From the law of conservation of momentum
m1u1+ m2u2= m1v1+ m2v2
110*8+ 110*-10= 110*-10 + 110* v2
v2= 8 m/sec
Answer:
power emitted is 1.75 W
Explanation:
given data
length l = 5 cm = 5 ×
m
diameter d = 0.074 cm = 74 ×
m
total filament emissivity = 0.300
temperature = 3068 K
to find out
power emitted
solution
we find first area that is π×d×L
area = π×d×L
area = π×74 ×
×5 ×
area = 1162.3892 ×
m²
so here power emitted is express as
power emitted = E × σ × area × (temperature)^4
put here all value
power emitted = 0.300× 5.67 × 1162.3892 ×
× (3068)^4
power emitted = 1.75 W
Answer:
f = 1 m
Explanation:
The magnification of the lens is given by the formula:
![M = \frac{q}{p}](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7Bq%7D%7Bp%7D)
where,
M = Magnification = 4
q = image distance = 5 m
p = object distance = ?
Therefore,
![4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m](https://tex.z-dn.net/?f=4%20%3D%20%5Cfrac%7B5%5C%20m%7D%7Bp%7D%5C%5C%5C%5Cp%20%3D%20%5Cfrac%7B5%5C%20m%7D%7B4%7D%5C%5C%5C%5Cp%20%3D%201.25%5C%20m)
Now using thin lens formula:
![\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7Bp%7D%2B%5Cfrac%7B1%7D%7Bq%7D%5C%5C%5C%5C%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%5Cfrac%7B1%7D%7B1.25%5C%20m%7D%2B%5Cfrac%7B1%7D%7B5%5C%20m%7D%5C%5C%5C%5C%5Cfrac%7B1%7D%7Bf%7D%20%3D%201%5C%20m%5E%7B-1%7D%5C%5C%5C%5C)
<u>f = 1 m</u>
The energy would be transferred from the objects. Also do not forget, add direction to your answer.
In an exothermic reaction, there is a transfer of energy to the surroundings in the form of heat energy. The surroundings of the reaction will experience an increase in temperature. Many types of chemical reactions are exothermic, including combustion reactions, respiration & neutralization reactions of bases & acids.