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Nady [450]
3 years ago
10

what is the orbital speed for a satellite 3.5 x 10^8m from the center of mars? Mars mass is 6.4 x 10^23 kg

Physics
1 answer:
adell [148]3 years ago
6 0

Answer:

v = 349.23 m/s

Explanation:

It is required to find the orbital speed for a satellite 3.5\times 10^8\ m from the center of mass.

Mass of Mars, M=6.4\times 10^{23}\ kg

The orbital speed for a satellite is given by the formula as follows :

v=\sqrt{\dfrac{GM}{r}} \\\\v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6.4\times 10^{23}}{3.5\times 10^8}} \\\\v=349.23\ m/s

So, the orbital speed for a satellite is 349.23 m/s.

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tester [92]

Answer:

Im gonna say it is answer A:) Hope this helps!

Explanation:

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3 years ago
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What is meant by global warming?
pychu [463]

The global warming is the uniform heating of the globe due to the effect of greenhouse gases.

Answer: Option B

<u>Explanation:</u>

Global warming is thus the theory that says that earth is warming up with an increase in temperature over time due to increment in the greenhouse gases.

These are greenhouse gases that are getting increased due to the human activity or say over exploitation of the natural resources. The heating of the globe has caused ozone layer depletion which is also causing warming.

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3 years ago
List these things in order of size, starting with the smallest thing:
jek_recluse [69]
Moon, earth, sun solar system, galaxy, universe
4 0
3 years ago
The block slides on a horizontal frictionless surface. The block has a mass of 1.0kg and is pushed 5.0N at 45°. What is the magn
Firlakuza [10]

Answer:

3.5\:\mathrm{m/s^2}

Explanation:

Newton's 2nd law is given as \Sigma F = ma.

To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.

Using trigonometry to find the horizontal component of the force:

\cos 45^{\circ}=\frac{x}{5},\\\frac{\sqrt{2}}{{2}}=\frac{x}{5},\\x=\frac{5\sqrt{2}}{2}

Use this horizontal component of the force to solve for for the acceleration of the object:

\frac{5\sqrt{2}}{2}=1.0\cdot a,\\a=\frac{5\sqrt{2}}{2}\approx \boxed{3.5\:\mathrm{m/s^2}}

8 0
2 years ago
39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
3 years ago
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