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Anettt [7]
3 years ago
9

In the example given below, Aaron applies a force of 300N and Bob applies a force of 450N :

Physics
1 answer:
garri49 [273]3 years ago
3 0

Answer:

Explanation:

This problem is all about torque. The "rules" are that in order for a system to be in rotational equilibrium, the sum of the torques on the system have to equal 0 (in other words, they have to equal each other {cancel each other out}). The equation for torque is

τ = F⊥r where τ is torque, F⊥ is the perpendicular force, and r is the lever arm length in meters. We also have to understand that in general Forces moving clockwise are negative and Forces moving counterclockwise are positive. Now we're ready for the problem:

A. The counterclockwise torque:

τ = 300(3) so

τ = 900N*m

B. The clockwise torque:

τ = -450(2.5) so

τ = -1100N*m

C. Obviously the system is not in roational equilibrium because one side is experiencing a greater torque than the other. This system will move clockwise as it currently exists.

D. In order for the system to be in rotational equilibrium, we have to move Bob's location from the fulcrum. Let's see to where.

The torques have to be the same on both sides of the fulcrum; mathematically, that looks like this:

F⊥r = F⊥r  Filling in:

300(3) = 450r and

900 = 450r so

2 = r. This means that Bob will have to move closer to the fulcrum by a half of a meter to 2 meters from the fulcrum in order for the system to be in balance.

Isn't this so much fun?!

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Free charges do not remain stationary when close together. To illustrate this, calculate the magnitude of the instantaneous acce
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a=2.304×10¹⁶m/s²

Explanation:

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charge of proton q=1.6×10⁻¹⁹C

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acceleration a

Solution

Apply the Coulombs Law

F=k\frac{q_{1}q_{2}  }{r^{2} }

Where k is coulombs constant (k=9×10⁹Nm²/C²)

q=q₁=q₂

r=d

So

F=k\frac{|q^{2} |}{d^{2} }\\ as \\F=ma\\ma=k\frac{|q^{2} |}{d^{2} }\\a=\frac{k}{m} \frac{|q^{2} |}{d^{2} }\\a=\frac{(9*10^{9} )*(1.6*10^{-19} )^{2} }{(1.6*10^{-27} )*(2.5*10^{-9} )^{2} }\\ a=2.304*10^{16}m/s^{2}  

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A total charge of Q coulomb is uniformly distributed along a rod 40cm in length.find find the electric field intensity 20cm away
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E1=1.598*10^11v/m

<h3>What is the electric field intensity 20cm away from the rod along its perpendicular bisector?</h3>

Generally, the equation for the  initial electric field intensity   is mathematically given as

dEp=\int{kd/r}cosd\theta

Therefore

e1=kd/r{sin\theta2+sinR1}

Hence

E1=(9*10^9/20*10^{-2})({sin45+sin45})*B/40*10^-2

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In conclusion, the electric field intensity

E1=1.598*10^11v/m

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