They get it from the experiment.
According to the law of reflection, when a ray of light reflects off a surface, the angle of incidence is equal to the angle of reflection.
So in the above situation, if the incident ray makes an angle of 65° with respect to the normal to the mirror's surface, the angle of reflected ray will also be equal to 65 degrees. The attached image will explain this better.
So the answer is,
c. 65 degrees
Answer:
0.103 m/s to the south
Explanation:
The total momentum of the launcher+ball system must be conserved before and after the launch, so we can write:
where
is the total initial momentum (before the launch)
is the mass of the launcher
is the velocity of the launcher after the launch
is the mass of the ball
is the velocity of the ball after the launch (we take the north direction as positive)
Solving for , we find
and the negative sign means that the direction is south.
No because an atom consists of <u>two</u> main parts <em>and</em> <u>three</u> subatomic particles - protons, neutrons, electrons. Each one is smaller than an atom, therefore they are subatomic particles. An atom only requires protons and electrons to be an atom - e.g. Hydrogen has 1 proton and 1 electron. Neutrons do not affect the overall charge of the atom, and only increase the atomic mass.
Answer:
F= 4788 N
Explanation:
Because the car moves with uniformly accelerated movement we apply the following formula:
vf²=v₀²+2*a*d Formula (1)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
d=36.9 m
v₀=14.0 m/s m/s
vf= 0
Calculating of the acceleration of the car
We replace dta in the formula (1)
vf²=v₀²+2*a*d
(0)²=(14)²+2*a*(36.9)
-(14)²= (73.8) *a
a= - (196) / (73.8)
a= - 2.66 m/s²
Newton's second law of the car in direction horizontal (x):
∑Fx = m*ax Formula (2)
∑F : algebraic sum of the forces in direction x-axis (N)
m : mass (kg)
a : acceleration (m/s²)
Data
m=1800 Fkg
a= - 2.66 m/s²
Magnitude of the horizontal net force (F) that is required to bring the car to a halt in a distance of 36.9 m :
We replace data in the formula (2)
-F= (1800 kg) * ( -2.66 m/s²
)
F= 4788 N