0 min.
1 min = 60 s.

The acceleration of the ship during the first 2.0 minutes is:

$a = \frac{\Delta v }{t} ) \frac{(1341m/s-0m/s)}{2.0min} \times \frac{1min}{60s} = 11 m/s^{2}$

Learn more: brainly.com/question/16274121

3 0

2 years ago