A planet is orbiting a nearby star. Suppose that the star’s gravity increases over time. What effect will this increase most lik
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Answer:
a) 1.6*10^6 V
b) 13.35*10^6 V
Explanation:
The electric potential at origin is the sum of the contribution of the two charges. You use the following formula:
(1)
q1 = 3.90µC = 3.90*10^-6 C
q2 = -2.4µC = -2.4*10^-6 C
r1 = 1.25 cm = 0.0125 m
r2 = -1.80 cm = -0.018 m
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
You replace all the parameters in the equation (1):
hence, the total electric potential is approximately 1.6*10^6 V
b) For the coordinate (1.50 cm , 0) = (0.015 m, 0) you have:
r1 = 0.0150m - 0.0125m = 0.0025m
r2= 0.015m + 0.018m = 0.033m
Then, you replace in the equation (1):
hence, for y = 1.50cm you obtain V = 13.35*10^6 V
A single photon carries an energy equal to
where h is the Planck's constant and f is the frequency of the photon.
This means that the higher the frequency of the light, the higher the energy. Among the 5 different options mentioned by the problem, the light with highest frequency is ultraviolet, which has frequencies in the range [3-30] PHz, while visible light (red, blue, green) and infrared have lower frequency, so ultraviolet light has the highest energy per photon.
where h is the Planck's constant and f is the frequency of the photon.
This means that the higher the frequency of the light, the higher the energy. Among the 5 different options mentioned by the problem, the light with highest frequency is ultraviolet, which has frequencies in the range [3-30] PHz, while visible light (red, blue, green) and infrared have lower frequency, so ultraviolet light has the highest energy per photon.
Answer:
a. μ 3 ± 1.8 = [1.2,4.8]
b. The correct answer is option D. No, because the sample size is large enough.
Explanation:
a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:
μ±(t*s)/sqrt(n)
where:
μ = is the 95% confidence interval estimate
x_ = mean of the sample = 3
s = standard deviation of the sample = 5.8
n = size of the sample = 41
t = the t statistic for 95% confidence and 40 (n-1) degrees of freedom = 2.021
substituting all the variable, we have:
μ 3 ± (2.021*5.8)/sqrt(41) = 3 ± 1.8 = [1.2,4.8]
b. The correct answer is option D. No, because the sample size is large enough.
Using the the Central Limit Theorem which states that regardless of the distribution shape of the underlying population, a sampling distribution of size which is ≥ 30 is normally distributed.