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kiruha [24]
3 years ago
8

What does vf stand for a.fringe velocity b.first velocity c.final velocity

Physics
1 answer:
Elza [17]3 years ago
5 0
The correct answer is C. Final Velocity

Hope this helped!
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Carter pushes a bag full of basketball jerseys across the gym floor. The he pushes with a constant force of 21 newtons. If he pu
myrzilka [38]
Power is the rate of work that is done per time. Work is calculated from the product of force and the displacement. So, by using the definition of power and work, we can relate power with force, displacement and the time. We do as follows:

Power = Work / time
Power = Force x displacement / time
Power = 21 N (9 m) / 3 s
Power = 63 J /s = 63 W
6 0
3 years ago
Calculate the gravitational potential energy of a 1200 kg car at the top of a hill that is 42 m high.
GalinKa [24]
It IS <span>PE = (1200 kg)(9.8 m/s²)(42 m) = 493,920 J </span>
6 0
3 years ago
Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe
Mice21 [21]

As per the question the distance of venus from sun is given as 0.723 AU

We have been asked to calculate the time period of the planet venus.

As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically

                                        T^{2} \alpha R^{3}

                                         ⇒ T^{2} = KR^{3} where is k is the proportionality  constant

We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days

Hence T_{1} =365.5 days

The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun

Hence R_{1} =1 AU

The distance of venus from sun is 0.723 AU i.eR_{2} =0.723

From keplers law we know that-\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{3} }

                            ⇒T_{2} ^{2} =T_{1} ^{2} *\frac{R_{2} ^{3} }{R_{1} ^{3} }

Putting the values mentioned above we get-

                                      T_{2} ^{2} =50,350.132851075

                                         ⇒ T_{2} =\sqrt{50,350.132851075}

                                        ⇒T_{2} = 224.388352752710 days.

Hence the time period of venus is 224.388352752710 days

                                         

                     






                           

7 0
4 years ago
Read 2 more answers
Gamma rays are waves of energy that have no charge. <br> True<br> False
Andre45 [30]
Gamma rays are waves of energy that have no charge. TRUE.
3 0
3 years ago
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A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?
Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length f=\frac{R}{2}=\frac{12}{2}=6cm

Now for mirror we know that \frac{1}{f}=\frac{1}{u}+\frac{1}{v}

So \frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}

16.66-0.333=\frac{1}{v}

v = 0.750 m

Now magnification of the mirror is m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025

5 0
4 years ago
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