Answer:
D) Higher Frequency
Explanation:
Higher frequency because it says a vehicle “towards” the officer
Answer:
r = 1.63×10^5 mi
Explanation:
Let r = distance of object from earth
Rs = distance between earth and sun
Ms = mass of the sun
= 3.24×10^5 Me (Me = mass of earth)
At a distance R from earth, the force Fs exerted by the sun on the object is equal to the force Fe exerted by the earth on the object. Using Newton's universal law of gravitation,
Fs = Fe
GmMs/(Rs - r)^2 = GmMe/r^2
This simplifies to
Ms/(Rs - r)^2 = Me/r^2
(3.24×10^5 Me)/(Rs - r)^2 = Me/r^2
Taking the reciprocal and then its square root, this simplifies further to
Rs - r = (569.2)r ----> Rs = 570.2r
or
r = Rs/570.2 = (9.3×10^7 mi)/570.2
= 1.63×10^5 mi
Answer:
The charge on the ball bearing 4.507 × 10^-8 C
Explanation:
From Coulomb's law
F = kq1q2/r²
make q2 the subject
q2 = Fr²/kq1
q2 = (1.8×10^-2 × 0.026²) ÷ (9×10^9 × 30×10^-9)
q2 = 4.507 × 10^-8 C
