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Doss [256]
3 years ago
10

Dominic is one of the tallest son of Mang Marco. He have a height of 180 cm and weight of 68 kg. find her BMI and weight status.

Physics
1 answer:
netineya [11]3 years ago
7 0

Answer:

BMI = 20.98 [kg/m^2]; Normal or Healthy Weight

Explanation:

Body mass index (BMI) is a person’s weight in kilograms divided by the square of height in meters. Therefore we must convert the height from centimeters to meters.

BMI = 68 / (1.8)^2

BMI = 20.98 [kg/m^2]

BMI Weight Status

Below 18.5                   Underweight

18.5 – 24.9                   Normal or Healthy Weight

25.0 – 29.9                   Overweight

30.0 and Above           Obese

So the weight status is Normal or Healthy Weight

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A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that
astraxan [27]

Answer:

A) k=34867.3384\ N.m^{-1}

B) \omega'\approx84\ Hz

Explanation:

Given:

mass of car, m=1380\ kg

A)

frequency of spring oscillation, f=1.6\ Hz

We knkow the formula for spring oscillation frequency:

\omega=2\pi.f

\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f

\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6

k_{eq}=139469.3537\ N.m^{-1}

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

<u>So, the stiffness of each spring is (as they are identical):</u>

k=\frac{k_{eq}}{4}

k=\frac{139469.3537}{4}

k=34867.3384\ N.m^{-1}

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }

\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }

\omega'\approx84\ Hz

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3 years ago
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Answer:

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At what speed do a bicycle and its rider, with a combined mass of 100 kg , have the same momentum as a 1600 kg car traveling at
noname [10]
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3 years ago
A 20kg object acceleration by a force of 200N with coefficient of kineticfriction is 0.4 what is acceleration of the object?​
Schach [20]

Answer:

<u>Given</u><em> </em><em>-</em><em> </em><u>M</u><u> </u><u>=</u><u> </u>20 kg

k = 0.4

F = 200 N

<u>To </u><u>find </u><u>-</u><u> </u> acceleration

<u>Solution </u><u>-</u><u> </u>

F= kMA

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5 0
2 years ago
You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee
Archy [21]

Answer:

V = 4.81 V

Explanation:

  • As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
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        V_{rint} = I* r_{int}

  • The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

       V = V_{b} - V_{rint}  = 5.03 V = 6.0 V - 0.383 A* r_{int}

  • We can solve for rint, as follows:

         r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega

  • When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

       V = V_{b} - V_{rint}  = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V

  • As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.
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