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love history
2 months 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:
Sav [3.1K]2 months ago
8 0
To calculate the rate, first convert units properly. Since 1 kilogram equals 1,000,000 micrograms, 1.6 kilograms is 1,600,000 micrograms. One week has 604,800 seconds. Therefore, dividing 1,600,000 micrograms by 604,800 seconds gives the rate. Simplifying, this results in 2.65 µg/s. I hope this answers your question.
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An infinite charged wire with charge per unit length lambda lies along the central axis of a cylindrical surface of radius r and
inna [3103]
The electric flux through the cylindrical surface surrounding the infinite charged wire is given by the formula ∅E = E x 2πrl. To analyze this, we consider an infinitely long straight wire with a uniform linear charge density of λ Cm⁻¹. The electric field at a distance r from this charge can be evaluated using a cylindrical Gaussian surface of radius r and length l, oriented along the wire. Only the curved surface of the cylinder contributes to the total flux since the other surfaces are perpendicular to E.
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11 days ago
When Royce was 10 years old, he had a mass of
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The gravitational force between Royce and Earth would double by the age of 16. Newton’s law of universal gravitation states that the gravitational force is proportional to the masses involved and inversely proportional to the square of the distance between them. At age 10, Royce weighed 30 kg, and by age 16, he weighed 60 kg. Since his mass doubled from 10 to 16 years, this results in a corresponding doubling of the gravitational force.
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23 days ago
A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or
Softa [3030]

Answer:

 Baseball mass:  m_b=3.992*10^{14}kg  

Explanation:

 Circumference of a baseball is calculated using 2πr = 23 cm

 Thus, the radius comes out to be 3.66 cm, which equals 3.66*10^{-2} m

 The mass density of the baseball matches that of a neutron or proton.

 Proton mass = 10^{-27} kg  

 Proton diameter = 10^{-15} m

 Proton radius =  5*10^{-16} m

 Volume of the baseball is \frac{4}{3} \pi r^3

 Now by substituting all values into the mass per unit volume equation for the baseball, we get:  

         \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

       Therefore, the baseball mass amounts to m_b=3.992*10^{14}kg              

5 0
1 month ago
Read 2 more answers
A good quarterback can throw a football at 27 m/s (about 60 mph). If we assume that the ball is caught at the same height from w
ValentinkaMS [3465]

Response:

The ball remained airborne for 3.896 seconds

Explanation:

Given that

g = 9.8 m/s², representing gravitational acceleration,

If the angle of launch is 45°, the horizontal range will be maximized.

Both horizontal and vertical launch velocities are equal, each equating to

v_h  =  v cos θ

v_h  =  27 × cos 45°

         = 19.09 m/s.

The duration to reach maximum height is half of the flight time.

v = u + at   ∵ v = 0 (at maximum height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The total time in the air equals twice the time to reach maximum height

2 t₁ = 3.896 s

The horizontal distance covered is

D = v × t

D = 3.896×19.09

   = 74.375 m

The ball was in the air for 3.896 seconds

8 0
1 month ago
How long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘C? Note that t
Softa [3030]
The required duration is 16.1 minutes. To determine the heat needed to raise the temperature, we must calculate the following amounts, where Q represents the required heat, m stands for mass, V represents the volume, C signifies specific heat, and ΔT indicates temperature change. After substituting the provided values into the formula and calculating, the next step is determining the required time based on the formula t = Q/P, where P is given as 1500 W. Ultimately, we find that the time needed is 16.1 minutes.
5 0
28 days ago
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