Answer:
twice
Explanation:
From magnification = height of image / height of object
Distance of image/ distance of object = magnification
If the distance and height of the object represents the initial light distance and the exposed surface respectively.
And similarly the distance and height of the image represents the final light distance and the exposed surface respectively.
Hence the new image exposure would be twice as large.
If we use the formula our point of investigation is Height of image,
H2= D2/D1× H1
H2 = 2D2/D1 × H1
H2 = 2H1
Answer:
Explanation:
(a) Work done, W = 1.82 x 10^4 J
(b) internal energy, U = - 4.07 x 10^4 J ( as it decreases)
(c) According to the first law of thermodynamics
Q = W + U
Q = 1.82 x 10^4 - 4.07 x 10^4
Q = - 2.25 x 10^4 J
Here are the planets listed in order of their distance from the Sun: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. An easy mnemonic for remembering the order is “My Very Educated Mother Just Served Us Noodles
Hope this helps.
Answer:
If you ring the doorbell and no one opens the door, you'll infer that no one is home rather than continuing to ring the doorbell to an empty house. Being able to understand this and look for another solution is another example of mental flexibility.
Explanation:
Answer:
Its period if its length is increased by a factor of four is 5 s.
Explanation:
The period of a simple pendulum is given by;
![T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{l}{g} } \\\\ \frac{T^2}{4\pi^2} = \frac{l}{g}\\\\\frac{T^2}{l} = \frac{4\pi^2}{g} \\\\let \ \frac{4\pi^2}{g} \ be \ constant \\\\\frac{T_1^2}{l_1} = \frac{T_2^2}{l_2} \\\\](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%20%7D%20%5C%5C%5C%5C%5Cfrac%7BT%7D%7B2%5Cpi%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%20%7D%20%5C%5C%5C%5C%20%5Cfrac%7BT%5E2%7D%7B4%5Cpi%5E2%7D%20%3D%20%5Cfrac%7Bl%7D%7Bg%7D%5C%5C%5C%5C%5Cfrac%7BT%5E2%7D%7Bl%7D%20%3D%20%5Cfrac%7B4%5Cpi%5E2%7D%7Bg%7D%20%5C%5C%5C%5Clet%20%5C%20%5Cfrac%7B4%5Cpi%5E2%7D%7Bg%7D%20%20%5C%20be%20%5C%20constant%20%5C%5C%5C%5C%5Cfrac%7BT_1%5E2%7D%7Bl_1%7D%20%20%3D%20%5Cfrac%7BT_2%5E2%7D%7Bl_2%7D%20%5C%5C%5C%5C)
Given;
initial period, T₁ = 2.5
initial length, = L₁
new length, L₂ = 4L₁
the new period, T₂ = ?
![\frac{T_1^2}{l_1} = \frac{T_2^2}{l_2} \\\\T_2^2 = \frac{T_1^2 l_2}{l_1} \\\\T_2 = \sqrt{\frac{T_1^2 l_2}{l_1}} \\\\ T_2 = \sqrt{\frac{(2.5)^2 \ \times \ 4l_1}{l_1}}\\\\ T_2 =\sqrt{(2.5)^2 \ \times \ 4}\\\\T_2 = \sqrt{25} \\\\T_2 = 5\ s](https://tex.z-dn.net/?f=%5Cfrac%7BT_1%5E2%7D%7Bl_1%7D%20%20%3D%20%5Cfrac%7BT_2%5E2%7D%7Bl_2%7D%20%5C%5C%5C%5CT_2%5E2%20%3D%20%5Cfrac%7BT_1%5E2%20l_2%7D%7Bl_1%7D%20%5C%5C%5C%5CT_2%20%3D%20%5Csqrt%7B%5Cfrac%7BT_1%5E2%20l_2%7D%7Bl_1%7D%7D%20%5C%5C%5C%5C%20%20T_2%20%3D%20%5Csqrt%7B%5Cfrac%7B%282.5%29%5E2%20%5C%20%5Ctimes%20%5C%204l_1%7D%7Bl_1%7D%7D%5C%5C%5C%5C%20%20T_2%20%3D%5Csqrt%7B%282.5%29%5E2%20%5C%20%5Ctimes%20%5C%204%7D%5C%5C%5C%5CT_2%20%3D%20%5Csqrt%7B25%7D%20%5C%5C%5C%5CT_2%20%3D%205%5C%20s)
Therefore, its period if its length is increased by a factor of four is 5 s.