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2 years ago

If 44 ± 3 beats are counted (in exactly 30 s) what is the percent uncertainty in the measurement of their heartbeats per minute?

Physics

1 answer:

Anika [276]2 years ago

5 0

Answer:

6.82%

Explanation:

If 44 ±3 beats are counted in 30 seconds, twice the number of beats will be counted in 1 minute or 60 seconds, i.e. 88 beats per minute. The absolute uncertainty in 1 minute will also be twice the uncertainty in 30 seconds, so 88 ±6 beats per minute.

The percentage uncertainty can be found by below:

$\%_{uncertainty} = \frac{absolute\ uncertainty}{absolute\ value}*100$

$\%_{uncertainty} = \frac{6}{88}*100$

$\%_{uncertainty} = 6.82\%$

The percentage uncertainty can also be found for the 30 second interval

$\%_{uncertainty} = \frac{absolute\ uncertainty}{absolute\ value}*100$

$\%_{uncertainty} = \frac{3}{44}*100$

$\%_{uncertainty} = 6.82\%$

While the absolute uncertainty varies depending on the time interval, the percentage uncertainty remains the same: 6.82%

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1 year ago

Consider two children sitting on a merry-go-round, with one closer to the outer edge and one closer to the center. show answer N

dolphi86 [110]

Answer:

They both have the same angular speed.

Explanation:

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What we can see from the formula is that, since the $2\pi$ does not change its value, the angular speed depends only on the period T.

In this case for both the children closer to the outher edge and for the children closer to the center, the time to complete a lap is the same, because the time does not depend on where they are sitting in the marry go round. This means that the period for both is the same.

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3 years ago

A car accelerates from rest at a rate of 6.50 m/s2. Determine thevelocity of the car at t = 4.00 s.

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Answer:

1.62 m/s2

Explanation:

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2 years ago

A student team is to design a human powered submarine for a design competition. The overall length of the prototype submarine is

Allushta [10]

Answer:

a) The speed is 61.42 m/s

b) The drag force is 10.32 N

Explanation:

a) The Reynold´s number for the model and prototype is:

$Re_{m} =\frac{p_{m}V_{m}L_{m} }{u_{m} }$

$Re_{p} =\frac{p_{p}V_{p}L_{p} }{u_{p} }$

Equaling both Reynold's number:

$\frac{p_{p}V_{p}L_{p} }{u_{p} }=\frac{p_{m}V_{m}L_{m} }{u_{m} }$

Clearing Vm:

$V_{m} =\frac{p_{p}V_{p}L_{p} u_{m} }{u_{p} p_{m} L_{m} }=\frac{999.1*0.56*8*1.849x10^{-5} }{1.138x10^{-3}*1.184*1 } =61.42m/s$

b) The drag force is:

$\frac{F_{Dm} }{p_{m}V_{m}^{2}L_{m}^{2} } =\frac{F_{Dp} }{p_{p}V_{p}^{2}L_{p}^{2} } \\F_{Dp} =\frac{F_{Dp}p_{p}V_{p}^{2}L_{p}^{2} }{p_{m}V_{m}^{2}L_{m}^{2} } \\F_{Dp}=\frac{2.3*999.1*0.56^{2} *8^{2} }{1.184*61.42^{2}*1^{2} } =10.32N$

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2 years ago

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