Find the best answers to your questions with the help of IDNLearn.com's knowledgeable users. Our Q&A platform offers reliable and thorough answers to ensure you have the information you need to succeed in any situation.
Sagot :
Certainly! Let's break down the problem step by step to find the speed of a point on the equator of the pulsar.
1. Understand the given information:
- The pulsar rotates 30 times per second.
- The diameter of the pulsar is 12 km.
2. Find the radius:
- The radius is half of the diameter. Thus, the radius is:
[tex]\[ \text{Radius} = \frac{\text{Diameter}}{2} = \frac{12 \text{ km}}{2} = 6 \text{ km} \][/tex]
3. Calculate the circumference of the circle (equator) of the pulsar:
- The circumference of a circle is given by the formula:
[tex]\[ \text{Circumference} = 2 \pi \times \text{Radius} \][/tex]
- Substituting the radius we found:
[tex]\[ \text{Circumference} = 2 \pi \times 6 \text{ km} = 37.69911184307752 \text{ km} \][/tex]
4. Determine the linear speed:
- Linear speed can be found by multiplying the circumference by the rotation rate:
[tex]\[ \text{Speed} = \text{Circumference} \times \text{Rotation rate} = 37.69911184307752 \text{ km} \times 30 \text{ rotations per second} \][/tex]
- Performing this calculation:
[tex]\[ \text{Speed} = 1130.9733552923256 \text{ km/s} \][/tex]
5. Round the result to 3 significant figures:
- Rounding 1130.9733552923256 to 3 significant figures:
[tex]\[ \text{Rounded speed} = 1130.973 \text{ km/s} \][/tex]
Therefore, the speed of a point on the equator of the pulsar, correct to 3 significant figures, is [tex]\( 1130.973 \text{ km/s} \)[/tex].
1. Understand the given information:
- The pulsar rotates 30 times per second.
- The diameter of the pulsar is 12 km.
2. Find the radius:
- The radius is half of the diameter. Thus, the radius is:
[tex]\[ \text{Radius} = \frac{\text{Diameter}}{2} = \frac{12 \text{ km}}{2} = 6 \text{ km} \][/tex]
3. Calculate the circumference of the circle (equator) of the pulsar:
- The circumference of a circle is given by the formula:
[tex]\[ \text{Circumference} = 2 \pi \times \text{Radius} \][/tex]
- Substituting the radius we found:
[tex]\[ \text{Circumference} = 2 \pi \times 6 \text{ km} = 37.69911184307752 \text{ km} \][/tex]
4. Determine the linear speed:
- Linear speed can be found by multiplying the circumference by the rotation rate:
[tex]\[ \text{Speed} = \text{Circumference} \times \text{Rotation rate} = 37.69911184307752 \text{ km} \times 30 \text{ rotations per second} \][/tex]
- Performing this calculation:
[tex]\[ \text{Speed} = 1130.9733552923256 \text{ km/s} \][/tex]
5. Round the result to 3 significant figures:
- Rounding 1130.9733552923256 to 3 significant figures:
[tex]\[ \text{Rounded speed} = 1130.973 \text{ km/s} \][/tex]
Therefore, the speed of a point on the equator of the pulsar, correct to 3 significant figures, is [tex]\( 1130.973 \text{ km/s} \)[/tex].
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.