Answer:
f = - 0.143 m = - 14.3 cm
Explanation:
First, we will calculate the power of the combination of lenses:
P = P₁ + P₂
where,
P = Power of Combination = ?
P₁ = Power of first lens = - 5D
P₂ = Power of second lens = - 2D
Therefore,
P = - 5D - 2D
P = - 7D
Now, the focal length can be given as:
<u>f = - 0.143 m = - 14.3 cm</u>
Negative focal length indicates that combination will act as diverging lens.
Fire is it that lives if it is fed, and dies if you give it a drink.
<h3><u>
Explanation:</u></h3>
Fire is very essential part of human life. It is used for cooking food and for other important activities. Without fire we cannot not survive. Something or the other should be heated before consumption and this can be achieved only with fir. It is also used in the darker places for viewing many things around us.
Thus, fire can survive if we give fuel or any wooden pieces and when water is poured on it it will turn off. Hence Fire is the one that survives when it is fed and dies when water is given as a drink to it.
Answer:
In hot gases , the atoms keeps colliding with each other and sometimes the energy liberated during collision takes the electron to a higher level,thus, .The object is a cloud of hot gas and finally the electron returns back emitting photon
Answer:
The activation energy is 
Explanation:
From the question we are told that
The rate constant is k
at the temperature 
The value of k is 
at temperature 
The value of k is 
The rate constant is mathematically represented as
Where Q is the activation energy
R is the ideal gas constant with a value of 
C is a constant
T is the temperature
For the first rate constant
For the second rate constant
Now the ratio between the two given rate constant is
![\frac{k_1 }{k_2} = e^{(\frac{Q}{R} [\frac{1}{\frac{T_2 - 1}{T_1} } ] )}](https://tex.z-dn.net/?f=%5Cfrac%7Bk_1%20%7D%7Bk_2%7D%20%20%3D%20%20e%5E%7B%28%5Cfrac%7BQ%7D%7BR%7D%20%5B%5Cfrac%7B1%7D%7B%5Cfrac%7BT_2%20-%201%7D%7BT_1%7D%20%7D%20%5D%20%29%7D)
=> ![ln [\frac{k_1}{k_2} ] = \frac{Q}{R} * [\frac{1}{\frac{T_2 -1}{T_1} } ]](https://tex.z-dn.net/?f=ln%20%5B%5Cfrac%7Bk_1%7D%7Bk_2%7D%20%5D%20%3D%20%20%5Cfrac%7BQ%7D%7BR%7D%20%20%2A%20%5B%5Cfrac%7B1%7D%7B%5Cfrac%7BT_2%20-1%7D%7BT_1%7D%20%7D%20%5D)
substituting values
![ln [\frac{1.05 *10^{-8}}{2.95 *10^{-4}} ] = \frac{Q}{8.314} * [\frac{1}{\frac{673 -1}{573} } ]](https://tex.z-dn.net/?f=ln%20%5B%5Cfrac%7B1.05%20%2A10%5E%7B-8%7D%7D%7B2.95%20%2A10%5E%7B-4%7D%7D%20%5D%20%3D%20%20%5Cfrac%7BQ%7D%7B8.314%7D%20%20%2A%20%5B%5Cfrac%7B1%7D%7B%5Cfrac%7B673%20-1%7D%7B573%7D%20%7D%20%5D)
=>
Answer:
Option C. 0.73 g/cm³
Explanation:
From the question given above, the following data were obtained:
Mass = 80 g
Area (A) = 10000 cm²
Thickness = 0.11 mm
Density =?
Next, we shall convert 0.11 mm to cm. This can be obtained as follow:
10 mm = 1 cm
Therefore,
0.11 mm = 0.11 mm × 1 cm / 10 mm
0.11 mm = 0.011 cm
Thus, 0.11 mm is equivalent to 0.011 cm.
Next, we shall determine the volume of the paper. This can be obtained as follow:
Area (A) = 10000 cm²
Thickness = 0.011 cm
Volume =?
Volume = Area × Thickness
Volume = 10000 × 0.011
Volume = 110 cm³
Finally, we shall determine the density of the paper. This can be obtained as follow:
Mass = 80 g
Volume = 110 cm³
Density =?
Density = mass / volume
Density = 80 / 110
Density = 0.73 g/cm³
Therefore the density of the paper is 0.73 g/cm³
