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devlian [24]
3 years ago
15

The very best atmospheric conditions on earth result in an angular resolution limit of about 0.4 arcseconds if an astronomy obse

rves under such conditions at visible wavelengths using a 0.2-meter telescope, the angular resolution is limited by
A) Turbulance in the earths atmosphere
B) Diffraction by dust in the interstellar medium
C)Variable changes in the physical size of the star being observed
D)Diffraction effects from the telescopes optics
Physics
1 answer:
Veseljchak [2.6K]3 years ago
3 0
The correct answer would be A
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uniformly charged conducting sphere of 1.2 m diameter has surface charge density 8.1 mC/m2. Find (a) the net charge on the spher
Stels [109]

Answer:

(a) = 3.7 × 10⁻⁵

(b) = 4.1 × 10⁻⁶ N.M²/C

Explanation:

(a) Diameter of the sphere, d = 1.2 m

Radius of the sphere, r = 0.6 m

Surface charge density, = 8.1 mC/m2 = 8.1 × 10⁻⁶ C/m²

Total charge on the surface of the sphere,

Q = Charge density × Surface area

    = 4πr²σ

   = 4 (3.14) (0.12²) (8.1 × 10⁻⁶)

   = 3.66 × 10⁻⁵C

   ≅ 3.7 × 10⁻⁵C

Therefore, the net charge on the sphere is  3.7 × 10⁻⁵C

(b)

Total electric flux (∅)

=Q / ε₀

ε₀ =  8.854 × 10⁻¹² N⁻¹C² m⁻²

Q =   3.66 × 10⁻⁵C

 =  3.66 × 10⁻⁵ / 8.854 × 10⁻¹²

= 4.1 × 10⁻⁶ N.M²/C

5 0
3 years ago
A current of 12 amps is measured in a circuit with a total resistance of 9.0 ohms. What is the size of the voltage source that s
Lynna [10]
A. 108 volts is the answer.
7 0
3 years ago
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28. Someone throws a rubber ball vertically upward from the roof of a building 8.00 m in height. The ball rises,
Debora [2.8K]

Answer:

v_o= 12 m/s

Explanation:

The ball experienced a constant acceleration motion, so we need to apply the following formula:

y=y_o+v_o*t+\frac{1}{2}*a*t^2\\\\vo=\frac{y-y_o-\frac{1}{2}*a*t^2}{t}\\\\v_o=\frac{0m-8m-\frac{1}{2}*(-9.81m/s^2)*(3.00s)^2}{3.00s}\\v_o=12m/s

Note: we set the acceleration of gravity as negative because it is going down.

the ball was thrown vertically upward with an initial velocity of 12 m/s

6 0
3 years ago
A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 41.5-gram mass is attached at the 28.5-cm
zzz [600]

Answer:

The value is  M  =  42.3 \  kg

Explanation:

From the question we are told that

    The first  position of the fulcrum  is  x = 49.7 cm

    The mass  attached is m  =  41.5 \  g

    The position of the attachment is  x_1 =  28.5 \  cm  

    The second position of the fulcrum is  x_2 = 39.2 \  cm

Generally the sum of clockwise torque =  sum of anti - clockwise torque

So  

       CWT  =  m (x_2 - x_1)

Here CWT  stands for clockwise torque

       ACWT  =  M  ( x - x_2)

So

      m (x_2 - x_1) =    M  ( x - x_2)

=>   41.5  (39.2 -  28.5 ) =    M  ( 49.7  -39.2 )

=>    M  =  42.3 \  kg

3 0
3 years ago
The sphere that refers to Earth's water is called what?
WARRIOR [948]

Answer:

Its called the hydrosphere UwU!

Explanation:

6 0
3 years ago
Read 2 more answers
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