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Mazyrski [523]
3 years ago
5

why is it incorrect to think that the more digits you include in your answer , the more accurate it is ?

Physics
1 answer:
Anestetic [448]3 years ago
8 0
Because the more that u answer with the extra digits the more u make ur answer wrong?
You might be interested in
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
Xelga [282]

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.

8 0
2 years ago
A digital stop-clock measures time in minutes and seconds.
ad-work [718]

Answer:

01:20 (1 minute 20 seconds)

Explanation:

subtract the starting time from the stopping time to get the time used: 02:10 - 00:50 = 01:20

4 0
3 years ago
A spring with a spring constant of 400 n / M has a mass hung on it so it stretches 8 cm. calculate how much mass the spring is s
I am Lyosha [343]

Answer:

3.3kg

Explanation:

Given parameters:

Spring constant = 400N/m

Extension  = 8cm  = 0.08m

Unknown:

Amount of mass the spring is supporting  = ?

Solution:

To solve this problem:

     F  = kE

F is the force

k is the spring constant

E is the extension

  So;

            F  = 400 x 0.08  = 32N

Mass;

       Force  = mass x acceleration due to gravity

          32  = mass x 9.8

  Mass = 3.3kg

5 0
3 years ago
It is important to organize your data because
serg [7]
I would say B. hope it helps 
5 0
3 years ago
Read 2 more answers
A box of negligible mass rests at the left end of a 2.00-m, 25.0-kg plank (Fig. P11.43). The width of the box is 75.0 cm, and sa
Sati [7]

Answer:

Required mass of sand is 20 kg

Explanation:

Given:

Mass of the plank = 25 kg

Distance of the Center of gravity of the Plank from the fulcrum = \frac{2}{2}-0.50 = 0.5m

Distance of the Center of gravity of the sand box from the fulcrum = \frac{2}{2}-\frac{0.75}{2}= 0.625m

Balancing the torque due to the plank and the sand box with respect to the fulcrum

Torque = Force × perpendicular distance

thus, we get

(25 × g) × 0.5 = weight of sand × 0.625

where, g is the acceleration due to gravity

or

(25 × g) × 0.5 = (mass of sand × g) × 0.625

or

mass of sand = 20 kg

<u>Hence, the required mass of the sand is </u><u>20 kg</u>

8 0
3 years ago
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