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Irina18 [472]
3 years ago
8

I need help with all the blanks

Physics
1 answer:
levacccp [35]3 years ago
6 0
There are only two out of the four squares that contain bb. So, 50% of the offspring will be brown. The other 50% of the offspring will be a hybrid because their combination is Bb. There is a 50% chance that all of the offspring could be born black.
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Although all sports require a person to be in shape to some degree, some
Alexandra [31]

Answer:

Chess

Explanation:

Chess is considered a sport

3 0
3 years ago
The Hubble Space Telescope in orbit above the Earth has a 2.4 m circular aperture. The telescope has equipment for detecting ult
strojnjashka [21]

Answer:

Option d

The minimum angular separation between two objects that the Hubble Space Telescope can resolve is 4.8x10^{-8}rad.  

Explanation:

The resulting image in a telescope that will be gotten from an object is a diffraction pattern instead of a perfect point (point spread function (PSF)).

That diffraction pattern is gotten because the light encounters different obstacles on its path inside the telescope (interacts with the walls and edges of the instrument).

 

The diffraction pattern is composed by a central disk, called Airy disk, and diffraction rings.    

 

The angular resolution is defined as the minimal separation at which two sources can be resolved one for another, or in other words, when the distance between the two diffraction pattern maxima is greater than the radius of the Airy disk.

The angular resolution can be determined in analytical way by means of the Rayleigh criterion.          

\theta = 1.22\frac{\lambda}{D}  (1)

Where \lambda is the wavelength and D is the diameter of the telescope.

Notice that it is necessary to express the wavelength in the same units than the diameter.

\lambda = 95nm \cdot \frac{1x10^{-9}m}{1nm} ⇒ 9.5x10^{-8}m

Finally, equation 1 can be used.

\theta = 1.22(\frac{9.5x10^{-8}m}{2.4m})

\theta = 4.8x10^{-8}rad              

Hence, the minimum angular separation between two objects that the Hubble Space Telescope can resolve is 4.8x10^{-8}rad.    

5 0
3 years ago
Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the
Airida [17]

Complete Question:

Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

Potential V is given as;

V =\frac{Kq}{r}

where;

K is coulomb's constant = 9x10⁹ N.m²/C²

r is the distance of the charge

q is the magnitude of the charge

The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V

Total potential due to this charges  = 4500 V + 6750 V = 11250 V

6 0
4 years ago
An iron wire has a cross-sectional area equal to 5.00×10⁻⁶ m² . Carry out the following steps to determine the drift speed of th
Doss [256]
  1. In mass, there are 55.85 × 10⁻³ kg/mol in in 1 mole of iron.
  2. The molar density of iron is equal to 1.41 × 10⁵ mol/m³.
  3. The density of iron atoms is equal to 8.49 × 10²⁸ atoms/m³.
  4. The number density of conduction electrons is equal to 1.70 × 10²⁹ conduction electrons/m³.
  5. The drift speed of conduction electrons is equal to 2.21 × 10⁻⁴ m/s.

<h3>How to calculate the drift speed of the conduction electrons?</h3>

Mathematically, the drift speed of the conduction electrons can be calculated by using this formula:

V = (m × σ × V)/ρ × e × f × l)

V = I/(n × A × Q)

Where:

  • U represents the drift speed of the conduction electrons, in m/s.
  • m represents the molecular mass of the metal, in kg.
  • e represents the elementary charge, in C.
  • f represents the number of free electrons per atom.
  • σ represents the electric conductivity of the medium at a particular temperature in S/m.
  • ρ represents the density of the conductor, in kg/m³.
  • ℓ represents the length of the conductor, in m.
  • ΔV represents the voltage applied or potential difference across the conductor in V.

<h3>How many kilograms are there in 1 mole of iron? </h3>

Molar mass of iron = 55.85 g/mol.

In Kilograms, we have:

Mass = 55.85 × 1/1000

Mass = 55.85 × 10⁻³ kg/mol.

For the molar density of iron, we have:

Molar density = density/molar mass

Molar density = 7874/0.056

Molar density = 1.41 × 10⁵ mol/m³.

For the density of iron atoms, we have:

Density of iron atoms = Avogadro's constant × molar density

Density of iron atoms = 6.023 × 10²³ × 1.406 × 10⁵

Density of iron atoms = 8.49 × 10²⁸ atoms/m³.

For the number density of conduction electrons, we have:

Fe ---> Fe²⁺ + 2e⁻

Number density of conduction electrons = 2 conduction electrons/1 atom of iron

Number density of conduction electrons = 2 × 8.49 × 10²⁸

Number density of conduction electrons = 1.70 × 10²⁹ conduction electrons/m³.

For the drift speed of conduction electrons, we have:

V = I/(n × A × Q)

V = 30/(1.70 × 10²⁹ × 1.602 × 10⁻¹⁹ × 5 × 10⁻⁶)

Drift speed, V = 2.21 × 10⁻⁴ m/s.

Read more on drift speed here: brainly.com/question/15219891

#SPJ4

Complete Question:

An iron wire has a cross-sectional area of 5.00 x 10-6 m2. Carry out steps (a) through (e) to compute the drift speed of the conduction electrons in the wire.

(a) How many kilograms are there in 1 mole of iron?

(b) Starting with the density of iron and the result of part (a), compute the molar density of iron (the number of moles of iron per cubic meter).

(c) Calculate the number density of iron atoms using Avogadro’s number.

(d) Obtain the number density of conduction electrons given that there are two conduction electrons per iron atom.

(e) If the wire carries a current of 30.0 A, calculate the drift speed of conduction electrons.

4 0
2 years ago
Please help me with Part B &amp; help me correct the rest of the questions
AleksAgata [21]

Answer:

Magnetic activity and sunspots are in a proportional relationship.

Explanation:

From the graph, as one increases, so does the other, thus we describe the relationship as proportional.

Other than that, all of your other answers seem correct.

5 0
3 years ago
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