Answer:
45.89m/s²
Explanation:
Given
Distance S = 305m
Time t = 3.64s
To get the acceleration during this run, we will apply the equation of motion:
S = ut+1/2at²
Substitute the given parameters into the formula and calculate the value of a
305 = 0+1/2 a(3.64)²
304 = 1/2(13.2496)a
304 = 6.6248a
a = 304/6.6248
a = 45.89m/s²
Hence the average acceleration during this run is 45.89m/s²
Answer:
The final acceleration of the car, v = 70 m/s
Explanation:
Given,
The initial velocity of the car, u = 20 m/s
The acceleration of the car, a = 10 m/s²
The time period of travel, t = 5 s
Using the I equations of motion
v = u + at
= 20 + 10(5)
= 20 + 50
= 70 m/s
Hence, the final acceleration of the car, v = 70 m/s
It has 99 electrons because an element has the same number of protons and electrons
(1 parsec) is the distance at which an object has a parallax of 1 arcsecond. The distance is about 3.26 light years.
Another way to understand it is: The distance from which the Earth's orbit appears 1 arcsecond across.
For a parallax angle of 1/2 arcsecond, the distance is <em>2 parsecs </em>(about 6.52 light years).
1 arcsecond is 1/3600 of a degree, 0.00028 degree.
These objects would be classified as extreme trans Neptunian object (ETNO).
Explanation:
ETNO’s are the objects lying beyond the planet Neptune and orbiting the Sun. They follow a highly eccentric path which is tilted. ETNO has been grouped into three major according to their respective perihelia.
Within this region (beyond Neptune’s orbit), a hypothetical planet has been discovered. It was discovered following its gravitational effect on the other objects of Kuiper Belt (region beyond the orbit of the Neptune- the last planet of our Solar system)
The Planet is assumed to be around 2 times the Earth’s size and around 10 times heavier than Earth.
