Answer:
20 cm
Explanation:
We can solve the problem by using the magnification equation:

where
is the size of the image
is the height of the real object (the man)
is the distance of the image from the lens
is the distance of the object (the man) from the lens
Solving the formula for
, we find

And the negative sign means the image is inverted.
The heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.
Explanation:
The heat energy required to convert a substance or to heat up or increase the temperature of a substance can be obtained from the specific heat formula.
As per this formula, the heat energy applied should be equal to the product of mass of the substance with temperature gradient and also with specific heat of the substance. Basically, the heat provided to increase or convert a substance should be more than the specific heat of the substance.

Since, here the mass of the substance X is given as m = 20g and the temperature change is given from -10°C to 70°C.
Then ΔT = (70-(-10))=70+10=80°C.
As the substance is unknown, the specific heat of that substance can also not be determined. Hence keep it as C.

Q = 1600C J
Thus, the heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.
I think it’s c. if not i’m sorry!!
The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.
The formula used for the radius of the Airy disk is given by,

Where,
Range of the radius
wavelength
f= focal length
Our values are given by,
State 1:



State 2:



Replacing in the first equation we have:


And also for,


Therefor, the airy disk radius ranges from
to 
Answer:
v_y = 12.54 m/s
Explanation:
Given:
- Initial vertical distance y_o = 10 m
- Initial velocity v_y,o = 0 m/s
- The acceleration of object in air = a_y
- The actual time taken to reach ground t = 3.2 s
Find:
- Determine the actual speed of the object when it reaches the ground?
Solution:
- Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:
y = y_o + v_y,o*t + 0.5*a_y*t^2
0 = 10 + 0 + 0.5*a_y*(3.2)^2
a_y = - 20 / (3.2)^2 = 1.953125 m/s^2
- Use the principle of conservation of total energy of system:
E_p - W_f = E_k
Where, E_p = m*g*y_o
W_f = m*a_y*(y_i - y_f) ..... Effects of air resistance
E_k = 0.5*m*v_y^2
Hence, m*g*y_o - m*a_y*(y_i - y_f) = 0.5*m*v_y^2
g*(10) - (1.953125)*(10) = 0.5*v_y^2
v_y = sqrt (157.1375)
v_y = 12.54 m/s