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AlekseyPX
3 years ago
7

What information can be determined by observing dark lines within the spectrum of a given star?

Physics
1 answer:
Scrat [10]3 years ago
6 0
I believe the answer is D.  
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In dim light
faltersainse [42]

The Answer is A, the iris dilates the pupil.

4 0
3 years ago
Read 2 more answers
A real (non-Carnot) heat engine, operating between heat reservoirs at temperatures of and performs 4.1 kJ of net work, and rejec
Sati [7]
There are some missing data in the problem. The full text is the following:
"<span>A </span>real<span> (</span>non-Carnot<span>) </span>heat engine<span>, </span>operating between heat reservoirs<span> at </span>temperatures<span> of 710 K and 270 K </span>performs 4.1 kJ<span> of </span>net work<span>, and </span>rejects<span> 9.7 </span>kJ<span> of </span>heat<span>, in a </span>single cycle<span>. The </span>thermal efficiency<span> of a </span>Carnot heat<span> engine, operating between the same </span>heat<span> reservoirs, in percent, is closest to.."

Solution:
The efficiency of a Carnot cycle working between cold temperature </span>T_C and  hot temperature T_H is given by
\eta = 1 - \frac{T_C}{T_H}
and it represents the maximum efficiency that can be reached by a machine operating between these temperatures. If we use the temperatures of the problem, T_C=270 K and T_H=710 K, the efficiency is
\eta = 1 - \frac{270 K}{710 K}=0.62 = 62%

Therefore, the correct answer is D) 62 %.
6 0
3 years ago
In a single-slit diffraction experiment, a coherent light source illuminates a slit in a barrier, and the resulting pattern is p
UkoKoshka [18]

Answer:

hsjdsdddwqdqwdd

Explanation:

4 0
3 years ago
If you need 40.0 Nm of torque in order to loosen a nut on a wn
KonstantinChe [14]

Answer:

0.301 m

Explanation:

Torque = Force × Radius

τ = Fr

40.0 Nm = 133 N × r

r = 0.301 m

The mechanic must apply the force 0.301 m from the nut.

6 0
3 years ago
A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere?
11111nata11111 [884]

Answer:

\sigma=0.014\ C/m^2

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2

So, the surface charge density on the sphere is 0.014\ C/m^2.

7 0
3 years ago
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