Answer:
(a): The car's relative position to the base of the cliff is x= 32.52m.
(b): The lenght of the car in the ir is tfall= 1.78 sec.
Explanation:
Vo= 0
V= ?
d= 50m
h= 30m
a= 4 m/s²
t= √(2*d/a)
t= 5 sec
V= a*t
V= 20 m/s
Vx= V * cos(24º)
Vx= 18.27 m/s
Vy= V* sin(24º)
Vy= 8.13 m/s
h= Vy*t + g*t²/2
clearing t:
tfall= 1.78 sec (b)
x= Vx * tfall
x= 32.52 m (a)
The impulse required to decrease the speed of the boat is equal to the variation of momentum of the boat:

where
m=225 kg is the mass of the boat

is the variation of velocity of the boat
By substituting the numbers into the first equation, we find the impulse:

and the negative sign means the direction of the impulse is against the direction of motion of the boat.
Answer:
Depends.
Explanation:
Whether the object is going left or right, the speed will stay the same until friction eventually stops it. <em>However, </em>if, for example, we're talking about an object going straight before veering right, then yes, speed <em>does</em> matter. An object will normally have to speed up or slow down momentarily when changing direction to keep itself sustained on the ground.
So, honestly? It really depends on what we're talking about!
Hope this helped!
Source(s) used: None.
Answer:
h = 3.3 m (Look at the explanation below, please)
Explanation:
This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy
Kinetic energy =
m
Plug in the numbers =
(4.0)(
)
Solve = 2(64) = 128 J
Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.
Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)
Kinetic energy = Potential Energy
128 J = 39.2h
h = 3.26 m
h= 3.3 m (because of significant figures)
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.