<span>To relate or measure the by the quantity of something, not against the quantity</span>
Answer:
I am pretty sure it is C
Explanation:
It can be found all over the universe
The correct answer is<span> gases, energy, temperature, phases
Gravity and nuclear forces are not encompassed in the kinetic molecular theory as it deals with movement and behavior of gas molecules. It does not include their conversion to other types of energy or anything similar. </span>
The correct answer is
C. Light can pass through Object B faster than it can pass through Object A.
In fact, the index of refraction of a material is defined as:
where c is the speed of light in vacuum and v is the speed of light in the material. Rearranging the equation, we can write the speed of light in the material as:
So we that, the smaller the refractive index n, the greater the speed of light in the material, v. In this problem, object B has lower refractive index than object A, so light travels faster in object B.
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = / vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m