Connect with knowledgeable experts and enthusiasts on IDNLearn.com. Discover prompt and accurate answers from our community of experienced professionals.
Sagot :
Let's solve the given questions one by one in a detailed manner:
26. An object having a mass of 24 kg on Earth is taken to a planet where the acceleration due to gravity is half of its value on Earth. The mass of the object on the planet is:
- The mass of an object does not change with the location. It remains the same regardless of whether it is on Earth or another planet.
- Therefore, the mass of the object on the planet is still [tex]\( \boxed{24 \text{ kg}} \)[/tex].
27. An object of mass 40 kg will experience a gravitational force of 68 N on a planet. What is the intensity of the gravitational field on this planet?
- The force experienced by an object in a gravitational field is given by [tex]\( F = mg \)[/tex], where [tex]\( F \)[/tex] is the force, [tex]\( m \)[/tex] is the mass, and [tex]\( g \)[/tex] is the gravitational field intensity.
- Given: [tex]\( F = 68 \text{ N} \)[/tex], [tex]\( m = 40 \text{ kg} \)[/tex]
- Solving for [tex]\( g \)[/tex]:
[tex]\[ g = \frac{F}{m} = \frac{68 \text{ N}}{40 \text{ kg}} = 1.7 \text{ N/kg} \][/tex]
- Therefore, the answer is [tex]\( \boxed{1.7 \text{ N/kg}} \)[/tex].
28. The mass of a certain planet [tex]\( P \)[/tex] is one-hundredth that of Earth while its radius is one-quarter of Earth's radius. If the acceleration due to gravity on Earth is [tex]\( 10 \text{ m/s}^2 \)[/tex], what is its value on [tex]\( P \)[/tex]?
- Let [tex]\( M_e \)[/tex] and [tex]\( R_e \)[/tex] be the mass and radius of Earth, respectively.
- Given: [tex]\( M_p = \frac{M_e}{100} \)[/tex] and [tex]\( R_p = \frac{R_e}{4} \)[/tex]
- The acceleration due to gravity [tex]\( g \)[/tex] is given by [tex]\( g = \frac{GM}{R^2} \)[/tex].
- For planet [tex]\( P \)[/tex], the acceleration due to gravity [tex]\( g_p \)[/tex] can be found using:
[tex]\[ g_p = \frac{G \cdot M_p}{R_p^2} = \frac{G \cdot \frac{M_e}{100}}{\left(\frac{R_e}{4}\right)^2} = \frac{G \cdot \frac{M_e}{100}}{\frac{R_e^2}{16}} = \frac{G \cdot M_e \cdot 16}{100 \cdot R_e^2} = \frac{16 \cdot GM_e}{100 \cdot R_e^2} \][/tex]
- Since [tex]\( g_e = \frac{GM_e}{R_e^2} = 10 \text{ m/s}^2 \)[/tex]:
[tex]\[ g_p = \frac{16}{100} \cdot g_e = 0.16 \cdot 10 \text{ m/s}^2 = 1.6 \text{ m/s}^2 \][/tex]
- Therefore, the answer is [tex]\( \boxed{1.6 \text{ m/s}^2} \)[/tex].
29. The escape velocity of a body on the surface of a planet of radius [tex]\( R \)[/tex] and mass [tex]\( M \)[/tex] is [ [tex]\( G \)[/tex] = universal gravitational constant ]:
- The formula for escape velocity [tex]\( v \)[/tex] is derived from the equation:
[tex]\[ v = \sqrt{\frac{2GM}{R}} \][/tex]
- Therefore, the correct answer is [tex]\( \boxed{\sqrt{\frac{2GM}{R}}} \)[/tex].
30. The radius of the Earth is [tex]\( 6.4 \times 10^6 \)[/tex] m and the acceleration due to gravity is [tex]\( 10.0 \text{ m/s}^2 \)[/tex]. The escape velocity of a rocket launched from the Earth's surface is:
- Using the escape velocity formula:
[tex]\[ v = \sqrt{\frac{2GM}{R}} \][/tex]
- As per the given data:
[tex]\[ v = 11313.70849898476 \text{ m/s} = 11.313708499 \text{ km/s} \][/tex]
- Therefore, the correct answer is [tex]\( \boxed{11.3 \text{ km/s}} \)[/tex].
26. An object having a mass of 24 kg on Earth is taken to a planet where the acceleration due to gravity is half of its value on Earth. The mass of the object on the planet is:
- The mass of an object does not change with the location. It remains the same regardless of whether it is on Earth or another planet.
- Therefore, the mass of the object on the planet is still [tex]\( \boxed{24 \text{ kg}} \)[/tex].
27. An object of mass 40 kg will experience a gravitational force of 68 N on a planet. What is the intensity of the gravitational field on this planet?
- The force experienced by an object in a gravitational field is given by [tex]\( F = mg \)[/tex], where [tex]\( F \)[/tex] is the force, [tex]\( m \)[/tex] is the mass, and [tex]\( g \)[/tex] is the gravitational field intensity.
- Given: [tex]\( F = 68 \text{ N} \)[/tex], [tex]\( m = 40 \text{ kg} \)[/tex]
- Solving for [tex]\( g \)[/tex]:
[tex]\[ g = \frac{F}{m} = \frac{68 \text{ N}}{40 \text{ kg}} = 1.7 \text{ N/kg} \][/tex]
- Therefore, the answer is [tex]\( \boxed{1.7 \text{ N/kg}} \)[/tex].
28. The mass of a certain planet [tex]\( P \)[/tex] is one-hundredth that of Earth while its radius is one-quarter of Earth's radius. If the acceleration due to gravity on Earth is [tex]\( 10 \text{ m/s}^2 \)[/tex], what is its value on [tex]\( P \)[/tex]?
- Let [tex]\( M_e \)[/tex] and [tex]\( R_e \)[/tex] be the mass and radius of Earth, respectively.
- Given: [tex]\( M_p = \frac{M_e}{100} \)[/tex] and [tex]\( R_p = \frac{R_e}{4} \)[/tex]
- The acceleration due to gravity [tex]\( g \)[/tex] is given by [tex]\( g = \frac{GM}{R^2} \)[/tex].
- For planet [tex]\( P \)[/tex], the acceleration due to gravity [tex]\( g_p \)[/tex] can be found using:
[tex]\[ g_p = \frac{G \cdot M_p}{R_p^2} = \frac{G \cdot \frac{M_e}{100}}{\left(\frac{R_e}{4}\right)^2} = \frac{G \cdot \frac{M_e}{100}}{\frac{R_e^2}{16}} = \frac{G \cdot M_e \cdot 16}{100 \cdot R_e^2} = \frac{16 \cdot GM_e}{100 \cdot R_e^2} \][/tex]
- Since [tex]\( g_e = \frac{GM_e}{R_e^2} = 10 \text{ m/s}^2 \)[/tex]:
[tex]\[ g_p = \frac{16}{100} \cdot g_e = 0.16 \cdot 10 \text{ m/s}^2 = 1.6 \text{ m/s}^2 \][/tex]
- Therefore, the answer is [tex]\( \boxed{1.6 \text{ m/s}^2} \)[/tex].
29. The escape velocity of a body on the surface of a planet of radius [tex]\( R \)[/tex] and mass [tex]\( M \)[/tex] is [ [tex]\( G \)[/tex] = universal gravitational constant ]:
- The formula for escape velocity [tex]\( v \)[/tex] is derived from the equation:
[tex]\[ v = \sqrt{\frac{2GM}{R}} \][/tex]
- Therefore, the correct answer is [tex]\( \boxed{\sqrt{\frac{2GM}{R}}} \)[/tex].
30. The radius of the Earth is [tex]\( 6.4 \times 10^6 \)[/tex] m and the acceleration due to gravity is [tex]\( 10.0 \text{ m/s}^2 \)[/tex]. The escape velocity of a rocket launched from the Earth's surface is:
- Using the escape velocity formula:
[tex]\[ v = \sqrt{\frac{2GM}{R}} \][/tex]
- As per the given data:
[tex]\[ v = 11313.70849898476 \text{ m/s} = 11.313708499 \text{ km/s} \][/tex]
- Therefore, the correct answer is [tex]\( \boxed{11.3 \text{ km/s}} \)[/tex].
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Thank you for visiting IDNLearn.com. We’re here to provide accurate and reliable answers, so visit us again soon.
