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n200080 [17]
3 years ago
6

A 10,000kg space ship is orbiting the moon with a radius of 25km from the ship to the center of the moon. It's tangential speed

is 125m/s. What is the ships centripetal acceleration?
Is this the correct set-up for this problem.

ac=10,000kg(125m/s)2/25,000m

True or False?
Physics
1 answer:
galina1969 [7]3 years ago
3 0

Answer:

False

Explanation:

ac = v^2/r

acceleration is not dependent on the mass of the orbiting object.

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What is the force measured in Newtons between a proton and an electron in the ground state?
goldenfox [79]
The answer is zero hope I helped
5 0
3 years ago
What is the phase of the moon when the moon is positioned between the sun and earth?
vladimir2022 [97]

Answer:

Solar eclipse

Explanation:

A solar eclipse is when the moon passes between the sun and Earth, causing it to go dark and to give the moon a halo effect :D hope this helped!

8 0
2 years ago
Which of the following is not part of the process known as oxidative phosphorylation?(a) Molecular oxygen serves as a final elec
Andre45 [30]

Answer:

(d) ATP molecules are produced in the cytosol as glucose is converted into pyruvate.

5 0
3 years ago
9. A football punter attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in th
vichka [17]

Answer:

Angle is 55.52°

and Initial Speed is v=26.48 m/s

Explanation:

Given data

x_{o}=0m\\ y_{o}=1.23m\\a_{oy}=a_{1y}=g=-9.8m/s^{2} \\x_{1}=67.0m\\y_{1}=0m\\t_{o}=0\\a_{ox}=m/s^{2} \\t_{1}=4.50s

Applying the kinematics equations for motion with uniform acceleration in x and y direction

So

x_{1}=x_{o}+v_{ox}t_{1}=67.0m\\0+4.50v_{o}Cos\alpha =67.0m\\v_{o}Cos\alpha =14.99\\v_{o}=14.99/Cos\alpha.....(1) \\and\\y_{1}=y_{o}+v_{oy}t_{1}+(1/2)a_{oy}t_{1}^{2} =0m\\ 1+4.50v_{o}Sin\alpha+(-9.8/2)(4.5)^{2}=0\\  v_{o}Sin\alpha=21.828.....(2)

Put the value of v₀ from equation (1) to equation (2)

So

\frac{14.99}{Cos\alpha }(Sin\alpha ) =21.828\\as\\tan\alpha =Sin\alpha /Cos\alpha \\So\\14.99tan\alpha =21.828\\tan\alpha =21.828/14.99\\\alpha =tan^{-1}(21.828/14.99) \\\alpha =55.52^{o}

Put that angle in equation (1) or equation (2) to find the initial velocity

So from equation (1)

v_{o}=(\frac{14.99}{Cos\alpha } ) \\v_{o}=(\frac{14.99}{Cos(55.52) } ) \\v_{o}=26.48m/s

7 0
3 years ago
The doctor writes a prescription for na heparin 20,000 units in 500 ml n.s. infuse over 8 hours. what is the flow rate in ml/hr?
Ne4ueva [31]
I think this type of equation could be conducted in simple division equation since it does not involve drop rate.

we know that there is 500 ml of substance and should be infused within 8 hours period.

So the flow rate in ml/hr would be: 

500/8 = 62.5 ml/hr                                                                                    
8 0
3 years ago
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