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Angelina_Jolie [31]
3 years ago
5

Does anyone want to join an LGBTQ squad?

Physics
1 answer:
Korvikt [17]3 years ago
6 0
Mooooooooooooooooooo
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Jj thomas' model contained electrons. 

The current model contains a nucleus with electrons orbiting around it. :)
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Identify Label these cell parts
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If many household substances classify as acids or bases what is the ph scale
bekas [8.4K]

Answer: Some household items that contain acids include: yogurt, vinegar,lemon juice, citric acid, apples, jelly, pineapples, cranberry sauce, milk, and batteries and In chemistry, pH is a scale used to specify how acidic or basic a water-based solution is. Acidic solutions have a lower pH, while basic solutions have a higher pH. At room temperature, pure water is neither acidic nor basic and has a pH of 7.

4 0
3 years ago
What linear speed must an earth satellite have to be in a circular orbit at an altitude of 159 km?
IgorLugansk [536]
Gravitational acceleration, g = GM/r^2. Additionally, for a satellite in a circular orbit, g = v^2/r

Where, G = Gravitational constant, M = Mass of earth, r = distance from the center of the earth to the satellite, v = linear speed of the satellite.

Equating the two expressions;
v^2/r = GM/r^2
v = Sqrt (GM/r);
But GM = Constant = 398600.5 km^3/sec^2
r = Altitude+Radius of the earth = 159+6371 = 6530 km

Substituting;
v = Sqrt (398600.5/6530) = 7.81 km/sec = 781 m/s
8 0
4 years ago
A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee
Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2}  = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\

where,

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Me = Mass of Earth = 5.97 x 10²⁴ kg

r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m

v = orbital speed = ?

Therefore,

v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\

<u>v = 7.69 x 10³ m/s</u>

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v =  \frac{2\pi r}{T}\\\\

where,

T = Time Period of Satellite = ?

Therefore,

T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\

<u>T = 5500 s = 91.67 min = 1.53 h</u>

7 0
3 years ago
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