Answer:
Explanation:
Image of distant object will be made at far point or at 52.5 so
object distance u = infinity
image distance v = - 52.5 cm
focal length required = f
Lens formula
1 / v - 1 / u = 1 / f
1 / - 52.5 - 0 = 1 / f
f = -52.5 cm
= -.525 m
Power P = 1 / f = - 1 / .525
= - 1.90
now , for eye with glass we shall find new near point .
v = ?
u = - 17.2 cm
f = - 52.5 cm
1 / v - 1 / u = 1 / f
1 / v + 1 / 17.2 = - 1 / 52.5
1 / v = - 1 / 17.2 - 1 / 52.5
= - .05813 - .019
= - .07713
u = - 12.96 cm
so new near point will be 12.96 cm
The total work is
(mass of the elevator, kg) x (9.8 m/s²) x (9.0 m) Joules .
Thank you for posting your question here at brainly. I hope the answer will help. Below are the choices that can be found elsewhere:
<span>A. 1.5 * 10^3 Watts
B. 7.3 * 10^2 Watts
C. 3.5 * 10^2 Watts
D. 2.5 * 10^2 Watts
</span>
<span>Work = force*displacement = 10^2*87 = 8,700 joule
Power = work/time = 8,700/6 = 1.45*10^3 (rounded up to 1.5 kw). The answer is A. </span>
Answer:
Explanation:
Bohr's energy expression is as follows
E_n = 13.6 z² /n² where z is atomic no and n is principal quantum no of the atom .
z for helium is 2 and for ionised atom is 5 . Let energy of n₁ level of He is equal to energy level n₂ of ionised atom
so
13.6 x 2² / n₁² = 13.6 x 5² / n₂²
n₁ / n₂ = 2/5 , ie 2nd energy level of He matches 5 th energy level of ionised atom .
For quantum numbers less than or equal to 9 , If we take n₁ = 8 for He
Putting it in the equation above
2² / 8² = 5² / n₂²
n₂ = 5 x 8 / 2
= 20 .
energy
= - 13.6 x2² / 8²
= - 0.85 eV .
Answer:
Explanation:
When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.
When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other .So both the shells lose their charges .The positive half shell also loses all its charges
When we separate the half shells , there will be no deflection in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.
