Answer:
what should be done io protect forests
Lets say sphere 1 has a charge of 12 + and sphere 2 has a charge of 0 +. After they are touched Sphere 1 becomes 6 + and sphere 2 6 +. So 6 - 12 = a change of 6 -, while 6 - 0 = a change of 6 + Therfore,
Answer: The sign of the charge change / transfered are opposites.
Answer:
13.4cm
Explanation:
According to Rayleigh’s criterion the angular resolution to distinguish two objects is given by:
θ = 50.0*10^-7 rad
λ: wavelength of the light = 550nm
b = diameter of the objective
By doing b the subject of the formula and replacing the values of the angle and wavelength you obtain:
hence, the smallest diameter objective lens is 13.4cm
Answer:
a) a geostationary satellite is that it is always at the same point with respect to the planet,
b) f = 2.7777 10⁻⁵ Hz
c) d) w = 1.745 10⁻⁴ rad / s
Explanation:
a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet
- T = 10 h (3600 s / 1h) = 3.6 104 s
b) the period the frequency are related
T = 1 / f
f = 1 / T
f = 1 / 3.6 104
f = 2.7777 10⁻⁵ Hz
c) the distance traveled by the satellite in 1 day
The distance traveled is equal to the length of the circumference
d = 2pi (R + r)
d = 2pi (69 911 103 + 120 106)
d = 1193.24 m
d) the angular velocity is the angle traveled between the time used.
.w = 2pi /t
w = 2pi / 3.6 10⁴
w = 1.745 10⁻⁴ rad / s
how fast is
v = w r
v = 1.75 10-4 (69.911 106 + 120 106)
v = 190017 m / s
Answer:
The peak emf generated by the coil is 2.67 V
Explanation:
Given;
number of turns, N = 940 turns
diameter, d = 24 cm = 0.24 m
magnetic field, B = 5 x 10⁻⁵ T
time, t = 5 ms = 5 x 10⁻³ s
peak emf, V₀ = ?
V₀ = NABω
Where;
N is the number of turns
A is the area
B is the magnetic field strength
ω is the angular velocity
V₀ = NABω and ω = 2πf = 2π/t
V₀ = NAB2π/t
A = πd²/4
V₀ = N x (πd²/4) x B x (2π/t)
V₀ = 940 x (π x 0.24²/4) x 5 x 10⁻⁵ x (2π/0.005)
V₀ = 940 x 0.04524 x 5 x 10⁻⁵ x 1256.8
V₀ = 2.6723 V = 2.67 V
The peak emf generated by the coil is 2.67 V
