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3 years ago

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

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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The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo

Jobisdone [24]

(a) 4.0 m/s

We can solve this part just by analyzing the vertical motion of the froghopper.

The initial vertical velocity of the froghopper as it jumps from the ground is given by

$u_y = u_0 sin \theta$ (1)

where

$u_0$ is the takeoff speed

$\theta=58.0^{\circ}$ is the angle of takeoff

The maximum height reached by the froghopper is

h = 58.7 cm = 0.587 m

We know that at the point of maximum height, the vertical velocity is zero:

$v_y = 0$

Since the vertical motion is an accelerated motion with constant (de)celeration $g=-9.8 m/s^2$ , we can use the following SUVAT equation:

$v_y^2 - u_y^2 = 2gh$

Solving for $u_y$ ,

$u_y = \sqrt{v_y^2-2gh}=\sqrt{-2(-9.8)(0.587)}=3.4 m/s$

And using eq.(1), we can now find the initial takeoff speed:

$u_0 = \frac{u_y}{sin \theta}=\frac{3.4}{sin 58.0^{\circ}}=4.0 m/s$

(b) 1.47 m

For this part, we have to analyze the horizontal motion of the froghopper.

The horizontal velocity of the froghopper is

$u_x = u_0 cos \theta = (4.0) cos 58.0^{\circ} =2.1 m/s$

And this horizontal velocity is constant during the entire motion.

We now have to calculate the time the froghopper takes to reach the ground: this is equal to twice the time it takes to reach the maximum height.

The time needed to reach the maximum height can be found through the equation

$v_y = u_y + gt$

Solving for t,

$t=-\frac{u_y}{g}=-\frac{3.4}{9.8}=0.35 s$

So the time the froghopper takes to reach the ground is

$T=2t=2(0.35)=0.70 s$

And since the horizontal motion is a uniform motion, we can now find the horizontal distance covered:

$d=u_x T = (2.1)(0.70)=1.47 m$

7 0

3 years ago

When waves travel through water, why do the moving water particles continue to return to their starting position?

Marizza181 [45]

Answer:

Waves transfer energy, not matter

Explanation:

7 0

3 years ago

A rocket is launched from rest and moves in a straight line at 30.0 degrees above the horizontal with an acceleration of 35.0 m/

klemol [59]

Answer:

t = 123.59s

Explanation:

For the launch pad section:

Vf = Vo + a*t where Vo=0.

Vf = 35*25 = 875m/s

The distance traveled during the launch:

$d = Vo*t+\frac{a*t^2}{2} = 0+\frac{35*25^2}{2} = 10937.5m$

Now the projectile motion, we know that its initial speed is the speed calculated previously and the initial height is the y-component of the previously calculated distance.

$\Delta Y = Vo*sin(30)*t - \frac{g*t^2}{2}$