Answer:
In the clarification portion elsewhere here, the definition of the concern is mentioned.
Explanation:
So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.
Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:
- Mostly at night would they have been seen.
- Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.
Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.
Answer: V = 15 m/s
Explanation:
As stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,
F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz
Using doppler effect formula
F = C/ ( C - V) × f
Where
F = observed frequency
f = source frequency
C = speed of light = 3×10^8
V = speed of the car
Substitute all the parameters into the formula
2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10
2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)
1.000000049 = 3×10^8/(3×10^8 - V)
Cross multiply
300000014.7 - 1.000000049V = 3×10^8
Collect the like terms
1.000000049V = 14.71429
Make V the subject of formula
V = 14.71429/1.000000049
V = 14.7 m/s
The speed of the car is 15 m/s approximately
Pressure
= Force/Area
Area = π(d^2)/4
= π(0.4^2)/4
=0.126 m2
Pressure
= 50/0.126
= 396.825 Pa
Answer:
d = 76.5 m
Explanation:
To find the distance at which the boats will be detected as two objects, we need to use the following equation:

<u>Where:</u>
θ: is the angle of resolution of a circular aperture
λ: is the wavelength
D: is the diameter of the antenna = 2.10 m
d: is the separation of the two boats = ?
L: is the distance of the two boats from the ship = 7.00 km = 7000 m
To find λ we can use the following equation:
<u>Where:</u>
c: is the speed of light = 3.00x10⁸ m/s
f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz
Hence, the distance is:

Therefore, the boats could be at 76.5 m close together to be detected as two objects.
I hope it helps you!
Answer:
1.19 m/s²
Explanation:
The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so
f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)
Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²
Equating (1) and (2) we ave
2(√mg/μ)/f = T²g/4π²
Making g subject of the formula
g = 2π√(2√(m/μ)/f)/T
The period T = 316 s/100 = 3.16 s
Substituting the other values into , we have
g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16
g = 2π√(2 × 35.877/200 Hz)/3.16
g = 2π√(71.753/200 Hz)/3.16
g = 2π√(0.358)/3.16
g = 2π × 0.599/3.16
g = 1.19 m/s²