Answer:
Hello some part of your question is missing below is the missing part
2. What is the force on the charged particle if it is now located at the 0V potential difference line? (mN) (hint: The electric field can be obtained as above using the 0V and -10V equipotential lines.)
answer :
1) 0.8 mN
2) 0.8 mN
Explanation:
Given data:
1) Calculate the force on the charged particle
q = 80 μC , Va = 30v , Vb = 40v, ∝ = 1 m
E = ( Δv ) / ∝
= ( Vb - Va ) / ∝
F = qE
= 80 μC * ( 40 - 30 ) / 1 m
= 800 μC
F = 0.8 mN
<u>2) Calculate the force on the charged particle when it is located at 0V</u>
Va = -10V , Vb = 0V, q = 80 μC, ∝ = 1 m
F = qE
where E = ( 0 - ( -10 ) / 1
F = 80 μC * ( 0 - ( -10 ) / 1
= 800 μC = 0.8 mN
Answer:
Distance ran by basketball player = 4.5 m
Explanation:
We have equation of motion, v = u+at
Substituting
6 = 0 + a x 1.5
a = 4m/s²
Now we need to find distance traveled, we have the equation of motion
v² = u² + 2as
Substituting
6² = 0² + 2 x 4 x s
8 s = 36
s = 4.5 m
Distance ran by basketball player = 4.5 m
B!
This is because gravitational pull energy needs a height. The higher the height the higher the energy.
The tiny particles that make up elements are called "molecules"
Answer:
<u>A kangaroo hops 60 m to the east in 5 s. What is the kangaroo's average velocity? ... The kangaroo stops at a lake for a drink of water and then starts hopping again to the south. Each second, the kangaroo's velocity increases 2.5 m/s.</u>