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Allushta [10]
3 years ago
5

what times are the moon phases visible? My science teacher said that the phases of the moon are only visible at sertant times of

the day. Did i hear him right?
Physics
1 answer:
klio [65]3 years ago
6 0
Your teacher is right. The moon can be seen early in the morning sometimes and late at night. Different phases are only visible on certain days as one day might be full quarter, the next full moon, the next first quarter, etc.
You might be interested in
How many Joules of energy are required to run a 100W light bulb for one day?
Vlada [557]

The following are the answers to the questions presented:

a. The joules of energy required to run a 100W light bulb for one day is 8640000J
b. The amount of coals that has to be burned to light that light bulb for one day is 0.96kg

The solution would be like this for this specific problem:

<span>P=<span>W/s</span>→W=Pt=100W1day <span><span>24h/</span><span>1day </span></span><span><span>3600s/</span><span>1h</span></span>=8640000J</span>

<span>W=<span>30/100</span>wm→m=<span><span>100W/</span><span>30w</span></span>=<span><span>100×8640000J/</span><span>30×30×<span>10in thepowerof6 </span><span>J/<span>kg</span></span></span></span>=0.96kg</span>

<span>I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.</span>

4 0
3 years ago
ASAP
nikitadnepr [17]

Answer:

A. 59.4

Explanation:

The refractive index of the glass, n₁ = 1.50

The angle of incidence of the light, θ₁ = 35°

The refractive index of air, n₂ = 1.0

Snell's law states that n₁·sin(θ₁) = n₂·sin(θ₂)

Where;

θ₂ = The angle of refraction of the light, which is the angle the light will have when it passes from the glass into the air

Therefore;

θ₂ = arcsin(n₁·sin(θ₁)/n₂)

Plugging in the values of n₁, n₂ and θ₁ gives;

θ₂ = arcsin(1.50 × sin(35°)/1.0) ≈ 59.357551° ≈ 59.4°

The angle the light will have when it passes from the glass into the air, θ₂ ≈ 59.4°.

6 0
3 years ago
Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R.
Dennis_Churaev [7]

Answer:

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

Explanation:

The rotational kinetic energy when the cylinder is with the rope is:

E_k=\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^2

where we used the fact that both rope and cylinder hast the same w. This E_k must conserve, that is, E_k must equal E_k when the rope leaves the cylinder. Hence, the final w is given by:

E_{k1}=E_{k2}\\\\\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^{2}=\frac{1}{2}I_c\omega^2\\\\\omega=\sqrt{\omega_0^2(\frac{I_c+I_r}{I_c})} (1)

For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:

I_c=\frac{1}{2}MR^2\\\\I_r=mR^2

Finally, by replacing in (1):

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

hope this helps!!

7 0
3 years ago
A 2N and 6N force pull on an object to the right and a 4N force pulls to the left a 0.5kg object. What is the net force on the o
Angelina_Jolie [31]

Explanation:

F net = 2+6-4 ( 2 and 6 N are in same direction so they get added, 4N in opposite direction so it will be subtracted)

F net=4 N

3 0
3 years ago
Please answer me my question​
Angelina_Jolie [31]
3 - b) weight decreases
4 -a) 50 kg as mass is same everywhere
5 - b) 200 dynes
5 0
3 years ago
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