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2 years ago

Planet with an atmosphere that rains sulfuric acid

Physics

1 answer:

Paladinen [302]2 years ago

3 0

Answer:

Venus

Explanation:

Venus is the second plate in the solar system. It is a terrestrial planet and it is part of the inner rocky planets.

In Venus, it rains sulfuric acid but the rain never reaches the surface before it becomes evaporated. The acid forms from the combination of sulfur oxide and water in the atmosphere at a height of about 42km. As it condenses and falls, it becomes evaporated back at lower elevations. The surface is therefore protected from the sulfuric acid rain.

The sulfur oxide and water vapor must have been derived from volcanic activities in geologic times past.

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Essay about why people should not join a gang 300 word

Salsk061 [2.6K]

Answer:

I don't know if it's fair for me to write an entire essay for you

but if you would like I can list some reasons you can incorporate into an essay.

-------------------------------------------------------------------------------------------------------

-If you are joining a gang you are most likely going to have a greater chance of being targeted, even if it seems like you would be protected.

-Most of the time when someone joins a gang a child is being separated from a parent and will face most aggression.

-This can also cause academic failure in students, from lack of education and probably a pretty bad background.

-For teens when joining a gang they will do things that aren't suited for their age (I would list things but I'm not trying to get myself reported if you know what I mean)

-You will also have a great risk of being imprisoned and facing bad consequences, and these consequences can even be having the death penalty.

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I hope this is good enough, sorry I didn't give you a full essay. I think this should help with that though.

6 0

1 year ago

PLEASE HELPPPPPPPPPP

tekilochka [14]

Answer:

sorry me computer is blocking the picture but i will try to unblock so i can answer

Explanation:

7 0

2 years ago

Read 2 more answers

A diffraction grating is placed 1.00 m from a viewing screen. Light from a hydrogen lamp goes through the grating. A hydrogen sp

Colt1911 [192]

Answer:

λ = 396.7 nm

Explanation:

For this exercise we use the diffraction ratio of a grating

d sin θ = m λ

in general the networks works in the first order m = 1

we can use trigonometry, remembering that in diffraction experiments the angles are small

tan θ = y / L

tan θ = $\frac{sin \theta}{cos \theta}$ = sin θ

sin θ = y / L

we substitute

$d \ \frac{y}{L}$ = m λ

with the initial data we look for the distance between the lines

d = $\frac{m \lambda \ L}{y}$

d = 1 656 10⁻⁹ 1.00 / 0.600

d = 1.09 10⁻⁶ m

for the unknown lamp we look for the wavelength

λ = d y / L m

λ = 1.09 10⁻⁶ 0.364 / 1.00 1

λ = 3.9676 10⁻⁷ m

λ = 3.967 10⁻⁷ m

we reduce nm

λ = 396.7 nm

8 0

2 years ago

If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind th

Sedbober [7]

Answer:

c = 4,444.44

Explanation:

You have the following expression for the acceleration of the projectile:

$a=cs$ (1)

s: distance to the ground of the projectile

To find the value of the constant c you use the following formula:

$v^2=v_o^2+2a \Delta s$ (2)

vo: initial velocity = 0 m/s

v: final speed = 200 m/s

Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m

You replace the expression (1) into the expression (2):

$v^2=2(cs)\Delta s$

You do the constant c in the last equation, then you replace the values of v, s and Δs:

$c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44$

6 0

2 years ago

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

2 years ago