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agasfer [191]
2 years ago
11

Calculate the momentum of a 4 kg Bolling ball being thrown at a speed of 3 m/s

Physics
1 answer:
bonufazy [111]2 years ago
7 0

We are given –

  • Mass of boiling ball is, m = 4 kg
  • Speed is, v = 3 m/s
  • Momentum, P =?

As we know –

↠Momentum = Mass × Speed(Velocity)

↠Momentum = 4 × 3 kgm/s

↠Momentum = 12 kgm/s

  • Henceforth,Momentum will be 12 kgm/s.
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Answer:

Explanation:

Given

N_1=1 rev/s

angular velocity \omega =2\pi N_1=6.284 rad/s

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\omega _2=2\pi N_2  

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I_1\omega _1=I_2\omega _2

6\times 6.284=2\times \omega _2

\omega _2=18.85\ rad/s

N_2=3 rev/s

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3 years ago
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3 years ago
It is very important that experiments are documented and<br> conducted with a procedure that can be?
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Repeated

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A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.
marishachu [46]

Answer:

a) the magnitude of the force is

F= Q(\frac{kqs}{r^3}) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b)  the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = \frac{kq}{r^{2} }

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = \frac{kp}{r^{3} } , where p = q × s

E(r) = \frac{kqs}{r^{3} } where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q(\frac{kqs}{r^3})

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall  F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

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3 years ago
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