1. What did you observe about the magnitudes of the forces on the two charges? Were they the same or different? Does your answer
1 answer:
Answer:
Following are the solution to the given question:
Explanation:
Its strength from both charges is equivalent or identical. The power is equal. And it is passed down
Therefore, the extent doesn't rely on the fact that charges are the same or different. Newton's third law complies with Electrostatic Charges due to a couple of charges. They are similar in magnitude, and they're in the other way.
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Number 3.
When you have two parallel wire and the charges are moving, they create a magnetic field around the wire (right-hand rule, thumb points in the direction of current).
If you want the math, here ya go:
F: magnetic force on the wire, L: length of the wire I: current in one wire I': current in other wire r: distance of the wire.
As we can see, two wires that has current through it will generate a force on each other.
When you have two parallel wire and the charges are moving, they create a magnetic field around the wire (right-hand rule, thumb points in the direction of current).
If you want the math, here ya go:
F: magnetic force on the wire, L: length of the wire I: current in one wire I': current in other wire r: distance of the wire.
As we can see, two wires that has current through it will generate a force on each other.
Answer:
From 0.75 to 1.25 hours
Explanation:
Given
See attachment for graph
Required
Point where they didn't move
This means that we identify the point where the distance didn't change.
Given that the distance is plotted on the y-axis, we simply check for the end points of any horizontal line on the graph
The horizontal line on the graph represents 30km and the time interval is: 0.75 to 1.25 hours.
<em>Hence, (c) is correct</em>
Answer:
A. α = 94.4 rad/s
B. a = 28.32 m/s
C. N = 34N
D. α = 94.4 rad/s
a = 28.32 m/s
N = 44.4 N
Explanation:
part A:
using:
∑T = Iα
where T is the torque, I is the moment of inertia and α is the angular momentum.
firt we will find the moment of inertia I as:
I =
Where M is the mass and R is the radius of the wheel, then:
I =
I = 0.36 kg*m^2
Replacing on the initial equation and solving for α, we get::
∑T = Iα
Fr = Iα
34 N = 0.36α
α = 94.4 rad/s
part B
we need to use this equation :
a = αr
where a is the aceleration of the cord that has already been pulled off and r is the radius of the wheel, so replacing values, we get:
a = (94.4)(0.3 m)
a = 28.32 m/s
part C
Using the laws of newton, we know that:
N = T
where N is the force that the axle exerts on the wheel part and T is the tension of the cord
so:
N = 34N
part D
The anly answer that change is the answer of the part D, so, aplying laws of newton, it would be:
-Mg + N +T = 0
Then, solving for N, we get:
N = -T+Mg
N = -34 + (8 kg)(9.8)
N = 44.4 N