The average velocity can be calculated using the formula:
v = d / t
For the 1st car, the velocity is calculated
as:
v1 = 8.60 m / 1.80 s = 4.78 m / s
While that of the 2nd car is:
v2 = 8.60 m / 1.66 s = 5.18 m / s
Now we can solve for the acceleration using the formula:
v2^2 = v1^2 + 2 a d
Rewriting in terms of a:
a = (v2^2 – v1^2) / 2 d
a = (5.18^2 – 4.78^2) / (2 * 8.6)
a = 0.23 m/s
Therefore the train has a constant acceleration of about
0.23 meters per second.
The water formed on the surface of the water evaporation loss (evaporation), consisting of plant transpiration water loss (transpiration) is called. Soil near the plant and the resulting water loss is called by evapotranspiration.
57 nuetrons and 44 elctrons just incase you need it there are 44 protons too.Hope this halps
Answer:
Earth: 22.246 N
Moon: 3.71 N
Jupiter: 58.72 N
Explanation:
The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.
To calculate anything, we need to convert to standard measurements.
5.00 lbs = 2.27 kg
On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N
Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N
Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N
The question is incomplete.
The distance between the Moon and Earth influences: 1) the attractive gravitational force between them, 2) the tides, 3) the eclipses, 4) the period of each full turn of the moon around the Earth.
Assuming the question refers to the gravitational attraction, we must use the fact that, as per, Newton's Universal Gravitaional Law, the attractive force between the two bodies is inversely related to the square distance that separates them.
Then, if the Moon were twice as far, the gravitational pull would be one fourth (1/4) of actual pull.
