Answer:
Therefore the ratio of diameter of the copper to that of the tungsten is
Explanation:
Resistance: Resistance is defined to the ratio of voltage to the electricity.
The resistance of a wire is
- directly proportional to its length i.e
- inversely proportional to its cross section area i.e
Therefore
ρ is the resistivity.
The unit of resistance is ohm (Ω).
The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m
The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m
For copper:
......(1)
Again for tungsten:
........(2)
Given that 
and 
Dividing the equation (1) and (2)
[since and ]
Therefore the ratio of diameter of the copper to that of the tungsten is
KE = 1/2mv^2
KE= 1/2(2)(5)^2
KE= 25 J
Complete Question
The complete question is shown on the first uploaded image
Compare the percent error between your value and the accepted value of 1.26 times 10-6 T · m/A. (Use the accepted value given to three significant figures in your calculation.). 100% % error = %
Answer:
The permeability of free space is 
The percentage error is % error = 5.25%
Explanation:
From the question we are told that
The slope is 
, and
Generally 
So 
From the relation given in the question
![\mu_0 = \frac{2R}{N} [\frac{B}{I} ]](https://tex.z-dn.net/?f=%5Cmu_0%20%3D%20%5Cfrac%7B2R%7D%7BN%7D%20%5B%5Cfrac%7BB%7D%7BI%7D%20%5D)
Where R is the radius of the coil 
N is the number of loops of the coil = 10
Now from the question we are told that
substituting into the equation above
Substituting values
Generally the % error is mathematically represented as
%
Given that the accepted value is 
Hence substituting values
%
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: Period = 0.2 seconds; frequency = 5Hz
Explanation:
Number of oscillations = 50
Time required = 10 seconds
Period (T) = ?
Frequency of the oscillations (F) = ?
A) Recall that frequency is the number of oscillations that the mass spring system completes in one second.
i.e Frequency = (Number of oscillations / time taken)
F = 50/10 = 5Hz
B) Period, T is inversely proportional to frequency. i.e Period = 1/Frequency
T = 1/5Hz
T = 0.2 seconds
Thus, the the period and frequency of the oscillations are 0.2 seconds and 5Hz respectively.
