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Snowcat [4.5K]
3 years ago
10

What latitudes experience both summer and winter?

Physics
1 answer:
sergiy2304 [10]3 years ago
7 0

There are only two times of the year when the Earth's axis is tilted neither toward nor away from the sun, resulting in a equal amount of daylight and darkness at all latitudes. These events are referred to as Equinoxes.The word equinox is derived from two Latin words - aequus (equal) and nox (night). At the equator, the sun is directly overhead at noon on these two equinoxes. How do the climatological temperature patterns follow the four seasons? The temperatures are pretty much the same when comparing early winter to late winter and pretty much the same when comparing early summer to late summer. However, late Fall is much colder than early Fall and late Spring is much warmer than early Spring. Why is that? The reason is linked to the changing of the solar sun angle during each season. In the Fall, the sun angle is always getting lower and in the Spring it is always getting higher. From this it would make sense that late Fall is much colder than early Fall and late Spring is much warmer than early Spring.  In the winter, the sun angle is INCREASING and in the summer the sun angle is DECREASING. This fact causes temperatures to be pretty much the same when comparing early winter to late winter and pretty much the same when comparing early summer to late summer. The cumulative cooling of the ground and ocean is partially offset by an increasing sun angle in the winter and the cumulative warming of the ground and ocean is partially offset by a decreasing sun angle in the summer. This causes the Spring and Fall seasons to seem much quicker in duration than the Summer and Winter seasons.

<h2><u>I know this is long... but does this help??</u></h2>
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S ship maneuvers to within 2.50x10^3 m of an islands 1.80x10^3 m high mountain peak and fires a projectile at an enemy ship 6.10
lilavasa [31]

velocity of the projectile is given as

v = 2.50 \times 10^3 m/s

now the angle is given as 75 degree

so here we will have

v_x = 2.50 \times 10^3 cos75 = 0.65 \times 10^3 m/s

v_y = 2.50 \times 10^3 sin75 = 2.4 \times 10^3 m/s

now the time taken to reach the peak is given as

t = \frac{x}{v_x}

t = \frac{2.50 \times 10^3}{0.65 \times 10^3} = 3.86 s

now the height moved by the projectile in same time is given as

y = v_y t + \frac{1}{2}at^2

now we have

y = (2.4 \times 10^3)(3.86) - \frac{1}{2}(9.81)(3.86)^2

y = 9.2 \times 10^3 m

so distance between the peak and projectile is given as

d = 9.2 \times 10^3 - 1.80 \times 10^3 = 7.4 \times 10^3 m

5 0
3 years ago
A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The t
azamat

Answer:

(a) x'(t)= 142

(b) 142

(c) y'(t)= -32t+5

(d) 96.8 mph

(e) 0.426 s

(f) 0.061 rad

Explanation:

Velocity is a time-derivative of position.

(a) x(t) = 142t

x'(t)= 142

(b) Since x'(t)= 142 is independent of t, it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.

(c) y(t) = - 16t^2+5t+5

y'(t)= -32t+5

(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to x(t).

x(t)= 142t = 60.5

t=\dfrac{60.5}{142}= 0.426.

In this time, the vertical velocity, y'(t) is

y(t)= -32\times0.426+5 = -8.632

The speed of the ball at thus point is s=\sqrt{142^2+(-8.632)^2}=142 ft/s

To convert this to mph, we multiply the factor 3600/5280

s=142\times\dfrac{3600}{5280}=96.8 \text{ mph}

(e) The time has been determined from (d) above.

t= 0.426

(f) This angle is given by

\theta=\tan^{-1}\dfrac{y'(t)}{x'(t)}

\theta=\tan^{-1}\dfrac{-8.632}{142}=\tan^{-1}-0.0607=3.47 (Note here we are considering the acute angle so we ignore the negative sign)

In radians, this is

\theta=3.47\times\dfrac{\pi}{180}=0.061 \text{ rad}

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Answer:

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