Answer:
The answer is C)The force of gravity from Earth acting on the spacecraft decreased because the distance from Earth increased.
Explanation:
Gravity, a force, is dependent on the mass of the object exerting the gravity and the distance of an outside object from that object. The larger the object, the more gravity it will exert on an outside object. This force decreases as you move away from the object, but it will always still exist and never be equal to 0.
Actually Welcome to the Concept of the Projectile Motion.
Since, here given that, vertical velocity= 50m/s
we know that u*sin(theta) = vertical velocity
so the time taken to reach the maximum height or the time of Ascent is equal to
T = Usin(theta) ÷ g, here g = 9.8 m/s^2
so we get as,
T = 50/9.8
T = 5.10 seconds
thus the time taken to reach max height is 5.10 seconds.
Answer:
option a.
Explanation:
We can think of an atom as a nucleus (where the protons and neutrons are) and some electrons orbiting it.
We also know that the mass of an electron is a lot smaller than the mass of a proton or the mass of an electron.
So, if all the protons and electrons of an atom are in the nucleus, we know that most of the mass of an atom is in the nucleus of that atom.
Then we define the mass number, which is the total number of protons and neutrons in an atom. Such that the mass of a proton (or a neutron) is almost equal to 1u
Then if we define A as the total number of protons and neutrons, and each one of these weights about 1u
(where u = atomic mass unit)
Then the weight of the nucleus is about A times 1u, or:
A*1u = A atomic mass units.
Then the correct option is:
The mass of the nucleus is approximately EQUAL to the mass number multiplied by __1__ Atomic Mass unit.
option a.
KE = 1/2mv^2
KE= 1/2(2)(5)^2
KE= 25 J
Answer:
From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km
So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr
So runner towards the west will be
distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8
So equating east and west time we have
x/9= (11-x)/8
8x=99-9x
17x=99
x=5.92 km
That is the distance covered by runner towards the east and he will meet the runner toward the west at
6-5.92=0.08 km west of the flagpole.
