Distance of fall from rest,
without air resistance = (1/2) (gravity) (time)²
= (1/2) (9.8 m/s²) (95 sec)²
= (4.9 m/s²) (9,025 sec²)
= 44,222.5 meters .
The depth of the mine shaft is five times the height of Mt. Everest !
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)

q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
Answer:
6
Explanation:
because I did this assignment, :) your welcome
Next time do it by yourself, but here's the answer kid
1 nanowatt = 1 nanojoule/sec
1 watt = 1 joule/sec
10 watts = 10 joules/sec
100 watts = 100 joules/sec
742.914 watts = 742.914 joules/sec
1,000 watts = 1,000 joules/sec
10,000 watts = 10,000 joules/sec
100,000 watts = 100,000 joules/sec
1 megawatt = 1 megajoule/sec
1 gigawatt = 1 gigajoule/sec
1 petawatt = 1 petajoule/sec
We don't care what frequency the transmission is using,
or who their morning DJ is.
Answer:0.43
Explanation:
Given
mass of car 
Speed of car 
Distance traveled before coming to halt 
Let
the coefficient of friction
Maximum deceleration road can provide during motion is

using 


