This question involves the concepts of the thin lens formula, focal length, and image distance.
The image of the house is "16 ft" away from the lens.
According to the thin lens formula:
![\frac{1}{f}=\frac{1}{p}+\frac{1}{q}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7Bp%7D%2B%5Cfrac%7B1%7D%7Bq%7D)
where,
f = focal length = 8 ft
p = object distance = 16 ft
q = image distance = ?
Therefore,
![\frac{1}{8\ ft}=\frac{1}{16\ ft}+\frac{1}{q}\\\\\frac{1}{q}=\frac{1}{8\ ft}-\frac{1}{16\ ft}\\\\\frac{1}{q}=0.125\ ft^{-1}-0.0625\ ft^{-1}\\\\q=\frac{1}{0.0625\ ft^{-1}}\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B8%5C%20ft%7D%3D%5Cfrac%7B1%7D%7B16%5C%20ft%7D%2B%5Cfrac%7B1%7D%7Bq%7D%5C%5C%5C%5C%5Cfrac%7B1%7D%7Bq%7D%3D%5Cfrac%7B1%7D%7B8%5C%20ft%7D-%5Cfrac%7B1%7D%7B16%5C%20ft%7D%5C%5C%5C%5C%5Cfrac%7B1%7D%7Bq%7D%3D0.125%5C%20ft%5E%7B-1%7D-0.0625%5C%20ft%5E%7B-1%7D%5C%5C%5C%5Cq%3D%5Cfrac%7B1%7D%7B0.0625%5C%20ft%5E%7B-1%7D%7D%5C%5C%5C%5C)
<u>q = 16 ft</u>
Learn more about the thin lens formula here:
brainly.com/question/3074650
Answer: The force you exert on the handle of a screen door as you open the door is on the order of 10 N. The force exerted by a honeybee as it lands on a flower is on the order of 0.001 N. The thrust force produced by the engines of a passenger jet as it takes off is on the order of 100,000 N. The weight of the largest aircraft carriers is on the order of 1,000,000,000 N.
Explanation:
In the first case, force made by hands and arms movement is in the range of 1
or 10N. In the second case, we are talking of a honeybee, an insect that weighs grames. The third case is the force required to make a plane take off (it is %10 of the thrust force of a plain with a weight of 300000N to 1000000N). In the last case, we are talking of the largest aircraft carriers, steel monsters that weight thousand of tons.
Answer:
C
Explanation:
The fume hood's ventilation system pulls fumes and fire away from you and the lab, it is safer.
Explanation:
The given data is as follows.
m = 5000 kg, h = 800 km = ![0.8 \times 10^{6} m](https://tex.z-dn.net/?f=0.8%20%5Ctimes%2010%5E%7B6%7D%20m)
, r = R + h = ![7.17 \times 10^{6} m](https://tex.z-dn.net/?f=7.17%20%5Ctimes%2010%5E%7B6%7D%20m)
kg, G = ![6.67 \times 10^{-11} Nm^{2}/kg^{2}](https://tex.z-dn.net/?f=6.67%20%5Ctimes%2010%5E%7B-11%7D%20Nm%5E%7B2%7D%2Fkg%5E%7B2%7D)
As we know that,
![\frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E%7B2%7D%7D%7Br%7D%20%3D%20%5Cfrac%7BGmM_%7Be%7D%7D%7Br%5E%7B2%7D%7D)
v = ![\sqrt{\frac{GM_{e}}{r^{2}}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7BGM_%7Be%7D%7D%7Br%5E%7B2%7D%7D%7D)
And, it is known that formula to calculate angular velocity is as follows.
![\omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7Bv%7D%7Br%7D%20%3D%20%5Csqrt%7B%5Cfrac%7BGM_%7Be%7D%7D%7Br%5E%7B3%7D%7D%7D)
v = ![\sqrt{\frac{GM_{e}}{r^{3}}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7BGM_%7Be%7D%7D%7Br%5E%7B3%7D%7D%7D)
= ![\sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B6.67%20%5Ctimes%2010%5E%7B-11%7D%20Nm%5E%7B2%7D%2Fkg%5E%7B2%7D%20%5Ctimes%205.98%20%5Ctimes%2010%5E%7B-24%7D%20kg%5E%7B-2%7D%7D%7B%287.17%20%5Ctimes%2010%5E%7B6%7D%20m%29%5E%7B3%7D%7D%7D)
= ![1.0402 \times 10^{-3} rad/s](https://tex.z-dn.net/?f=1.0402%20%5Ctimes%2010%5E%7B-3%7D%20rad%2Fs)
Thus, we can conclude that speed of the satellite is
.
Answer:
d) 12 V
Explanation:
Due to the symmetry of the problem, the potential (relative to infinity) at the midpoint of the square, is the same for all charges, provided they be of the same magnitude and sign, and be located at one of the corners of the square.
We can apply the superposition principle (as the potential is linear with the charge) and calculating the total potential due to the 4 charges, just adding the potential due to any of them:
V = V(Q₁) + V(Q₂) +V(Q₃) + V(Q₄) = 4* 3.0 V = 12. 0 V