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-Dominant- [34]
3 years ago
12

A person on a diet loses 1.6 kg in a week. How many micrograms/second (µg/s) are lost?

Physics
1 answer:
Ivan3 years ago
8 0
Based on the given values above, in order for us to get the answer, we need to convert the units first. So in 1 kilogram, there is 1,000,000 micrograms. In this case, 1.6 kilograms is 1,600,000 micrograms. For the week to seconds, 1 week is equivalent to 604,800 seconds. Therefore, 1,600,000 micrograms/604,800 seconds. So we are going to simplify this. So it would be 2.65<span>µg/s. Hope this answers your question.</span>
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A force of 45 N is applied tangentially to the rim of a solid disk of radius 0.12 m. The disk rotates about an axis through its
bearhunter [10]

Answer:

Mass of the disk will be 2.976 kg

Explanation:

We have given force F = 45 N

Radius of the disk r = 0.12 m

Angular acceleration \alpha =140rad/sec^2

We know that torque \tau =I\alpha

And \tau =Fr

So Fr=I\alpha , here I is moment of inertia

So 50\times 0.12=I\times 140

I=0.0428kgm^2

We know that moment of inertia I=\frac{1}{2}mr^2

So 0.0428=\frac{1}{2}\times m\times 0.12^2

m = 2.976 kg

6 0
3 years ago
Brainly malfunction?|<br><br> I'm not sure why but...
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Answer:

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Explanation:

6 0
3 years ago
Add the following displacement vectors: <br> 12 m south and 15 m 55° E of N??
daser333 [38]
12m S=0m E, -12m N
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4 0
3 years ago
Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

       Sphere A : ma = 80 kg , ( 0 , 0 )

       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

                  vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

                  Fc = vector Fac + vector Fbc

                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

3 0
3 years ago
A soft, silvery-white metal combines with a yellow gas to form a white crystal-like solid. What can be said about this change? T
Mandarinka [93]

Answer: it is D 2022 edge

Explanation:

5 0
2 years ago
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