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Alex17521 [72]
3 years ago
7

If you had 8 balls and 7 of them were a certain weight, and 1 of them was heavier, how could you find the heaviest ball. All the

balls look the same. How can you find the heaviest ball?
Physics
1 answer:
dybincka [34]3 years ago
7 0

Answer:

There are two method of comparing the balls 1) using a balance  2) by  only 2 weighings.

Explanation:

There are two method of comparing the balls 1) using a balance  2) by  only 2 weighings.

Make the following groups - --- (1,2,3),(4,5,6),(7,8)

Step 1. compare the Weigh (1,2,3) and (4,5,6)

there are 2 possible outcomes:

1---both the group are of same weight. and named as (Case A)

2--- one of the group is heavier than other and named as  (Case B)

Step 2. Let examine both case

In Case A --in this case, now compare the weight of 7th and 8th ball. By this you have recognize the heavier ball by 2 weighing method.

In Case B -- considered the heaviest group (assume group (1,2,3) is heavy), from this group take randomly two ball and compare the their weight. out of these two ball, one  is heavy else the third ball is.

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Answer:

a_{cp}=7.77m/s^2

Explanation:

The equation for centripetal acceleration is a_{cp}=\frac{v^2}{r}.

We know the wheel turns at 45 rpm, which means 0.75 revolutions per second (dividing by 60), so our frequency is f=0.75Hz, which is the inverse of the period T.

Our velocity is the relation between the distance traveled and the time taken, so is the relation between the circumference C=2\pi r and the period T, then we have:

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Putting all together:

a_{cp}=\frac{(2\pi r f)^2}{r}=4 \pi^2f^2r=4 \pi^2(0.75Hz)^2(0.35m)=7.77m/s^2

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3 years ago
A pilot flew a jet from City A to City B, a distance of 2400 mi. On the return trip, the average speed was 20% faster than the o
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Answer:

Explanation:

Let the velocity be Va for outbound journey and 1.2 Va for return journey .

Total time taken = 2400 / Va + 2400 / 1.2 Va

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1 / Va ( 2400 + 2000 ) = 20 / 3

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A duck swimming on the surface of a pond has an
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From the given information:

  • Taking the movement of the Duck in the North as the x-direction
  • The movement of the Duck in the East direction as the y-direction

However, we will have to compute the initial velocity and the acceleration of the duck in their vector forms.

<h3>In vector form;</h3>

The initial velocity is:

\mathbf{u ^{\to} = 0.7 m/s ( -cos 25^0 \hat x + sin 25^0 \hat y ) \ m/s}

The acceleration is:

\mathbf{a ^{\to} = 0.5 m/s ( cos 41^0 \hat x - sin 41^0 \hat y ) \ m/s^2}

The objective of this question is to determine the speed of the duck at a certain time. Since it is not given, let's assume we are to determine the Duck speed after 4 seconds of accelerating;

Then, it implies that time (t) =  4 seconds.

Using the first equation of motion:

v^{\to} = u ^{\to} + a^{\to} t

Then, we can replace their values into the equation of motion in order to determine the speed:

\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+4 \times 0.5 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}

\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+2.0 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}

\mathbf{v^{\to} =\Big( ( -0.7 cos 25^0 \hat x + 0.7 sin 25^0 \hat y )+( 2.0cos 41^0 \hat x - 2.0sin 41^0 \hat y )\Big)}

Collect like terms:

\mathbf{v^{\to} =\Big( (2.0cos 41^0 -0.7 cos 25^0   )\hat x+(  0.7 sin 25^0 - 2.0sin 41^0 )\Big)\hat y}

\mathbf{v^{\to} =0.87500   \hat x- 1.01629 \hat y}

Thus, the magnitude is:

\mathbf{v^{\to} =\sqrt{(0.87500 )^2 +( 1.01629 )^2}}

\mathbf{v^{\to} =\sqrt{0.76563 +1.03285}}

\mathbf{v^{\to} =\sqrt{1.79848}}

\mathbf{v^{\to} =1.34 \ m/s}

Therefore, we can conclude that the speed of the duck after 4 seconds is 1.34 m/s

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brainly.com/question/17108011?referrer=searchResults

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