Answer:
0.3659
Explanation:
The power (p) is given as:
P = AeσT⁴
where,
A =Area
e = transmittivity
σ = Stefan-boltzmann constant
T = Temperature
since both the bulbs radiate same power
P₁ = P₂
Where, 1 denotes the bulb 1
2 denotes the bulb 2
thus,
A₁e₁σT₁⁴ = A₂e₂σT₂⁴
Now e₁=e₂
⇒A₁T₁⁴ = A₂T₂⁴
or
substituting the values in the above question we get
or
=0.3659
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!
Explanation:
Band of stability for atomic nuclei refer to a narrow region in the graph of number of neutrons to the number of protons for stable nuclei. It is the band of nuclei stability.
We know that radioactive elements achieve stability be ejecting nucleons( neutrons and protons). The regions( combination of number of neutrons and protons) in which the nucleus is most is called band of stability.
V = Ed
E = v/d
so electric field strength and distance are inversely propotional. if we increase the distance electric field will decrease. if we decrease the distance electric field will increase.
Answer:
option d) is correct
Explanation:
Given:
C = (5/9)(F-32)
For statement I
C = (5/9)F - (5/9)32
let the initial temperature in Fahrenheit be 0
thus,
C = (5/9)(0) - (5/9)32 = - (5/9)32
for 1° increase in Fahrenheit
we get
C = (5/9)(1) - (5/9)32 = 1 - (5/9)32
thus,
on comparing the results, statement I is true
now for statement II
let the initial temperature in Celsius be 0
thus,
0 = (5/9)(F) - (5/9)32
or
F = (9/5)(5/9)32 = 32
for 1° increase in Celsius
we get
1 = (5/9)(F) - (5/9)32
or
(5/9)F = 1 + (5/9)32
or
F = (9/5) + 32
or
F = 1.8 + 32
thus,
on comparing the results, statement II is true
Now, for statement III
let the initial let the initial temperature in Fahrenheit be 0
thus,
C = (5/9)(0) - (5/9)32 = - (5/9)32 = - 17.77
for (5/9)° increase in Fahrenheit
we get
C = (5/9)(5/9) - (5/9)32 = (0.3086) - 17.77 = -17.469
thus,
on comparing the results, statement III is false
Hence, option d) is correct