Answer:
Explanation:
For an electric force, F the formula:
F = kQq/r^2
Given:
r2 = 1/2 × r1
F1 × r1 = k
F1 × r1 = F2 × r2
F2 = (F1 × r1^2)/(0.5 × r1)^2
= (F1 × r1^2)/0.25r1^2
= 4 × F1.
Answer:
central nervous system
peripheral nervous system
Explanation:
The nervous system has two main parts: The central nervous system is made up of the brain and spinal cord. The peripheral nervous system is made up of nerves that branch off from the spinal cord and extend to all parts of the body.Oct 1, 2018
Answer:
From the narrative in the question, there seem to have been a break failure and the ordered step of response to this problem is to
1) Put on the hazard light to inform other road users of a problem or potential fault with your car and so they should continue their journey with caution.
2) Avoid pressing on the acceleration pedal as this might cause the car to gradually slow down due to friction and gravity
3)Try navigate the car to the service lane. This is the less busy lane where cars are sometimes parked briefly.
4) Continuously pump the breaks to try stop the car. Continuously pumping the breaks might just help you build enough pressure to stop the car because often time, there are some pressure left in the break.
5) At this point, the speed of the car should be relatively slow. So at this point, you could try apply the emergency hand break. Do not pull the emergency hand breaks if the car is on high speed. Doing this may cause the car to skid off the road.
2m/s because the hockey puck is traveling at a constant speed ( acceleration is 0 ). Unless something acts on the hockey puck it will travel 2 m/s forever.
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )
