KE = 1/2mv^2
m=10 v=3
KE=1/2*10*3^2
KE= 45 Joules
That's true.
In fact, the resistance of a wire is given by:
where
is the resistivity of the material
L is the length of the wire
A is the cross-sectional area of the wire
We see that the resistance of the wire is inversely proportional to the cross-sectional area: A. Therefore, the narrower the wire, the smaller A, the larger the resistance. But higher resistance means that the current flowing through the wire is lower, therefore the flow of electrons in the circuit is slower, and the initial sentence is true.
We know that AD is the shortest way but ABCD is connected in series
Equivalent Resistance =R1+R2+R3
=(10+10+10)ohm
=30ohm
Since BC and AD is connected in parallel
1/Req=1/R1 + 1/R2
=1/30 + 1/10
=3+1/30
=4/30
So, Req=30/4 ohm
Now,Current drawn from the battery
=> I = V/R
=> I = 30/4 ohm/3v
=> I = 30/4*3
So, I = 10/4 A
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Given that the arc length is:
s = ∫ √[1² + (dy/dx)²] dx
<span>So arc length between the two points is then: </span>
<span>s = 2*20sinh(x/20)
= 40sinh(x/20) </span>
<span>The straight distance between the two points is : d = 2x </span>
<span>So, x = d/2.
= 40/2
= 20 m </span>
<span>Plug this into the arc length equation to get:
s = 40sinh[20/20)]
=40* ½ (e - 1/e) </span>
<span> = 47 m</span>
Answer:
The magnitude is 63 kg m²/s, and the direction is -k.
Explanation:
Plug the values into the equation:
L = m v×r
L = (3.0 kg) (<5 i + 3j> m/s × <2i − 3j> m)
Take the cross product. The cross product of two dimensional vectors is:
<v₁ i + v₂ j> × <r₁ i + r₂ j> = <(v₁ r₂ − v₂ r₁) k>
Therefore:
L = (3.0 kg) <((5)(-3) − (3)(2)) k m²/s>
L = (3.0 kg) <-21 k m²/s>
Multiply:
L = <-63 k kg m²/s>
The magnitude is 63 kg m²/s, and the direction is -k.