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Gnesinka [82]
1 year ago
14

A tortoise can move with a speed of 10.0cm/s, while a rabbit can move 10 times faster. In a race, both of them started at the sa

me time, but the rabbit stopped to rest for three minutes. The tortoise wins by a distance of 10cm from the rabbit. How long is the race?
A. 125.2s
B. 159.8s
C. 199.9s
D. 205.7s​
Physics
1 answer:
Xelga [282]1 year ago
8 0

Answer:

C. 199.9 s

Explanation:

3 minutes = 3×60 = 180 seconds.

the turtle moves in that time 180×10 = 1800 cm.

in other words the rabbit gave it that much head-start (it does not matter if that was at the begin of in the middle of the race).

the rabbit moves with 10×10cm/s = 100cm/s.

the rabbit needs therefore 1800/100 = 18 seconds for the

1800 cm.

at that time the turtle has added another 18×10 = 180 cm.

for which the rabbit needs 180/100 = 1.8 seconds.

during that time the turtle has added 1.8×10 = 18 cm.

and so on.

in formal mathematics this looks like this :

1800 + 10x = 100x

after x seconds of the rabbit running both will have run the same distance, and it is a tie.

1800 = 90x

x = 20 seconds

so, at that point, the rabbit was actively running for 20 seconds and raced 20×100 = 2000 cm

and the turtle was actively running for 180 + 20 = 200 seconds, and also covered 200×10 = 2000 cm.

but our question tells us that the turtle won by 10 cm.

so, the race was over a little bit before these 200 seconds (for a tie).

this means, the rabbit could not run the last 10 cm for the tie (because the race was over and the turtle had won).

the rabbit would have needed 10/100 seconds for these 10 cm.

as speed = distance/time

we need to divide distance by speed

distance/1 / distance/time

to get time.

so,

10cm/1 / 100cm/s = 10s/100 = 1/10 s

so, we need to deduct this 1/10 s from the 200 seconds of the turtle (and also from the 20 seconds for the rabbit).

the race lasted of course the whole time the turtle was running (while the rabbit was resting, officially still participating in the race with speed 0 for 3 minutes).

and so, the race was 199.9 s long.

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what is required to bring about a phase change 1.an increase or decrease in pressure 2.an increase or decrease in energy 3.an in
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Answer

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Explanation

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2 years ago
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If the caffeine concentration in a particular brand of soda is 2.97 mg/oz, 2.97 mg/oz, drinking how many cans of soda would be l
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Explanation:

The given data is as follows.

     Concentration of caffeine = 2.97 mg/oz

     Number of oz in a can = 12 oz

Therefore, the concentration of caffeine in one can is calculated as follows.

                 = (12 \times 2.97) mg

                 = 35.64 mg

                 = 35.64 \times 10^{-3} g

Since, it is given that lethal dose is 10.0 g. Hence, number of cans are calculated as follows.

     No. of cans = \frac{\text{Lethal dose}}{\text{concentration in one can}}

                         = \frac{10 g}{35.64 \times 10^{-3} g}

                         = 280.58

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Thus, we can conclude that 281 cans of soda would be lethal.

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Answer:

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We have to find the total negative charge on the electrons in one mole of Helium.

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We know that q=ne

Where n =Number of fundamental units

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Using the formula

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Total negative charge in 1 mole=3.2\times 10^{-19}\times 6.02\times 10^{23}=1.92\times 10^5C

Hence, the total negative charge on the electrons in 1 mole of Helium=1.92\times 10^5 C

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