Answer:
See below
Explanation:
Vertical position is given by
df = do + vo t - 1/2 a t^2 df = final position = 0 (on the ground)
do =original position = 2 m
vo = original <u>VERTICAL</u> velocity = 0
a = acceleration of gravity = 9.81 m/s^2
THIS BECOMES
0 = 2 + 0 * t - 1/2 ( 9.81)t^2
to show t =<u> .639 seconds to hit the ground </u>
During this .639 seconds it flies horizontally at 10 m/s for a distance of
10 m/s * .639 s =<u> 6.39 m </u>
Answer:
Object should be placed at a distance, u = 7.8 cm
Given:
focal length of convex lens, F = 16.5 cm
magnification, m = 1.90
Solution:
Magnification of lens, m = -
where
u = object distance
v = image distance
Now,
1.90 = 
v = - 1.90u
To calculate the object distance, u by lens maker formula given by:
u = 7.8 cm
Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.
Answer:
Electrical resistivity and its inverse, electrical conductivity, is a fundamental property of a material that quantifies how strongly it resists or conducts electric current. A low resistivity indicates a material that readily allows electric current. Resistivity is commonly represented by the Greek letter ρ.
Answer:
There are 756.25 electrons present on each sphere.
Explanation:
Given that,
The force of repression between electrons, 
Let the distance between charges, d = 0.2 m
The electric force of repulsion between the electrons is given by :




Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :
q = ne


n = 756.25 electrons
So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.
Answer:
10 feet.
Explanation:
Conductors are considered outside the building when they installed. The requirement of NEC is above 10 feet for the clearance after the final grade for service conductors and cables where they are connected with the building, above the sidewalks and the to other areas which are accessible to the pedestrians.