Answer:
Because of the knowledge of <u>relative size</u>, it will be assumed that the smaller jetliner is farther away.
Explanation:
According to the theory of relative size, the distance that an object has to the viewing individual affects the perception of the individual regarding the size of the object.
As stated in this case, one of the jetliners is farther away from the other. Therefore, even if the jets are of equal size, the one that is at a greater distance is perceived to be smaller as it is at a greater viewing range. The one that is nearer to the individual seems bigger in comparison to the one farther away due to a closer viewing range.
Therefore, the jet that is nearer appears larger.
To know more about relative size, refer to:
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<span>The surface charge density = q/A
So q = surface charge density x Area
The surface area of a sphere of radius R is 4*Pi*R^2. R = d/2 where d is diameter. This leaves us with 1.3/2 = 0.65. Area = 4 * pie * (0.65)^2 = 5.30998.
So the net charge q = 8.1 * 10^(-6) * 5.30998 = 42.47998 * 10^(-6)
The Total electric flux = Q/e_0 where , 8.854 Ă— 10â’12, e_0 is permitivity of free space.
So Flux = 42.47998 * 10^(-6) / 8.854 * 10(â’12) = 4.833 * 10^(-6 - (-12)) = 4.833 * 10^(6)</span>
1,000 W = 1 kW
100 W = 0.1 kW
(0.1 kW) x (6 h) = 0.6 kWh <=== energy
(0.6 kWh) x (£0.1359/kWh) = £0.0815 <=== cost of it
A. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π
b. vy = 4π cos (4πt + π/2)
vy = 0
c. m = sin(4πt + π/2) / [<span>πt + cos(4πt + π/2)]
d. m = </span>sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]
e. t = -1.0
f. t = -0.35
g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax
h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax
i. s(t) = [<span>x(t)^2 + y</span>(t)^2]^(1/2)
h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt
k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.
Answer:
h = R/4 = 1.5925 x 10⁶ m = 1592.5 km
Explanation:
The variation in the value of acceleration due to gravity value with respect to the altitude is given by the following general formula:
![g' = g(1-2\frac{h}{R})](https://tex.z-dn.net/?f=g%27%20%3D%20g%281-2%5Cfrac%7Bh%7D%7BR%7D%29)
where,
h = altitude from surface of earth = ?
g = acceleration due to gravity on surface of earth = 9.81 m/s²
R = Radius of Earth = 6.37 x 10⁶ m
g' = acceleration due to gravity at given altitude = g/2
Therefore,
![\frac{g}{2} = g(1-2\frac{h}{R})\\\\2\frac{h}{R} = 1-\frac{1}{2}\\\\2\frac{h}{R} = \frac{1}{2}\\\\\frac{h}{R} = \frac{1}{4}\\\\](https://tex.z-dn.net/?f=%5Cfrac%7Bg%7D%7B2%7D%20%20%3D%20g%281-2%5Cfrac%7Bh%7D%7BR%7D%29%5C%5C%5C%5C2%5Cfrac%7Bh%7D%7BR%7D%20%3D%201-%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C2%5Cfrac%7Bh%7D%7BR%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%5Cfrac%7Bh%7D%7BR%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%5C%5C%5C%5C)
<u>h = R/4 = 1.5925 x 10⁶ m = 1592.5 km</u>