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Finger [1]
4 years ago
9

Approximately how fast is Jupiter orbiting the Sun? Approximately how fast is Jupiter orbiting the Sun? 10 km/skm/s a little les

s than 15 km/skm/s 20 km/skm/s cannot be determined from the information provided
Physics
1 answer:
Anna007 [38]4 years ago
5 0

Answer:

The correct answer is

a little less than 15 km/s.

Explanation:

The distance between the sun and Jupiter varies by about 75 million km between the perihelion and the aphelion with an average distance of 778 million km from the sun for which it takes Jupiter about 12 years to complete its orbit round the sun giving it an orbital speed of about 13.07 km/s

The size of Jupiter is more than the twice the combined size of all the other planets, which is about 1.300 times the size of earth.

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At a construction site, the site manager notices that a crane takes 20 seconds to lift a 500kg steel beam up to a height of 15 m
posledela
The work done is equal to the change in potential energy which is:
P.E = mgh
P.E = 500 x 9.81 x 15
P.E = 73,575 J

Power = work / time
Power = 73,575 / 20
Power = 3,700 Watts
5 0
4 years ago
Suppose electrons in a TV tube are accelerated through a potential difference of 2.00 104 V from the heated cathode (negative el
Ivanshal [37]

Answer:

83816746.4254 m/s

Explanation:

m = Mass of electron = 9.11\times 10^{-31}\ kg

q = Charge of electron = 1.6\times 10^{-19}\ C

V = Voltage = 2\times 10^4\ V

The kinetic energy of the electron is

K=\dfrac{1}{2}mv^2

Energy is given by

E=qV

Balancing the energy

qV=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2qV}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}}\\\Rightarrow v=83816746.4254\ m/s

The velocity of the electrons is 83816746.4254 m/s

5 0
3 years ago
The maximum lift-to-drag ratio of the World War I Sopwith Camel was 7.7. If the aircraft is in flight at 5000 ft when the engine
andre [41]

The related concepts to solve this problem is the Glide Ratio. This can be defined as the product between the height of fall and the lift-to-drag ratio. Mathematically, this expression can be written as,

R = h (\frac{L}{D})_{max}

Replacing,

R = 5000ft (7.7)

R = 38500ft

Converting this units to miles.

R = 38500ft (\frac{1mile}{5280ft})

R = 7.2916miles

Therefore the glide in terms of distance measured along the ground is 7.2916miles

3 0
4 years ago
A 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 N · m. What is the angle (assumed to
o-na [289]

Answer:

angle = 18.40 degree

Explanation:

given data

force = 95 N

distance = 0.50 m

torque = 15 N · m

to find out

angle between the wrench handle and the direction of the applied force

solution

we will apply here torque equation that is express as

torque =  distance × force × sin(θ)   ...................1

put here value we will get angle that is

15 = 0.50 × 95 × sin(θ)

sin(θ) = 0.315789

θ = 18.40 degree

6 0
3 years ago
Read 2 more answers
Please help
lukranit [14]
A ray box is the answer I give
3 0
3 years ago
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