B Longitude
Explanation:
The line of longitude or longitude does not affects climates.
Climate is the weather condition over a place for a long period of time. It usually takes several years to delineate the climate of an area.
- Longitude is the distance on the earth surface that runs form east to west.
- All lines of longitude are all great circles.
- Latitude are lines from north to south on the earth surface. Only the equator is a great circle.
- Latitude affects climate. In fact based on the division of lines of latitude on can predict climatic patterns on earth.
- Proximity to water bodies also affects climatic condition because of the effect of land and sea breeze and other climatic factors.
- Elevation affects climate a whole lot. Physiography of a place determines the weather to a very large extent.
learn more:
Climate brainly.com/question/10856870
#learnwithBrainly
true true tire true tire ture
please mark brainlist
Answer:
Vibrations of the eardrum are in turn translated into oscillations of tiny bones (ossicles) found within the middle ear. The Hammer, Anvil, and Stirrup magnify the oscillations and relay this information to the chamber of the inner ear.
Multiply by (1000 meters / 1 km).
Then multiply by (1 hour / 3600 seconds).
Both of those fractions are equal to ' 1 ', because the top
and bottom numbers are equal, so the multiplications
won't change the VALUE of the 72 km/hr. They'll only
change the units.
(72 km/hour) · (1000 meters / 1 km) · (1 hour / 3600 seconds)
= (72 · 1000 / 3600) (km·meter·hour / hour·km·second)
= 20 meter/second
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