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2 years ago

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

1 answer:

Ivan2 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 2.65µg/s. Hope this answers your question.

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How would you solve for x I cant remember right now 4x+6x=9x-10

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Combine all of the x's on one side of the equation and then finish the problem!

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2 years ago

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A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or

Svetlanka [38]

Answer:

Mass of base ball $m_b=3.992*10^{14}kg$

Explanation:

Circumference of baseball = 2πr = 23 cm

So radius of baseball = 3.66 cm = $3.66*10^{-2}$ m

Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

Mass of proton = $10^{-27}$ kg

Diameter of proton = $10^{-15}$ m

Radius of proton = $5*10^{-16}$ m

Volume of ball = $\frac{4}{3} \pi r^3$

Now substituting all values in Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

$\frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}$

$\frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}$

$m_b=3.992*10^{14}kg$

So mass of base ball $m_b=3.992*10^{14}kg$

5 0

2 years ago

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Assume that the particle has initial speed viviv_i. Find its final kinetic energy KfKfK_f in terms of viviv_i, MMM, FFF, and DDD

NeX [460]

Answer:

KE= 1/2mv²

Explanation:

The kinetic energy of a body is the energy possessed by virtue of the body in motion

Given the parameters

m which is the mass of the body

v which is the velocity of the body too

K.E = kinetic energy

The expression for the kinetic energy of a body is given as

KE= 1/2mv²

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2 years ago

A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm

Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

$mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g$

The mass of the meter stick is 126.99115 g

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2 years ago

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A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume

dalvyx [7]

Explanation:

It is given that,

Mass of the car 1, $m_1=900\ kg$

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Mass of the car 2, $m_2=750\ kg$

Initial speed of car 2, $u_1=20j\ m/s$ (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V$