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
The answer would be D, negative acceleration
Answer:
<em> 17.656 m</em>
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Explanation:
Initial speed u = 20 m/s
angle of projection α = 60°
at the top of the trajectory, one fragment has a speed of zero and drops to the ground.
we should note that the<em> top of the trajectory will coincide with halfway the horizontal range of the the projectile travel. This is because the projectile follows an upward arc up till it reaches its maximum height from the ground, before descending down by following a similar arc downwards.</em>
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To find the range of the projectile, we use the equation
R =
where g = acceleration due to gravity = 9.81 m/s^2
Sin 2α = 2 x (sin α) x (cos α)
when α = 60°,
Sin 2α = 2 x sin 60° x cos 60° = 2 x 0.866 x 0.5
Sin 2α = 0.866
therefore,
R = = 35.31 m
<em>since the other fraction with zero velocity drops a top of trajectory, distance between the two fragments assuming level ground and zero air drag, will be 35.31/2 = 17.656 m</em>
- Mass=m=10kg
- Radius=r=2m
- Speed=v=50m/s
Force:-
Now using Newton's second law
You would add all of them together to make 4. hope this helps((: