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stepan [7]
2 years ago
10

A small block slides without friction along a track toward a circular loop. The block has more than enough speed to remain firml

y in contact with the track as it goes around the loop. The magnitude of the block’s acceleration at the top of the loop isa) zero b) greater than zero but less than g c) equal to g d) greater than g

Physics
1 answer:
lapo4ka [179]2 years ago
3 0

Answer:

d) greater than g

Explanation:

You have to apply Newton's Second Law to have an idea about the value of <em>the block's aceleration.</em>

∑F = ma

Where m is the mass, a is the acceleration and F represents all the forces involved.

You have to draw a free body diagram for the block at the top of the loop. (See the attachment)

Applying Newton's Second Law:

y: N+mg=ma

Where N is the <em>normal force</em> (the contact force that the loop's surface applies to the block) and mg is the <em>weight.</em>

Therefore a= \frac{N}{m} + g

So, you can notice that <em>the magnitude of the block's acceleration</em> is greater than the acceleration of gravity because of the effect of the magnitude N/m.

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Answer:

The 6 fingers allele is dominant

Explanation:

We are told that the the individual is genotypically heterozygous, that is the have both types of the finger allele: the 5 finger allele and the 6 fingers allele however phenotypically, 6 fingers are observed. From this we can conclude that the 6 fingers allele is the one that is dominant because it is the one that is expressed phenotypically.

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6 0
2 years ago
A child pushes a 75 N toy car across the floor. What is the mass of the car?
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Answer:

7.6 kg

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w=75N

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3 0
3 years ago
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
Read 2 more answers
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