Answer:
18 grams
Explanation:
The mass of an object is the object's fundamental property. It is the amount or quantity of mater in the object. Irrespective of the object's location or the gravitational force applied on the object, the mass of an object remains the same. In other words, the mass of an object is the same everywhere.
The mass should however not be confused with its weight which depends on the mass and the gravitational force applied on it.
In summary, whether on earth or in the moon, the mass of an object remains the same. If the mass of the object is 18 grams on earth, it will also be 18 grams in the moon.
Answer:
20.1 m/s
Explanation:
Since You are later than usual getting to the stop and see the shuttle pulling away from the stop while you are still 3.9 m behind the bus stop. And In 40.9 m you will reach a barrier and you must catch the shuttle before that point.
Given that the shuttle has a constant acceleration of 4.5 m/s2.
The total distance to cover is:
Total distance = 40.9 + 3.9 = 44.8 m
Assuming you are starting from rest. Then initial velocity U = 0
Using the 3rd equation of motion to calculate the minimum velocity.
V^2 = U^2 + 2as
V^2 = 0 + 2 × 4.5 × 44.8
V^2 = 403.2
V = sqrt (403.2)
V = 20.1 m/s
Therefore, the minimum velocity you have to run at to catch the bus before it reaches the barrier is 20.1 m/s
Answer:
Air Resistance
Explanation:
If you were to drop both items on a plant without an atmosphere, they would both hit the ground at the same time. Since a feather doesn't have much mass compared to the hammer, it takes more time for the feather to "push" itself through and overcome the opposite push from the air
Answer:
A) The work done by the engine is: 6.8MJ/L
B) The fuel efficiency is ![11.53{\frac{km}{L}](https://tex.z-dn.net/?f=11.53%7B%5Cfrac%7Bkm%7D%7BL%7D)
Explanation:
A)
We know that the gasoline releases about 3.4*10^7 J of energy for each liter, and about 80% of that energy is lost as heat; it means that the other 20% of the energy released is taken for the engine to do work. In that sense, the work done by the engine is 20% of the 3.4*10^7 J that the gasoline releases for 1 liter, so:
![W_E=3.4*10^7J*\frac{20}{100}=6.8*10^6J =6.8MJ](https://tex.z-dn.net/?f=W_E%3D3.4%2A10%5E7J%2A%5Cfrac%7B20%7D%7B100%7D%3D6.8%2A10%5E6J%20%3D6.8MJ)
This last can be seen as a conversion factor, where we multiply the energy released by the gasoline by the factor (20 J taken for do work for each 100 J released).
B) We know that the car requires 5.9*10^5 J of work <u>for each km traveled</u>. That is the energy that the car requires, but it is not the energy that you have to give to the car; take in mind that the energy that you put in the car in gasoline liters will be not taken all, but just 20%. Also we know that the work done by the engine for 1 liter of gasoline is 6.8MJ, and that is just the work taken for do work (the useful energy), so we can connect both data:
![{\frac{1km}{5.9*10^5J}*\frac{6.8*10^6J}{1L}=11.53\frac{km}{L}](https://tex.z-dn.net/?f=%7B%5Cfrac%7B1km%7D%7B5.9%2A10%5E5J%7D%2A%5Cfrac%7B6.8%2A10%5E6J%7D%7B1L%7D%3D11.53%5Cfrac%7Bkm%7D%7BL%7D)
The first fraction,
is the ratio or the proportion of (1 km requieres 5.9*10^5 J); and we multiply by the second fraction
, which is the ratio: 6.8*10^6 J of work done for each liter of gasoline.