Answer:
236.3 x 
C
Explanation:
Given:
B(0)=1.60T and B(t)=-1.60T
No. of turns 'N' =100
cross-sectional area 'A'= 1.2 x m²
Resistance 'R'= 1.3Ω
According to Faraday's law, the induced emf is given by,
ℰ=-NdΦ/dt
The current given by resistance and induced emf as
I = ℰ/R
I= -NdΦ/dtR
By converting the current to differential form(the time derivative of charge), we get
= -NdΦ/dtR
dq= -N dΦ/R
The change in the flux dФ =Ф(t)-Ф(0)
therefore, dq = 
(Ф(0)-Ф(t))
Also, flux is equal to the magnetic field multiplied with the area of the coil
dq = NA(B(0)-B(t))/R
dq= (100)(1.2 x )(1.6+1.6)/1.3
dq= 236.3 x C
Question
Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles travelled in a straight line and some were deflected at different angles.
Which statement best describes what Rutherford concluded from the motion of the particles?
A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.
B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.
C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.
D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.
Answer:
The right answer is C)
Explanation:
In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.
So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".
Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.
Cheers!
Answer:
A. 2.8 m/s
Explanation:
Suppose that at the height of 0 m, the path of the pendulum is lowest.
If we use law of conservation of energy, the pendulum will have zero kinetic energy or K.E when it is at highest point, because K.E happens during movement of object and at the highest point all the energy will be P.E
P.E= mgh
Similarly, when the pendulum reaches at the lowest point, the height becomes zero and the P.E also becomes zero. Now all the energy will be K.E
K.E= 1/2 m v^2
In question, we are asked about the speed as the pendulum it reaches the lowest point of its path. Like we mentioned P.E will be zero at lowest point because of zero height. And also we will use law of conservation of energy because no energy has been lost from system.
K.E= P.E
1/2 m v^2 = mgh
Taking sq.root at both sides
v= Under root 2 gh
v=Under root 2x 9.8 m/s x0.4 m
v=Under root 7.84
v=2.8 m/sec
Hope it helps!
Answer:
10cm^3
Explanation:
Given data
Mass of steel = 80grams
density of steel= 8 g/cm^3
We know that the formula for density is given as
density= mass/volume
make volume subject of formula
volume= mass/density
volume=80/8
volume= 10cm^3
Hence, the volume 10cm^3
