The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.
The given parameters;
- <em>Pressure on the 13 th floor, P₁ = 35 PSI</em>
- <em>Distance between each floor, d = 10 ft</em>
The vertical pressure of the water is calculated as follows;
The vertical height of the first floor from the 13th floor = 130 ft
The vertical height of the 13 ft floor = 10 ft
Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.
Learn more about vertical height and pressure here: brainly.com/question/15691554
<u>Answer</u>
1) A. 96 Candelas
2) A. Both of these types of lenses have the ability to produce upright images.
3) C. 5 meters
<u>Explanation</u>
Q1
The formula for calculation the luminous intensity is;
Luminous intensity = illuminance × square radius
Lv = Ev × r²
= 6 × 4²
= 6 × 16
= 96 Candelabra
Q2
For converging lenses, an upright image is formed when the object is between the lens and the principal focus while a diverging lens always forms and upright image.
A. Both of these types of lenses have the ability to produce upright images.
Q3
Luminous intensity = illuminance × square radius
square radius = Luminous intensity/ illuminance
r² = 100/4
= 25
r = √25
= 5 m
Answer:
Explanation:
Given
Two masses 
and 
is released and there is tension T in the string
Suppose a is the acceleration of the system
Therefore from Diagram
For
------1
for m_2 body
-------2
From above two Equation it is said that Tension is greater than m_1g and less than m_2g
Answer:
Explanation:
Givens
d = 8.5 meters
vi = 0
a = 9.81
t = ?
Formula
d = vi * t + 1/2 a t^2
Solution
8.5 = 0 + 1/2 9.81 * t^2 multiply both sides by 2
8.5 = 4.095 t^2 Divide both sides by 4.095
8.5/4.095 = t^2
1.7329 = t^2 Take the square root of both sides
t = 1.316
It takes 1.316 seconds to hit the ground.
Answer:
The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)
Explanation:
Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.
Step 2: I must calculate the magnitude of the forces acting on the third charge.
F13: Force exerted by charge 1 on charge 3.
F23: Force exerted by charge 2 on charge 3.
K: Constant of Coulomb's law.
d13: distance from charge 1 to charge 3.
d23: distance from charge 2 to charge 3
Fr: Resulting force.
q1=+2.06 x 10-9 C
q2= -3.27 x 10-9 C
q3= +1.05 x 10-9 C
K=9-10^9 N-m^2/C^2
d13= 0,20 m
d23= 0,10 m
F13= K * (q1 * q3)/(d13)^2
F13=9,7335*10^(-8) N
F23=K * (q2 * q3)/(d23)^2
F23= -3,09 * 10^(-7)
Step 3: We calculate the resultant force on charge 3.
Fr=F13+F23= -2,11665 * 10^(-7)
