Question:
The water molecules now in your body were once part of a molecular cloud. Only about onemillionth of the mass of a molecular cloud is in the form of water molecules, and the mass density of such a cloud is roughly 2.0×10−21 g/cm^3.
Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.
Answer:
The volume of cloud that has the same density as the amount of water in our body is 1.4×10²⁵ cm³
Explanation:
Here, we have mass density of cloud = 2.0×10⁻²¹ g/cm^3
Density = Mass/Volume
Volume = Mass/Density = If the mass is 40 kg and the body is made up of 70% by mass of water, we have
28 kg water = 28000 g
Therefore the Volume = 28 kg/ 2.0×10⁻²¹ g/cm^3 = 1.4×10¹⁹ m³ = 1.4×10²⁵ cm³.
Therefore, the volume of cloud that has the same density as the amount of water in our body = 1.4×10²⁵ cm³.
Answer:
A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t]y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t], where the origin is at the left end of the string, the x-axis is along the string, and the y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation y(x, t) for this string if it were vibrating in its eighth harmonic?
Answer:
If you apply a force to separate 2 opposite poles, the potential energy of the system increases.
Answer:
182.28 W
Explanation:
Here ,
m = 7.30 Kg
distance , d= 28.0 m
time , t = 11.0 s
average power supplied = change in potential energy/time
average power supplied = m×g×d/time
average power supplied = 7.30×9.81×28/11
average power supplied = 182.28 W
the average power supplied is 182.28 W
