The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.
- The force per unit length between two parallel thin current-carrying
and 
wires at distance ' r ' is given by 
....(1) .
- If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.
A schematic of the information provided in the question can be seen in the image attached below.
From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3
Using equation (1) , we get
I₃ = 2.4 A and the current is pointing in the downward direction
Learn more about the magnitude and direction of forces here:
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In order to answer this exercise you need to use the formulas
S = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s
Answer:
8 electrons in the third energy level
Explanation:
From the description,the third energy level has 8 electron (represented by the small green balls you describe)
Answer:
78 km/h
Explanation:
If I normally drive a 12 hour trip at an average speed of 100 km/h, my destination has a total distance of:
- 100 km/h · 12 h = 1,200 km
Today, I drive the first 2/3 of the distance at 116 km/h. Let's first calculate what 2/3 of the normal distance is.
I've driven 800 km already. I need to drive 400 km more to reach my final destination. I need to figure out my average speed during this last 1/3 of the distance.
To do this, I first need to calculate how much time I spent driving 116 km/h for the past 800 km.
- 116 km/1 h = 800 km/? h
- 800 = 116 · ?
- ? = 800/116
- ? = 6.89655172
I spent 6.89655172 hours driving during the first 2/3 of the distance.
Now, I need to subtract this value from 12 hours to find the remaining time I have left.
- 12 h - 6.89655172 h = 5.10344828 h
Using this remaining time and my remaining distance, I can calculate my average speed.
- ? km/1 hr = 400 km/5.10344828 h
- 5.10344828 · ? = 400
- ? = 400/5.10344828
- ? = 78.3783783148
My average speed during the last third of the distance is around 78 km/h.
<h2>K.E/P.E = m/k tan²φ x ω²</h2>
Explanation:
The given position of block x = x₀ cos(ωt + φ)
The velocity of block v = dx/dt = - x₀ sin(ωt + φ) x ω
The kinetic energy = 1/2 mv² = 1/2 m x₀² sin²(ωt + φ) x ω²
The potential energy of spring = 1/2 k x² , where k is the spring constant
Thus P.E = 1/2 x k x x₀² cos²(ωt + φ)
When t = 0
K.E = 1/2 m x₀²sin²φ x ω²
P.E = 1/2 k x₀² cos²φ
Dividing these , we have
K.E/P.E = m/k tan²φ x ω²
