1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of charge +4.00 μC and mass 10.0 mg moving along the y-axis. If Q3 starts from rest at y = 2.00 cm, what is its speed when it reaches y = 4.00 cm?

Answers

Answer 1
Answer:

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = \sqrt{3^(2) +2^(2)  } = √(13) = 3.61cm = 0.036m

r₂ = \sqrt{4^(2) + 3^(2)  } = √(25) = 5cm = 0.05m

electric potential V = (kq)/(r)

change in potential ΔV = V_(1) - V_(2)

ΔV = (2kq_(1) )/(r_(1)) - (2kq_(2) )/(r_(2) ) , where q_(1) = q_(2)=2.00μC

ΔV = 2kq((1)/(r_(1)) - (1)/(r_(2) ))

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × ((1)/(0.036) - (1)/(0.05) )

ΔV= 2.789×10⁵

(1)/(2)mv^(2) = ΔV × q₃

(1)/(2) ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

Answer 2
Answer:

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC  = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

Read more about point charges at;brainly.com/question/13914561


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Answers

Answer:

The answer is below

Explanation:

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Sum of horizontal force is zero, hence:

50(9.81)sin(15)-\mu N+50acos(15)=0\n\n50(9.81)sin(15)-0.5 N+50acos(15)=0\ \ \ (3)

Solving equation 1, 2 and 3 simultaneously gives :

N = 447.8 N, a = 2.01 m/s², x = 0.25 m

x is supposed to be 0.3 m (0.6/2)

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Answers

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Answers

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Answers

Answer:

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A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2?A 0.31 s
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C 4.3s
D 70s

Answers

B 0.56 s is the time period of a twirlers baton.

What is Centripetal Acceleration?

Centripetal acceleration is defined as the property of the motion of an object which traversing a circular path.

Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration.

The centripetal acceleration is given by:

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Given values are:

a = 47.8 m/s²

D = 0.76 m  so , R = 0.76/2 = 0.38m

Using this formula,

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T² = (4*3.14^2*0.38)/(47.8)

T = 0.56 s

Therefore,

A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2 which have  time period of 0.56 s.

Learn more about Centripetal acceleration here:

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.

Answer:

C. 4.3 seconds

Explanation: