The air pressure ( or atmospheric pressure ) is the force of air over a unit of area. Changes in the air pressure causes the weather changes. High pressure usually brings good weather with dry and cool air. But in a low pressure zone warm air is rising up. This vertical movements are caused by winds high in the troposphere. Water molecules stay as a gas in warmer air. After the vertical movement they condense and bring steady continuous rain. Therefore the low pressure brings cloudly and rainy weather. Answer: The air pressure is most likely low<span>. </span>
Answer: The height of the cloud = 394.55 m
Explanation:
The observer is 500m away from the spotlight.
Let x be the distance from the observer to the interception of the segment of the height, h with the floor. The equations are thus:
Tan 45° = h/x ... eq1
Tan 75° = h/(500- x ) ... eq2
From eq 1, Tan 45° = 1, therefore eq1 becomes:
h = x ... eq3
Put eq3 into eq2
Tan 75° = h/(500- h)
h = ( 500 - h ) Tan 75°
h = 500Tan 75° - hTan75°
h + h Tan 75° = 500 Tan 75°
h ( 1 + Tan 75° ) = 500 Tan75°
h = 500Tan75°/ (1 + Tan 75°)
h= 1866.02 / 4.73
h = 394.55m
<h2>
Answer: 7020.117 m/s</h2>
Explanation:
The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:
(1)
Where:
is the gravity constant
the mass of the massive body around which the satellite is orbiting, in this case, the Earth
.
the radius of the orbit (measured from the center of the planet to the satellite).
This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth 
and the altitude of the satellite above the Earth's surface 
.
Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).
Now, rewriting equation (1) with the known values:
<span>(a) 12.02 m/s
(b) 52.2 meters
This problem is an example of integral calculus. You've been given an acceleration vector which is usually known as the 2nd derivative. From that you need to calculate the velocity function (1st derivative) and position (actual function) by successively calculating the anti-derivative. So:
A(t) = 6.30 - 2.20t
V(t) = 6.30t - 1.10t^2 + C
We now have a velocity function, but need to determine C. Since we've been given the velocity at t = 0, that's fairly trivial.
V(t) = 6.30t - 1.10t^2 + C
3 = 6.30*0 - 1.10*0^2 + C
3 = 0 + 0 + C
3 = C
So the entire velocity function is:
V(t) = 6.30t - 1.10t^2 + 3
V(t) = -1.10t^2 + 6.30t + 3
Now for the location function which is the anti-derivative of the velocity function.
V(t) = -1.10t^2 + 6.30t + 3
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + C
Now we need to calculate C. And once again, we've been given the location for t = 0, so
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + C
7.3 = -0.366666667*0^3 + 3.15*0^2 + 3*0 + C
7.3 = 0 + 0 + 0 + C
7.3 = C
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + 7.3
Now that we have the functions, they are:
A(t) = 6.30 - 2.20t
V(t) = -1.10t^2 + 6.30t + 3
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + 7.3
let's answer the questions.
(a) What is the maximum speed achieved by the cyclist?
This can only happen at those points that meet either of the following criteria.
1. The derivative is undefined for the point.
2. The value of the derivative is 0 for the point.
As it turns out, the 1st derivative of the velocity function is the acceleration function which we have. So
A(t) = 6.30 - 2.20t
0 = 6.30 - 2.20t
2.20t = 6.30
t = 2.863636364
So one of V(0), V(2.863636364), or V(6) will be the maximum value. Therefore:
V(0) = 3
V(2.863636364) = 12.0204545454545
V(6) = 1.2
So the maximum speed achieved is 12.02 m/s
(b) Total distance traveled?
L(0) = 7.3
L(6) = 59.5
Distance traveled = 59.5 m - 7.3 m = 52.2 meters</span>
