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ohaa [14]
2 years ago
5

Orion Balatik Big Dipper Apparent Brightness Absolute Brightness Polaris Star Trail Malihe Gemini Buwaya 1. Local constellation

that signifies the start of rainy season in the Philippines. 2. Local constellation that appears in the month of March. 3. It is seen in the Philippines during the months of April and May. 4. The constellation where we can find Merak and Dubhe. 5. It is the brightes star in the constelation Ursa Minor (Little Dipper). 6. It is a type of photograph that utilizes long-exposure times to capture the apparent motion of stars in the night sky due to the rotation of the Earth. 7. The star's brightness as seen from the Earth. 8. The brightness of the stars if they are at the same standard distance from Earth. 9. This constellation means hunter and is prominent in the night sky all over the world during winter. 10. The local constellation that appears on the month of February and signifies start of planting and setting of traps to protect the crops from animals.​
Physics
1 answer:
Roman55 [17]2 years ago
4 0
I’m nit sure about this but I say it’s 4 the constellation where we can find Merak and Dubhe.
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Is it proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinde
Gelneren [198K]

Answer:

No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.

Explanation:

A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.

Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:

* When the diameter and length are comparable (i.e have the same measurement)

When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.

Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.

8 0
2 years ago
Calculate the standard electrode potential difference (e°) of the daniell cell (at 1 bar) if temperature is 473.15 k.
anzhelika [568]
Missing data in the text of the exercise: The molar concentration of Zinc is 10 times the molar concentration of copper.

Solution:

1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
E^0=E_{cat}^0-E_{an}^0
where E_{cat}^0 is the standard potential at the cathode, while E_{an}^0 is the standard potential at the anode. For a Daniel Cell, at the cathode we have copper: E_{Cu}^0=+0.34 V, while at the anode we have zinc: E_{Zn}^0=-0.76 V. Therefore, at standard temperature the electrode potential difference of the Daniel Cell is
E^0=+0.34 V-(-0.76 V)=+1.1 V

2) To calculate E^0 at any temperature T, we should use Nerst equation:
E^0(T)=E^0- \frac{R T}{z F} \ln  \frac{[Zn]}{[Cu]}
where 
R=8.31 J/(K mol)
T=473.15 K is the temperature in our problem
z=2 is the number of electrons transferred in the cell's reaction
F=9.65\cdot 10^4 C/mol is the Faraday's constant
[Zn] and [Cu] are the molar concentrations of zinc and in copper, and in our problem we have [Zn]=10[Cu].
Using all these data inside the equation, and using E^0=+1.1 V, in the end we find:
E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V
8 0
3 years ago
archer shoots his arrow towards a target at a distance of 90 m, and hits ‘bullseye’ . Calculate the acceleration and time taken
Cerrena [4.2K]
Answer:

Acceleration = 45m/s^2
Time = 2 s

Explanation:

6 0
3 years ago
Red laser light from a He-Ne laser (λ = 632.8 nm) creates a second-order fringe at 53.2° after passing through the grating. What
Svetlanka [38]

Explanation:

It is given that,

Wavelength of red laser light, \lambda=632.8\ nm=632.8\times 10^{-9}\ m

The second order fringe is formed at an angle of, \theta=53.2^{\circ}

For diffraction grating,

d\ sin\theta=n\lambda

d=\dfrac{n\lambda}{sin\theta}, n = 2

d=\dfrac{2\times 632.8\times 10^{-9}}{sin(53.2)}

d=1.58\times 10^{-6}\ m

The wavelength λ of light that creates a first-order fringe at 22 is given by :

\lambda=d\ sin\theta

\lambda=1.58\times 10^{-6}\ sin(22)

\lambda=5.91\times 10^{-7}\ m

\lambda=591\ nm

Hence, this is the required solution.

6 0
3 years ago
When an object oscillating in simple harmonic motion is at its maximum displacement from the equilibrium position, which of the
ziro4ka [17]

Answer:

E. Zero Maximum

Explanation:

At the point of maximum displacement, the speed is zero while the restoring force is maximum. In fact:

- The restoring force is given by F=kx, where k is the spring constant and x is the displacement - at the point of maximum displacement, x is maximum, so F is maximum as well

- the total energy of the system is sum of kinetic energy and elastic potential energy:

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

where m is the mass of the system and v is the speed. Since E (the total energy) is constant due to the law of conservation of energy, we have that when K increases, U decreases, and viceversa. As a result, when x increases, v decreases, and viceversa. At the point of maximum displacement, x is maximum, so v will have its minimum value (which is zero, since the system is changing direction of motion).

4 0
3 years ago
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