Power of the glasses required is given as +2.5 D
this shows that person is having Farsightedness in which he can objects placed far clearly but he is not able to see clearly to the near objects
For normal vision near point of eye is 25 cm
Let say person can see the objects clearly at distance "d" from the eye
now from the formula
![\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bd_i%7D%20%2B%20%5Cfrac%7B1%7D%7Bd_o%7D%20%3D%20%5Cfrac%7B1%7D%7Bf%7D)
here given that
![\frac{1}{f} = 2.5](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%202.5)
![f = 40 cm](https://tex.z-dn.net/?f=f%20%3D%2040%20cm)
also we know that
![d_o = 25 cm](https://tex.z-dn.net/?f=d_o%20%3D%2025%20cm)
from above equation now
![\frac{1}{d} + \frac{1}{25} = \frac{1}{40}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bd%7D%20%2B%20%5Cfrac%7B1%7D%7B25%7D%20%3D%20%5Cfrac%7B1%7D%7B40%7D)
![d = 66.6 cm](https://tex.z-dn.net/?f=%20d%20%3D%2066.6%20cm)
So the distance of object must be 66.6 cm
Answer:
f = 1 Hz
Explanation:
From the attached figure, we find that the time period of the wave is 1 second.
It is a longitudinal wave. It travel in the form of compression and rarefaction. The relation between time period T and frequency f is given by :
![f=\dfrac{1}{T}\\\\\text{As T = 1 s}\\\\f=1\ s^{-1}\\\\=1\ Hz](https://tex.z-dn.net/?f=f%3D%5Cdfrac%7B1%7D%7BT%7D%5C%5C%5C%5C%5Ctext%7BAs%20T%20%3D%201%20s%7D%5C%5C%5C%5Cf%3D1%5C%20s%5E%7B-1%7D%5C%5C%5C%5C%3D1%5C%20Hz)
Hence, the frequency of this wave is 1 Hz.
Answer:
Explanation:
Let q be the charges on each spheres , 2d be the distance between them in equilibrium , T be tension in the string and F be the force of repulsion between them
F = k q² /4d²
For equilibrium
T sin15 = F
T cos 15 = mg
tan15 = F / mg
F = mg tan15
k q² /4d² = mg tan15
k q² = 4d² x mg tan15
= 4 x( .7 sin15)² x 3.1 x 10⁻³ x 9.8 x .2679
= 1.068 x 10⁻³
q = .3444 x 10⁻⁶ C
no of electrons
= .3444 x 10⁻⁶ / 1.6 x 10⁻¹⁹
= 215.25 x 10¹⁰
evaporation (if that's to fill in the blanks question)
Answer:
1.129×10⁻⁵ N
1.295 m
Explanation:
Take right to be positive. Sum of forces on the 31.8 kg mass:
∑F = GM₁m / r₁² − GM₂m / r₂²
∑F = G (M₁ − M₂) m / r²
∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²
∑F = 1.129×10⁻⁵ N
Repeating the same steps, but this time ∑F = 0 and we're solving for r.
∑F = GM₁m / r₁² − GM₂m / r₂²
0 = GM₁m / r₁² − GM₂m / r₂²
GM₁m / r₁² = GM₂m / r₂²
M₁ / r₁² = M₂ / r₂²
516 / r² = 207 / (0.482 − r)²
516 (0.482 − r)² = 207 r²
516 (0.232 − 0.964 r + r²) = 207 r²
119.9 − 497.4 r + 516 r² = 207 r²
119.9 − 497.4 r + 309 r² = 0
r = 0.295 or 1.315
r can't be greater than 0.482, so r = 0.295 m.