<span>b. medium
</span>The substance that a wave moves through is called aMEDIUM
NOT:
a. fluid
c. track
<span>d. vibration</span>
Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by
<span>C = 2L = 2*pi*R ---> R = L/pi </span>
<span>Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. </span>
<span>we can define a small charge dq as </span>
<span>dq = l*ds = l*R*da </span>
<span>So the electric field can be written as: </span>
<span>dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) </span>
<span>E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)</span>
Electromagnetic waves can travel in vaccum but mechanical wave does not
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.
They are held together because of Strong Nuclear Force.