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olganol [36]
3 years ago
7

If a biker starts at rest and accelerates to 15 m/s in 7.5 seconds, what is his acceleration? *

Physics
1 answer:
sergeinik [125]3 years ago
6 0

Answer:

2 m/s^2

Explanation:

from the question

v=15 m/s

t=7.5

a=?

from the first equation of motion

v=u+at

where,

v=final velocity

u=initial velocity

a=acceleration

t=time

from the question (u) will be zero because the body started at rest

v=u+at

15=(0)+a×7.5

15=7.5a

a=15/7.5

a=2 m/s^2

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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.
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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} }

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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

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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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