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sergij07 [2.7K]
3 years ago
13

The atmosphere of Mercury and Mars are very thin. What effect does the thin atmosphere have on the temperature on the surface of

these planets?
Physics
1 answer:
KengaRu [80]3 years ago
3 0

Answer:

Very hot during the day and very cold at night.

Explanation:

Due to the thin atmosphere, they have very hot climate during the day time and very cold climate at night. This happens because they contain very low amounts of greenhouse gases. These gases retain the heat at night. The atmosphere also prevents excessive light and UV rays from entering. The thin  atmosphere leads to many asteroids and comets hitting the surface of the planet. On earth, these asteroids usually, burn up in the mesosphere layer of the atmosphere. These asteroid collisions cause massive fires. This in turn,  causes the temperature to increase during the day. During the night time, massive fires cannot burn due to the low temperature because of the lack of greenhouse gases.

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slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa
xenn [34]

Answer:

Maximum height reached by the rocket is

y_{max} = 308 m

total time of the motion of rocket is given as

T = 16.44 s

Explanation:

Initial speed of the rocket is given as

v_i = 50 m/s

acceleration of the rocket is given as

a = 2 m/s^2

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

v_f^2 - v_i^2 = 2 a d

v_f^2 - 50^2 = 2(2)(150)

v_f = 55.68 m/s

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

v_f^2 - v_i^2 = 2 a d

0 - 55.68^2 = 2(-9.81)(y - 150)

y_{max} = 308 m

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

v_f - v_i = at

55.68 - 50 = 2 t

t_1 = 2.84 s

2) time to reach ground from this height

\Delta y = v_y t + \frac{1}{2}gt^2

-150 = 55.68 t - \frac{1}{2}(9.81) t^2

t_2 = 13.6 s

so total time of the motion of rocket is given as

T = 13.6 + 2.84 = 16.44 s

3 0
3 years ago
Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0
Alisiya [41]

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

           

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

            x ’= 368.61 m

            y ’= 450 sin 35

            y ’= 258.11 m

4 0
3 years ago
A temperature of 200°F is equivalent to approximately A. 840°C. B. 93.3°C. C. 232°C. D. 37.8°C
Eduardwww [97]
The equivalent of 200F is 93.3C

5 0
3 years ago
Read 2 more answers
Which structures protect the cell? Check all that apply.
MAXImum [283]

Answer:

cell wall and cell membrane

8 0
3 years ago
Read 2 more answers
Radioactive plutonium−239 (t1/2 = 2.44 × 105 yr) is used in nuclear reactors and atomic bombs. If there are 3.50 × 102 g of the
madam [21]

Answer:

t = 4.41 10⁻⁴ years

Explanation:

For this exercise we must use the concept of average life time, which is the time in which the quantity and substance decays in half

    T_{1/2}  = ln2 / λ

Let's calculate the decay constant of plutonium

    λ  = ln2 /  T_{1/2}

    λ  = ln 2 / 2.44 10⁵  

    λ  = 2.84 10⁻⁶  s⁻¹

Radioactive decay is a first order process

     N = No e (-λ   t)

Where N is the number of nuclei, the mass is this by molecular weight

    m = N PM

    m / PM = m₀ / PM e (- λ  t)

    m / m₀ = e (- λ  t)

    -λ  t = ln (m / m₀)

     t = -1 /λ   ln (m/m₀)

     t = - 1 / 2.84 10⁻⁶   ln (0.1 / 0.35)

     t = 4.41 10⁻⁴ years

7 0
3 years ago
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