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jonny [76]
3 years ago
14

The radius of the earth's very nearly circular orbit around the sun is 1.5×1011m. find the magnitude of the earth's velocity. as

sume a year of 365 days.
Physics
1 answer:
Luden [163]3 years ago
7 0
<span>Step 1 of 3 Calculate the magnitude of the earth’s velocity. The radius of earth orbit around the sun is . Time taken to complete one revolution is . Convert time from days into seconds. The expression for magnitude of earth velocity is, Substitute for , for . Further solved, Explanation | Common mistakes | Hint for next step Time taken to complete one resolution is 365 days. Convert time from days into seconds. Radius of earth orbit around the sun is given. Velocity is circumference of orbit upon time taken in one resolution. Calculate the magnitude of earth velocity. Change unit of velocity to Step 2 of 3</span>
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4 years ago
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rodikova [14]

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3 years ago
Which type of macromolecules a cell brake down food?
kobusy [5.1K]

Answer:

Hope this helps you find the answer

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8 0
3 years ago
An event occurs in system K' at x' = 2 m, y' = 3.7 m, z' = 3.7 m, and t' = 0. System K' and K have their axes coincident at t =
fiasKO [112]

Answer:

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

Explanation:

To find the coordinates of event in system K ,we have to use inverse Lorentz transformation

So

x=\frac{x^{|}+vt^{|} }{\sqrt{1-\frac{v^{2} }{c{2} } } } \\x=\frac{2m+0.92c(0) }{\sqrt{1-\frac{(0.92c)^{2} }{c{2} } } }\\x=5.103m\\y=y^{|}\\ y=3.7m\\z=z^{|}\\ z=3.7m

for t

r=\frac{1}{\sqrt{1-v^{2} } } \\r=\frac{1}{\sqrt{1-(0.92)^{2} } } \\r=2.551\\t=r(t^{|}+vx^{|}/c^{2}   )\\t=2.551(0s+(0.92c)(2)/c^{2} )\\t=1.57*10^{-8}s

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

6 0
3 years ago
If a car is traveling on the highway at a constant velocity, the force that pushes the car forward must be A. equal to the weigh
Gala2k [10]

the correct answer is c

4 0
4 years ago
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