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joja [24]
2 years ago
12

How does the atmospheric temperature and size of inner planets compare with those of outer planets?

Physics
1 answer:
amid [387]2 years ago
6 0
The inner planets are rocky and have diameters of less than 13,000 kilometers. The outer planets include Jupiter, Saturn, Uranus, and Neptune. The smaller, inner planets include Mercury, Venus, Earth, and Mars. Inner planet's atmosphere is thin. (Mercury has no atmosphere). Outer Planets: Outer planets' atmosphere is very thick. The four inner planets, Mercury, Venus, Earth, and Mars, are warmer than the outer gas giants. However, the temperature of the planets does not follow a linear path from the Sun.

Hope this helps!
Please give Brainliest!
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Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t
Step2247 [10]

Answer:

\Delta p = 90.7 kPa

Explanation:

specific gravity of oil is = \frac{\rho_{oil}}{\rho_w}

\rho_{oil} = 0.85*1000 = 850 kg/m3

we know that

change in pressure  for oil is given as

\Delta p = \rho gh

here density and h is for oil

\Delta p = 850*5 *9.81 = 41,692.5 kPa

change in pressure  for WATER is given as

\Delta p = \rho gh

here density is for water and h is for water

\Delta p = 1000*5 *9.81 = 49,050 kPa

pressure change due to both is given as

\Delta p = 41692.3 + 49050 = 90742.5 N/m2

\Delta p = 90.7 kPa

8 0
3 years ago
Read 2 more answers
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Alika [10]
The dot would be the earthquake epicenter. That is where the earthquake originated.
6 0
3 years ago
What is the GPE of a 15,000 kg airplane sitting on the ground?
Vesna [10]

Answer:

C, it is not moving

it has no potential

7 0
2 years ago
A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant bef
inessss [21]

Answer:

29.396988 m/s

Explanation:

Really, it depends on where the child is when he drops the ball - e.g., which planet he is on, and his distance from the center of that planet.

I'll assume that the child is on Earth at sea level at the equator, so that his distance from the geocenter is 6378000 meters.

The acceleration, g, is found from

g = GM/r²

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

M = 5.9724e+24 kg

r = 6.378e+6 m

g = 9.799086 m sec⁻²

An approximate answer is found from an equation from constant acceleration kinematics:

v = gt

t = 3.0 sec

v = 29.397259 m/s

Now, the above method is an approximation that makes the technically incorrect assumption that the acceleration of gravity is a constant throughout the entire fall. You get away with it because the drop is very short. In another situation, it might not be. So it would be nice to develop a more accurate method that does not assume constant gravitational acceleration. For that, we begin with the Vis Viva equation:

v = √[GM(2/r − 1/a)]

Here,

a = the semimajor axis of a plunge orbit, which is equal to half of the apoapsis distance of 6378000+h, where

h = the altitude from which the ball is dropped

We can (using some math) develop the following equation:

t − t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

t − t₀ = 3 sec

r = 6378000 meters

d = r + h

Using an iterative method (e.g. Newton's or Danby's), we can determine that the altitude,

h = 44.0954 meters

So,

d = 6378044.09538 meters

a = d/2 = 3189022.04769 meters

Now we can calculate that

v = 29.396988 m/s

This is the more nearly correct answer because it takes into account the variability of the gravitational acceleration during the fall.

5 0
3 years ago
A house has well-insulated walls. It contains a volume of 105 m3 of air at 305 K.
REY [17]

Answer: 85.46\ kJ

Explanation:

Given

Volume of air V=105\ m^3

Temperature of air T=305\ K

Increase in temperature \Delta T=0.7^{\circ}C

Specific heat for diatomic gas is C_p=\dfrac{7R}{2}

Energy required to increase the temperature is

\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}]

Insert the values

\Rightarrow Q=\dfrac{7}{2}\times \dfrac{1.01325\times 10^5\times 105}{305}\times 0.7\\ \text{Assuming air pressure to be atmospheric P=}1.01325\times 10^5\ N/m^2\\\\\Rightarrow Q=0.8546\times 10^5\\\Rightarrow Q=85.46\ kJ

6 0
3 years ago
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