Answer:
Example:
A car applying brake.
A player catching a bowl.
We make use of a one-sample t-test for a population mean.
One-sample t-test for a population mean
Option B
<h3> Sample mean and Sample standard deviation</h3>
A Sample Standard Deviation is the root-mean square of the data minus the sample mean,
The sample mean is is the mean of the randomly selected sample
Therefore, For a data or sample where we have no information on population standard deviation and here only one sample group is compared, we make use of a one-sample t-test for a population mean
A one-sample t-test for a population mean
More on Probability
brainly.com/question/795909
Answer:
Induced EMF,
Explanation:
Given that,
Radius of the circular loop, r = 5 cm = 0.05 m
Time, t = 0.0548 s
Initial magnetic field, 
Final magnetic field, 
The expression for the induced emf is given by :

= magnetic flux





So, the induced emf in the loop is 0.0143 volts. Hence, this is the required solution.
Answer:
aaksj
Explanation:
a) the capacitance is given of a plate capacitor is given by:
C = \epsilon_0*(A/d)
Where \epsilon_0 is a constant that represents the insulator between the plates (in this case air, \epsilon_0 = 8.84*10^(-12) F/m), A is the plate's area and d is the distance between the plates. So we have:
The plates are squares so their area is given by:
A = L^2 = 0.19^2 = 0.0361 m^2
C = 8.84*10^(-12)*(0.0361/0.0077) = 8.84*10^(-12) * 4.6883 = 41.444*10^(-12) F
b) The charge on the plates is given by the product of the capacitance by the voltage applied to it:
Q = C*V = 41.444*10^(-12)*120 = 4973.361 * 10^(-12) C = 4.973 * 10^(-9) C
c) The electric field on a capacitor is given by:
E = Q/(A*\epsilon_0) = [4.973*10^(-9)]/[0.0361*8.84*10^(-12)]
E = [4.973*10^(-9)]/[0.3191*10^(-12)] = 15.58*10^(3) V/m
d) The energy stored on the capacitor is given by:
W = 0.5*(C*V^2) = 0.5*[41.444*10^(-12) * (120)^2] = 298396.8*10^(-12) = 0.298 * 10 ^6 J