Answer:
Rate of change of magnetic field is
Explanation:
We have given diameter of the circular loop is 13 cm = 0.13 m
So radius of the circular loop 
Length of the circular loop 
Wire is made up of diameter of 2.6 mm
So radius 
Cross sectional area of wire 
Resistivity of wire 
Resistance of wire 
Current is given i = 11 A
So emf 
Emf induced in the coil is 


Answer: D
Neither A nor B
Explanation:
In order to check the clearances for rod and main bearings, you need a set of micrometers and a dial-bore gauge
Measuring the inside diameter of a main or rod bearing will require a dial bore gauge. The best ones to use are accurate down to 0.0001-inch.
So, both technician A and B are incorrect
Because the Earth's axis is not "straight up and down" as we move
around the sun.
So when we're on one side of the sun, the top pole leans slightly toward
the sun. During that time the sun shines more directly on the top half
of the Earth, and less directly on the bottom half. The people on the
top half see the sun higher in the sky, and their weather is warmer,
while the people on the bottom half see the sun lower in the sky, and
their weather is cooler.
Then, when we're on the other side of the sun, the top pole leans slightly
away from the sun. During that time the sun shines more directly on the
bottom half
of the Earth, and less directly on the top half. The people on
the bottom half see the sun higher in the sky, and their weather is warmer,
while the people on the top half see the sun lower in the sky, and their
weather is cooler.
The Earth makes the complete trip around the sun in one year, so the
people on the Earth go through this cycle of higher/lower sun and
warmer/cooler weather every year.
Answer:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
(1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

B. The energy of the emitted photon is given by the following formula:
(2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:

Next, you use the equation (2) and solve for λ:

C. The radius of the orbit is given by:
(3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:

hence, the radius of the atom in its 5-th state is 59.5 anstrongs