Answer:
The Magnifying power of a telescope is 
Explanation:
Radius of curvature R = 5.9 m = 590 cm
focal length of objective 
= 
⇒ = 
⇒ = 295 cm
Focal length of eyepiece 
= 2.7 cm
Magnifying power of a telescope is given by,
therefore the Magnifying power of a telescope is
Answer:
a) v = 2,9992 10⁸ m / s
, b) Eo = 375 V / m
, B = 1.25 10⁻⁶ T,
c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz
, T = 1.05 10⁻¹⁵ s
, UV
Explanation:
In this problem they give us the equation of the traveling wave
E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]
a) what the wave velocity
all waves must meet
v = λ f
In this case, because of an electromagnetic wave, the speed must be the speed of light.
k = 2π / λ
λ = 2π / k
λ = 2π / 1.99 10⁷
λ = 3,157 10⁻⁷ m
w = 2π f
f = w / 2 π
f = 5.97 10¹⁵ / 2π
f = 9.50 10¹⁴ Hz
the wave speed is
v = 3,157 10⁻⁷ 9.50 10¹⁴
v = 2,9992 10⁸ m / s
b) The electric field is
Eo = 375 V / m
to find the magnetic field we use
E / B = c
B = E / c
B = 375 / 2,9992 10⁸
B = 1.25 10⁻⁶ T
c) The period is
T = 1 / f
T = 1 / 9.50 10¹⁴
T = 1.05 10⁻¹⁵ s
the wavelength value is
λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm
this wavelength corresponds to the ultraviolet
I think that the answer to that is true hope that helps
Answer:
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Answer:
E = 10⁵ J
Explanation:
given,
Power, P = 100 TW
= 100 x 10¹² W
time, t = 1 ns
= 1 x 10⁻⁹ s
The energy of a single pulse is:-
Energy = Power x time
E = P t
E = 100 x 10¹² x 1 x 10⁻⁹
E = 10⁵ J
The energy contained in a single pulse is equal to 10⁵ J
