Answer:
Explanation:
We shall find the final velocity of aircraft with respect to aircraft carrier using the following relation.
v² = u² + 2as
v² = 0² + 2 x 59 x 97
v² = 11446
v = 107 m /s
velocity of aircraft carrier = 1.852 x 26 = 48.152 km/h
= 48.152 x 1000 / (60 x 60) m/s
= 13.37 m /s
This velocity of aircraft carrier will be added to the velocity of aircraft .
So absolute velocity of aircraft = 107 m /s + 13.37 m/s
= 120.37 m/s
Answer
given,
k = 250 N/m
q = 900 N/m³
(FSp)s=−kΔs−q(Δs)^3
work done = Force x displacement
limits are x = 0 to x = 0.15 m
work done
![W = [\dfrac{kx^2}{2}+\dfrac{qx^4}{4}+ C]_0^0.15](https://tex.z-dn.net/?f=W%20%3D%20%5B%5Cdfrac%7Bkx%5E2%7D%7B2%7D%2B%5Cdfrac%7Bqx%5E4%7D%7B4%7D%2B%20C%5D_0%5E0.15)
W = 3.375 + 0.1139
W = 3.3488 J
b) % cubic term =
% cubic term =
Answer:
Explanation:
When a camera shifts focus from a faraway object to a nearby object, the lens-to-film distance must increase. Likewise, when it shifts focus from a nearby object to a distant object, there must be an increase in the lens to film distance (that is, the image distance).
Therefore, if the picture of an object that is far away, the lens must move towards the film.
The focal length cannot be changed because it is fixed for a lens. Nevertheless, in order to focus on an object, the image distance can be changed.
Λ= V/f
<span>but change it to represent the speed of light, c </span>
<span>λ= c/f </span>
<span>c = 3.00 x 10^8 m/s </span>
<span>Plug in your given info and solve for λ(wavelength) </span>
<span>λ= 3.00 x 10^8 m/s / 7.5 x 10^14 Hz
(3.00 x 10^8) / (7.5 x 10^14) = 300,000,000 / 750,000,000,000,000 = 0.0000004
Hope this helps :)
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