A lamp has the shape of a parabola when viewed from the side. The lamp is centimeters wide and centimeters deep. How far is the
light source from the bottom of the lamp if the light source is placed at the focus
1 answer:
The question is not complete so i have attached it.
Answer:
The light source is 2 cm from the bottom of the lamp
Explanation:
From the attached image, we can see that the parabola opens up with its vertex at the origin.
Now, the standard form of equation for a parabola is:
x² = 4ay
Looking at the parabola in the attachment, the top right edge of the lamp has a coordinate of (12,18)
Thus, we have;
12² = 4a(18)
144 = 72a
a = 144/72
a = 2
Looking at the parabola again, the line of symmetry is at x = 0
Thus, axis of symmetry is at x = 0.
Thus, focus is at (0, 2)
So, if the light source is placed at the focus, the distance of the light source from the bottom of the lamp is 2 cm
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So, the average speed of the Cheetah is 17.6 m/s.
<h3>Introduction</h3>Hello ! I'm Deva from Brainly Indonesia. This time, I will help regarding the average speed. The average speed is obtained from finding the average of the speeds that occur or can be detected from the division between distance and travel time. The average speed can be formulated by :
With the following condition :
= average speed (m/s)
- s = shift or distance objects from initial movement (m)
- t = interval of the time (s)
We know that :
- s = shift = 88 m
- t = interval of the time = 5 seconds
What was asked :
Step by step :
So, the average speed of the Cheetah is 17.6 m/s.
Answer:
a) the work (W) done during the process is -2043.25 kJ
b) the work (W) done during the process is -2418.96 kJ
Explanation:
Given the data in the question;
mass of water vapor m = 10 kg
initial pressure P₁ = 550 kPa
Initial temperature T₁ = 340 °C
steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4
from superheated steam table
specific volume v₁ = 0.5092 m³/kg
so the properties of steam at p₂ = 550 kPa, and dryness fraction
x = 0.4
specific volume v₂ = v + xv
v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )
v₂ = 0.1377 m³/kg
Now, work done during the process;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.1377 - 0.5092 )
W = 5500 × -0.3715
W = -2043.25 kJ
Therefore, the work (W) done during the process is -2043.25 kJ
( The negative, indicates work is done on the system )
b)
What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
x₂ = 100% - 80% = 20% = 0.2
specific volume v₂ = v + x₂v
v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )
v₂ = 0.06939 m³/kg
Now, work done during the process will be;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.06939 - 0.5092 )
W = 5500 × -0.43981
W = -2418.96 kJ
Therefore, the work (W) done during the process is -2418.96 kJ