To solve this problem we use an amplification formula for divergent lenses

Where:
i: distance of the image to the lens
o: Distance from the object to the lens
h = height of the object
h '= height of the image


h '= 6 mm
The height is 6 mm
B
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Answer:
Explanation:
Given
Original volume V1=30cm^3 converting to L
=30/1000=0.03L
Constant pressure P= 648 tors
Converting to atm; 648 tors*1atm/760 torr=0.853 atm
Work=984J= 984**1L/101.33=9.7L.atm
Note before
W= -P(Vfinal-Vinitial)
-9.7/0.853+0.03L=11.68L
To develop this problem it is necessary to use the expression that allows us to convert the degrees Celsius to Fahrenheit. The expression that allows to realize it is given mathematically by:

In this way for 33.2 °C:



In this way for 38.2°C



Expressed in a range term, we can say that the measure in degrees Fahrenheit is:
![[91.76\° F , 100.76\° F]](https://tex.z-dn.net/?f=%5B91.76%5C%C2%B0%20F%20%2C%20100.76%5C%C2%B0%20F%5D)
Answer:
3.16 m·s⁻¹ at an angle of 71.6°
Explanation:
Assume that the diagram is like Fig. 1 below.
The boat is heading straight across the river and the current is directed straight downstream.
We have two vectors at right angles to each other.
1. Calculate the magnitude of the resultant
We can use the Pythagorean theorem (Fig. 2).
R² = (3 m·s⁻¹)² + (1 m·s⁻¹)² = 9 m²·s⁻² + 1 m²·s⁻² = 10 m²·s⁻²
R = √(10 m²·s⁻²) ≈ 3.16 m·s⁻¹
2. Calculate the direction of the resultant
The direction of the resultant is the counterclockwise angle (θ) that it makes with due East
.
tanθ = opposite/adjacent = 3/1 = 3
θ = arctan 3 = 71.6°
To an observer at point O, the velocity of the boat is 3.16 m·s⁻¹ at an angle of 71.6°.