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
Answer:
<em>3924 Pa</em>
<em></em>
Explanation:
Volume of cylinder = 2 L = 0.002 m^3 (1000 L = 1 m^3)
diameter of the inner cylinder = 8 cm = 0.08 m (100 cm = 1 m)
radius of the inner cylinder = diameter/2 = 0.08/2 = 0.04 m
area of the inner cylinder = 
where 
= 3.142,
and r = radius = 0.04 m
area of inner cylinder = 3.142 x 
= 0.005 m^2
<em>height h of the water in this cylinder = volume/area</em>
h = 0.002/0.005 = 0.4 m
<em>pressure at the bottom of the cylinder due to the height of water = pgh</em>
where
p = density of water = 1000 kg/m^3
g = acceleration due to gravity = 9.81 m/s^2
h = height of water within this cylinder = 0.4 m
pressure = 1000 x 9.81 x 0.4 = <em>3924 Pa</em>
Answer:
Jesseca wanted to create a material that reflected most of the light that fell on it.
Explanation: The Graphite was the material in the passage that had reflected most of the light.
<span>E=hc/wav. len
E = (6.62 x 10^-34 x 3 x 10^8)/0.0275 x 10^-9
E = 7.22182 x 10^-15 J
To convert to eV divide by 1.6 x 10^-19
E = 7.22182 x 10^-15/1.6 x 10^-19 eV
E =45.36 x 10^3 eV
Th energy, E, of a single x-ray photon in eV is = 45.36keV.
Number of photons, n = total energy/ energy of photon
n = 3.85 x 10^-6/7.22182 x 10^-15
n = 5.33 x 10^8 photons </span>
That was a lucky pick.
Twice each each lunar month, all year long, whenever the Moon,
Earth and Sun are aligned, the gravitational pull of the sun adds
to that of the moon causing maximum tides.
This is the setup at both New Moon and Full Moon. It doesn't matter
whether the Sun and Moon are both on the same side of the Earth,
or one on each side. As long as all three bodies are lined up, we
get the biggest tides.
These are called "spring tides", when there is the greatest difference
between high and low tide.
At First Quarter and Third Quarter, when the sun, Earth, and Moon form a
right angle, there is the least difference between high and low tide. Then
they're called "neap tides".
