The Mercury's mass for the given acceleration due to gravity is 0.3152 x 10²⁴ kg.
The ratio of the calculated and accepted value of the Mercury's mass is 0.95.
<h3>What is mass?</h3>
Mass is the amount of matter present in the object.
The mass of the object is always constant, anywhere it is on the Earth or Moon or any other planet.
Given is the acceleration due to gravity of Mercury planet at North pole is g = 3.698 m/s² and the radius of Mercury planet is 2440 km.
The acceleration due to gravity is related with mass as
g = GM/R²
Substitute the values, we have
3.698 = 6.67 x 10⁻¹¹ x M/(2440 x1000)³
M = 2.2016 x 10¹³ / 6.67 x 10⁻¹¹
M = 0.3152 x 10²⁴ kg
Thus, the mercury's mass is 0.3152 x 10²⁴ kg.
(b) Accepted value of Mercury's mass is 3.301 x 10²³ kg
Ratio of the value of mass calculated and accepted is
Mcalc/M accep = 0.3152 x 10²⁴ kg / 3.301 x 10²³ kg
= 0.95
Thus, the ratio is 0.95
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Answer:11
Explanation: because is the number that repeted 2 times
- The mechanic did 5406 Joules of work pushing the car.
That's the energy he put into the car. When he stops pushing, all the energy he put into the car is now the car's kinetic energy.
- Kinetic energy = (1/2) (mass) (speed²)
And there we have it
- The car's mass is 3,600 kg.
- Its speed is 'v' m/s .
- (1/2) (mass) (v²) = 5,406 Joules
(1/2) (3600 kg) (v²) = 5406 joules
1800 kg (v²) = 5406 joules
v² = (5406 joules) / (1800 kg)
v² = (5406/1800) (joules/kg)
= = = = = This section is just to work out the units of the answer:
- v² = (5406/1800) (Newton-meter/kg)
- v² = (5406/1800) (kg-m²/s² / kg)
= = = = =
v = √(5406/1800) m/s
<em>v = 1.733 m/s</em>
To solve this problem we will apply the concepts related to gravitational potential energy.
This can be defined as the product between mass, gravity and body height.
Mathematically it can be expressed as
Therefore the change in the internal energy of the system is 255.78
When a body performs a uniform circular motion, the direction of the velocity vector changes at every moment. This variation is experienced by the linear vector, due to a force called centripetal, directed towards the center of the circle that gives rise to centripetal acceleration, the mathematical expression is given as,
Where,
v = Tangential Velocity
r = Radius
The linear velocity was 2010m/s in a radius of 0.159m, then the centripetal acceleration is
Therefore the centripetal acceleration of the end of the rod is 
