Answer:
v = 24 cm and inverted image
Explanation:
Given that,
The focal length of the object, f = +8 cm
Object distance, u = -12 cm
We need to find the position &nature of the image. Let v be the image distance. Using lens formula to find it :

Put all the values,

So, the image distance from the lens is 24 cm.
Magnification,

The negative sign of magnification shows that the formed image is inverted.
In general, the quantity of heat energy, Q, required to raise a mass m kg of a substance with a specific heat capacity of <span>c </span>J/(kg °C), from temperature t1 °C to t2 °C is given by:
<span>Q </span>= <span>mc(t</span><span>2 </span><span>– t</span>1<span>) joules</span>
<span>So:</span>
(t2-t1) =Q / mc
<span>As we know:
Q = 500 J </span>
<span>m = 0.4 kg</span>
<span>c = 4180 J/Kg </span>°c
<span>We can take t1 to be 0</span>°c
t2 - 0 = 500 / ( 0.4 * 4180 )
t2 - 0 = 0.30°c
Answer:
When they are babies. to get them familar with it
Explanation:
Pls choose brainliest.
Complete question:
Consider the hypothetical reaction 4A + 2B → C + 3D
Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?
Answer:
the final concentration of A is 0.992 M.
Explanation:
Given;
time of reaction, t = 4.0 s
rate of change of the concentration of B = -0.0760 M/s
initial concentration of A = 1.600 M
⇒Determine the rate of change of the concentration of A.
From the given reaction: 4A + 2B → C + 3D
2 moles of B ---------------> 4 moles of A
-0.0760 M/s of B -----------> x

⇒Determine the change in concentration of A after 4s;
ΔA = -0.152 M/s x 4s
ΔA = -0.608 M
⇒ Determine the final concentration of A after 4s
A = A₀ + ΔA
A = 1.6 M + (-0.608 M)
A = 1.6 M - 0.608 M
A = 0.992 M
Therefore, the final concentration of A is 0.992 M.