Answer:
<h2>
6.36 cm</h2>
Explanation:
Using the formula to first get the image distance
1/f = 1/u+1/v
f = focal length of the lens
u = object distance
v = image distance
Given f = 16.0 cm, u = 24.8 cm
1/v = 1/16 - 1/24.8
1/v = 0.0625-0.04032
1/v = 0.02218
v = 1/0.02218
v = 45.09 cm
To get the image height, we will us the magnification formula.
Mag = v/u = Hi/H
Hi = image height = ?
H = object height = 3.50 cm
45.09/24.8 = Hi/3.50
Hi = (45.09*3.50)/24.8
Hi = 6.36 cm
The image height is 6.36 cm
Answer: C and D
Explanation: I’m pretty sure it’s C and D I got the same question but cold water let me know please, thanks!
Answer:
V_f = 287.04 mL
Explanation:
We are given the initial/original volume of the glycerine as 285 mL.
Now, after it is finally cooled back to 20.0 °C , its volume is given by the formula;
V_f = V_i (1 + βΔT)
Where;
V_f is the final volume
V_i is the original volume = 285 mL
β is the coefficient of expansion of glycerine and from online tables, it has a value of 5.97 × 10^(-4) °C^(−1)
Δt is change in temperature = final temperature - initial temperature = 32 - 20 = 12 °C
Thus, plugging in relevant values;
V_f = 285(1 + (5.97 × 10^(-4) × 12))
V_f = 287.04 mL
The correct option is C.
When the temperature of an object that is giving off light is increased, the particles in the object will move at a faster rate and there will be increased vibration of these molecules. This will makes the object to emit more light and to shine more brightly. Thus, the higher the temperature, the brighter the light that will be emitted.
Answer:
The two answers are in the explanation
Explanation:
Please find the attached files for the solution
