Explanation:
Given that,
Size of object, h = 0.066 m
Object distance from the lens, u = 0.210 m (negative)
Focal length of the converging lens, f = 0.14 m
If v is the image distance from the lens, we can find it using lens formula as follows :
(a) Magnification,
(b) Magnification,
h' is image height
Hence, this is the required solution.
Answer:
Explanation:
Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC
Force F between them
4.8\times10^{-4} =
=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹
133.33 = 28q - q²
q²- 28q +133.33 = 0
It is a quadratic equation , which has two solution
q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C
The gravitational force between the two objects A) It increases.
Explanation:
The gravitational force between two objects is given by:
(1)
where
G is the gravitational constant
are the masses of the two objects
r is the separation between the objects
In this problem, object A and object B are initially at a distance of
r = 100 m
And at that distance, the force between them is
F
Later, object A gains some mass. We notice from eq.(1) that the gravitational force is directly proportional to the mass: therefore, if the mass of either of the two objects increases, then the gravitational force between them also increases. Therefore, the new force will be larger than the original force:
F' > F
Learn more about gravitational force:
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To solve this exercise we will use the concept related to heat loss which is mathematically given as
Where,
m = mass
= Specific Heat
Change in temperature
Replacing with our values we have that
Specific heat of mercury
Replacing
Therefore the heat lost by mercury is 0.09J