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mash [69]
3 years ago
14

If an astronomer wants to find and identify as many stars as possible in a star cluster that has recently formed near the surfac

e of a giant molecular cloud (such as the Trapezium cluster in the Orion Nebula), what instrument would be best for her to use
Physics
1 answer:
Andru [333]3 years ago
5 0

Answer:

<em>Infrared telescope and camera</em>

<em></em>

Explanation:

An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.

Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, <em>Infrared images is better used, since they are able to penetrate the surrounding clouds of dust,</em> and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.

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a trampoline launches a 50kg person 2m into the air. if the springs push with 1960N of force, how much displacement was there in
Lina20 [59]

Answer: 0.5 m

Explanation:

Given

Mass of the person is m=50\ kg

Trampoline launches the person into the air up to height of h=2\ m

Force experience by springs is F=1960\ N

Here, the work done on displacing the springs is equivalent to the Potential energy acquired by the person i.e.

\Rightarrow F\cdot x=mgh\quad [\text{x=displacement of the trampoline}]\\\\\text{Insert the values}\\\\\Rightarrow x=\dfrac{50\times 9.8\times 2}{1960}\\\\\Rightarrow x=\dfrac{980}{1960}\\\\\Rightarrow x=0.5\ m

6 0
3 years ago
Read 2 more answers
A 62.00-cm guitar string under a tension of 70.000 N has a mass per unit length of 0.10000 g/cm. What is the highest resonant fr
quester [9]

Answer:

17,947.02 Hz

Explanation:

length (L) = 62 cm = 0.62 m

tension (T) = 70 N

mass per unit length (μ) = 0.10000 g/cm = 0.010000 kg/m

maximum frequency = 18,000 Hz

f = \frac{n}{2L} x \sqrt{\frac{T}{μ}}

f = \frac{n}{2 x 0.62} x \sqrt{\frac{70}{0.01} }

f = n x 67.47

18,000 = n x 67.47

n = 266.8≈ 266

the 267th overtone is the highest overtone that can be heard by this person, and its frequency would be 26 x 67.47 = 17,947.02 Hz

8 0
3 years ago
Two sound waves (speed 343 m/s) have different wavelengths. The first has a wavelength of 5.72 m, and the second a wavelength of
lys-0071 [83]

Answer:

The beat frequency is 30 Hz

Explanation:

Given;

velocity of the two sound waves, v = 343 m/s

wavelength of the first wave, λ₁ = 5.72 m

wavelength of the second wave, λ₂ = 11.44 m

The frequency of the first wave is calculated as follows;

F₁ = v/λ₁

F₁ = 343 / 5.72

F₁ = 59.97 HZ

The frequency of the second wave is calculated as follows;

F₂ = v/λ₂

F₂ = 343 / 11.44

F₂ = 29.98 Hz

The beat frequency is calculated as;

Fb = F₁ - F₂

Fb =  59.97 HZ - 29.98 Hz

Fb = 30 Hz

6 0
3 years ago
A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el
Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C

4 0
4 years ago
A point charge with a charge q1 = 4.00 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC mo
GrogVix [38]

Answer:

W = -0.480 J

Explanation:

given,

q₁ = 4 μC

q₂ = -4.10 μC

W = kq_1q_2(\dfrac{1}{a}+\dfrac{1}{b})

b = \sqrt{(0.27-0)^2+(0.27-0)^2}

b = 0.381

k = 8.99 × 10⁹ Nm²/C²

W = 8.99\times 10^9\times 4\times 10^{-6}\times (-4.1 \times 10^{-6})(\dfrac{1}{0.17}+\dfrac{1}{0.381})

W = [-147.436\times (5.88-2.62)\times 10^{-3}]J

W = -0.480 J

Work done by the electric force W = -0.480 J

4 0
3 years ago
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