Explanation:
First, simplify the circuit. Then calculate the parallel and consecutive resistances to find the answer.
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.
The answer is d. Wavelength
Answer:
The weight lifter would not get past this sticking point.
Explanation:
Generally torque applied on the weight is mathematically represented as
T = F z
To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm
for z the perpendicular distance
So Elbow Torque is 

To obtain the torque required we substitute 300 N for F and 30cm 
So the Required Torque is 

Now since
it mean that the weight lifter would not get past this sticking point
Answer:

Explanation:
The rotation rate of the man is:



The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:
![(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega](https://tex.z-dn.net/?f=%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%5Ccdot%20%280.16%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%29%20%3D%20%5B%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%2B20000%5C%2Ckg%5Ccdot%20m%5E%7B2%7D%5D%5Ccdot%20%5Comega)
The final angular speed is:
