Answer:
Stephen hawking if his family were scientists
Me and you! enjoy your day girllllaa or boy
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.
The relationship between the resistance and the resistivity of a wire is

where

is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by

where r is the radius of the wire. Substituting in the previous equation ,we find

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is

Therefore, the new resistivity must be 4 times the original one.
The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s
<h3>
Motion Under Gravity</h3>
The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.
Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.
a. how much later does the ball hit the ground?
The time can be calculated by considering the vertical component of the motion with the use of formula below.
h = ut + 1/2gt²
Where
- Initial velocity u = 0 ( vertical velocity )
- Acceleration due to gravity g = 9.8 m/s²
Substitute all the parameters into the formula
75 = 0 + 1/2 × 9.8 × t²
75 = 4.9t²
t² = 75/4.9
t² = 15.30
t = √15.3
t = 3.9 s
b. how far from the building will it land?
The range can be found by using the formula
R = ut
Where u = 4.6 m/s ( horizontal velocity )
R = 4.6 × 3.9
R = 18 m
c. what is the velocity of the ball just before it hits the ground?
The final velocity will be
v = u + gt
v = 4.6 + 9.8 × 3.9
v = 4.6 + 38.22
v = 42.82 m/s
Therefore, the answers are 3.9 s, 18 m and 42.82 m/s
Learn more about Vertical motion here: brainly.com/question/24230984
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Answer:
1 Newton
Explanation:
F=9*10^9*q0q1/r^2]]
F=9*10^9*(q0q1)/ r^2
r=3cm
F=4N
F=9*10^9*(q0q1)/3^2
4=9*10^9*(q0q1)/9
4=10^9 q0q1
q0q1=4/10^9
q0q1=4*10^-9
To calculate the force between the forces at a distance of 6 cm
F=9*10^9*(q0q1)/ r^2
=9*10^9*(4*10^-9)/6^2
=9*10^9*(4*10^-9)/36
=10^9*4*10^-9/4
=10^9*10^-9
=1 Newton