Answer:
the diameter of the outside edge of the receiver is 
Explanation:
From the schematic free body diagram illustrating what the question is all about below;
Let represent A to be the vertex where the receiver is being placed
S to be the focus
BP to be equal to r (i.e radius of the outer edge)
BC to be 2 r (i.e the diameter)
Given that AS = 4 in and AP is 18 in
Let AP be x- axis and AY be y -axis
A=(0,0)
S=(4,0) = (0,0)
So that the equation of the parabolic path of the receiver will be:
B = (AP, BP)
B = (18, r)
B lies y² = 16 x
r² = 16 x
r² 16 × 18
Diameter BC = 2r
Answer:
The volume of water evaporated is 199mL
Explanation:
Concentration is calculated with the following formula
where n is the number of moles of solute and V is the volume of the solution (in this case is the same as the solvent volume) in liters.
So we isolate the variable n to know the amount of moles, using the volume given in liters
Now, we isolate the variable V to know the new volume with the new concentration given.
Finally, the volume of water evaporated is the difference between initial and final volume.
A peak = A Rms x Sq root 2
Therefore 3.6 x sq root of 2
A peak = 5.09
The location of the point F that partitions a line segment from D to E (
), that goes from <u>negative 4</u> to <u>positive 5,</u> into a 5:6 ratio is fifteen halves (option 4).
We need to calculate the segment of the line DE to find the location of point F.
Since point D is located at <u>negative -4</u> and point E is at <u>positive 5</u>, we have:
Hence, the segment of the line DE () is 9.
Knowing that point F partitions the line segment from D to E () into a <u>5:6 ratio</u>, its location would be:
Therefore, the location of point F is fifteen halves (option 4).
Learn more about segments here:
I hope it helps you!
m = mass of pellet = 2.10 x 10⁻² kg
x = compression of spring at the time launch = 9.10 x 10⁻² m
h = height gained by the pellet above the initial position = 6.10 m
k = spring constant
using conservation of energy
spring potential energy = gravitational potential energy of pellet
(0.5) k x² = m g h
inserting the values
(0.5) k (9.10 x 10⁻²)² = (2.10 x 10⁻²) (9.8) (6.10)
k = 303.2 N/m
