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steposvetlana [31]
2 years ago
14

Can u help me. thank you​

Physics
2 answers:
katrin2010 [14]2 years ago
6 0
I can give you a search engine that could help you with all ir hw its called socratic it uses everything on the internet to search for answers it’s literally a search engine
Gemiola [76]2 years ago
4 0
I’m not sure but it’s either c because it’s the one that makes more sense but it’s could also be them
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Danny Diver weighs 500 N and steps off a diving board 10 m above the water. Danny hits the water with kinetic energy of
Morgarella [4.7K]

Answer:

Danny hits the water with kinetic energy of 5000 J.

Explanation:

Given that,

The Weight of Danny Diver,

F = 500 N

m*g= 500 N

He steps off a diving board 10 m above the water.

h=10 m

when Danny diver hits water he generates the kinetic energy.

We need to find the kinetic energy of the water.

Let kinetic energy is K.

K = m*g*h

Where g is acceleration due to gravity.

that g= 9.8 m/s^2

now substituting the values in above equation

K= (500) * 10

K= 5000 J

Hence,

he hits the water with kinetic energy of 5000 J.

Learn more about Kinetic energy here:

<u>brainly.com/question/15587458</u>

<u />

#SPJ4

7 0
2 years ago
What factors contribute to global winds
serg [7]
Temperature and elevation, if it is cold in Idaho and warm on the eastern end of a mountain side in california (or if warm air is going in that direction) then the cold air, being more dense, will go towards california while the cold air in Idaho will become warm. Same goes for the rest of the world 
3 0
3 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
Most comets have short periods and orbit close to the ecliptic plane. (T/F)
Lapatulllka [165]

Answer:

False

Explanation:

Most comets are located outside the solar system, in part of the cloud that originated from dust and gas that has remained virtually untouchable for billions of years. The orbit of these comets can reach the order of a light year. Thus, they are called long-period comets.

3 0
3 years ago
What is the most important factor for the formation of our planets
ohaa [14]
The gravitational pull of the Sun the interstellar dust attracting heat away from the protosun the process of nuclear fusion the nebular cloud condensing.
7 0
3 years ago
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