2. The Great Pyramid of Cheops had a height of H =147 m before its pinnacle stone fell. Its base is a square with edge length L = 230 m. Its volume is equal to: %3D 3. It has uniform density of p = 1.8x 10 kg/m³. Find a) the original height of its center of mass, and b) the work required to lift the blocks into place from sea level.
2. The Great Pyramid of Cheops had a height of H =147 m before its pinnacle stone fell. Its base is a square with edge length L = 230 m. Its volume is equal to: %3D 3. It has uniform density of p = 1.8x 10 kg/m³. Find a) the original height of its center of mass, and b) the work required to lift the blocks into place from sea level.
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Please solve 2A and 2B
![**Physics Problem Set with Solutions**
1. **Center of Mass of Carbon Monoxide Molecule**
The distance between the centers of carbon and oxygen atoms in a carbon monoxide gas molecule is \(1.13 \times 10^{-10} \, \text{m}\).
- Mass of carbon = \(12.01 \, \text{u}\)
- Mass of oxygen = \(16.00 \, \text{u}\)
**Solution:**
Center of mass position is \(6.46 \times 10^{-11} \, \text{m}\) from the carbon atom.
2. **Volume and Work on the Great Pyramid of Cheops**
- Original height: \(H = 147 \, \text{m}\)
- Base length: \(L = 230 \, \text{m}\)
- Volume: \(V = \frac{L^2H}{3}\)
**Find:**
a) Original height of its center of mass
b) Work required to lift the blocks from sea level
**Answers:**
a) \(36.8 \, \text{m}\)
b) \(1.7 \times 10^{12} \, \text{J}\)
3. **Soda Can Dynamics**
A cylindrical can with mass \(M\) and height \(H\) has soda draining out. Holes are punched at the top and bottom.
**Find:**
a) Initial center of mass
b) Center of mass when full can drains
c) Change of center mass during drainage
d) Height \(y\) of soda in relation to the can at any time:
\[
y = \frac{\sqrt{M^2 - Mm} - M}{m} H
\]
**Answers:**
a) \(H/2\)
b) \(H/2\)
c) Results from previous parts
d) Formula above
4. **Center of Mass Speed in Two-Particle System**
System:
- Particle 1: \(3.0 \, \text{kg}, \, 6.0 \, \text{m/s}\) in \(-y\) direction
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F097e4ba6-b3f3-4468-a4b3-d5ca2a9f0e8d%2F091d688f-1a6f-4b6e-b26f-e675a54eb595%2F5507i1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Physics Problem Set with Solutions**
1. **Center of Mass of Carbon Monoxide Molecule**
The distance between the centers of carbon and oxygen atoms in a carbon monoxide gas molecule is \(1.13 \times 10^{-10} \, \text{m}\).
- Mass of carbon = \(12.01 \, \text{u}\)
- Mass of oxygen = \(16.00 \, \text{u}\)
**Solution:**
Center of mass position is \(6.46 \times 10^{-11} \, \text{m}\) from the carbon atom.
2. **Volume and Work on the Great Pyramid of Cheops**
- Original height: \(H = 147 \, \text{m}\)
- Base length: \(L = 230 \, \text{m}\)
- Volume: \(V = \frac{L^2H}{3}\)
**Find:**
a) Original height of its center of mass
b) Work required to lift the blocks from sea level
**Answers:**
a) \(36.8 \, \text{m}\)
b) \(1.7 \times 10^{12} \, \text{J}\)
3. **Soda Can Dynamics**
A cylindrical can with mass \(M\) and height \(H\) has soda draining out. Holes are punched at the top and bottom.
**Find:**
a) Initial center of mass
b) Center of mass when full can drains
c) Change of center mass during drainage
d) Height \(y\) of soda in relation to the can at any time:
\[
y = \frac{\sqrt{M^2 - Mm} - M}{m} H
\]
**Answers:**
a) \(H/2\)
b) \(H/2\)
c) Results from previous parts
d) Formula above
4. **Center of Mass Speed in Two-Particle System**
System:
- Particle 1: \(3.0 \, \text{kg}, \, 6.0 \, \text{m/s}\) in \(-y\) direction
-
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