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Rudik [331]
3 years ago
11

Era that we live in 1. precambrian time 2. mesozoic 3. paleozoic 4. cenozoic

Physics
1 answer:
guajiro [1.7K]3 years ago
8 0

I believe we live in the Cenozoic era

We live in the Holocene Epoch, of the Quaternary Period, in the Cenozoic Era (of the Phanerozoic Eon).

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You are driving your car at a speed of 60 miles per hour, when you cross into Canada. Canada measures their speed in kilometers
Flauer [41]

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8 0
3 years ago
If a body of mass 5og moves in a circular path of radius 10cm .find work done
Sindrei [870]

Answer:

0Nm, no work is done.

Explanation:

Work done is defined as the Force per distance meaning force times the distance moved in the direction of the force.

Now the body of mass 50g has a centripetal force acting on it directed towards the centre. Now in actuality the body stays along the circle it doesn't really move to the centre of the circle.

Hence the force doesn't move a distance, and so from the definition of work done;

F×d ; d =0

Hence work done = mv2/r × 0= 0Nm

3 0
3 years ago
In a ray diagram showing refraction, the incident ray and the refracted ray...
Amanda [17]
C because of the positioning of the initial ray of refraction
4 0
2 years ago
PLEASE HELP!! ILL MARK BRAINLYEST!!
topjm [15]

Answer: True

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Mass is the amount of matter in an object. Everything is made up of matter.

3 0
3 years ago
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The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
Margaret [11]

Let  us consider two bodies having masses m and m' respectively.

Let they are  separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -  F = G\frac{mm'}{r^{2} }   where G is the gravitational force constant.

From the above we see that F ∝ mm' and F\alpha \frac{1}{r^{2} }

Let the orbital radius of planet  A is r_{1}  = r and mass of planet is m_{1}.

Let the mass of central star is m .

Hence the gravitational force for planet A  is f_{1} =G \frac{m_{1}*m }{r^{2} }

For planet B the orbital radius  r_{2} =2r_{1} and mass m_{2} = 3 m_{1}

Hence the gravitational force f_{2} =G\frac{m m_{2} }{r^{2} }

                                                 f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }

                                                 = \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }

Hence the ratio is  \frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2}  }{Gmm_{1}/r_{1} ^2 }

                                      =\frac{3}{4}     [ ans]


                                                 

                           

3 0
3 years ago
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