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maks197457 [2]
2 years ago
9

Listed following are locations and times at which different phases of the moon are visible from earth’s northern hemisphere. mat

ch these to the appropriate moon phase.
Physics
1 answer:
Airida [17]2 years ago
7 0
For the answer tot he questions above, I know this one.
The answers are
 <span>-sets 2-3 hours after the sun sets - waxing crescent 
-occurs about 3 days before the new moon - waning crescent 
-occurs 14 days after the new moon - full moon 
-rises at about the time the sun sets - full moon 
-visible due south at midnight - full moon 
-visible near eastern horizon just before sunrise waning crescent 
-visible near western horizon about an hour after sunset waxing crescent</span>
You might be interested in
A dog is 60m away while moving at constant velocity of 10m/s towards you. Where is the dog after 4 seconds?
evablogger [386]
20m away

the dog was 60m away from. you subtract 40m since it is 10m/s x 4 seconds
4 0
2 years ago
Read 2 more answers
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water wit
melamori03 [73]

Answer:

Explanation:

Given that,

Height of the bridge is 20m

Initial before he throws the rock

The height is hi = 20 m

Then, final height hitting the water

hf = 0 m

Initial speed the rock is throw

Vi = 15m/s

The final speed at which the rock hits the water

Vf = 24.8 m/s

Using conservation of energy given by the question hint

Ki + Ui = Kf + Uf

Where

Ki is initial kinetic energy

Ui is initial potential energy

Kf is final kinetic energy

Uf is final potential energy

Then,

Ki + Ui = Kf + Uf

Where

Ei = Ki + Ui

Where Ei is initial energy

Ei = ½mVi² + m•g•hi

Ei = ½m × 15² + m × 9.8 × 20

Ei = 112.5m + 196m

Ei = 308.5m J

Now,

Ef = Kf + Uf

Ef = ½mVf² + m•g•hf

Ef = ½m × 24.8² + m × 9.8 × 0

Ef = 307.52m + 0

Ef = 307.52m J

Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw

7 0
2 years ago
Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o
earnstyle [38]

The correct answer is Option C) Sample C would be best, because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest.


As the coefficient of absorption would define the energy present in the reflected wave, the material C has the highest percentage of absorption i.e. 62% and would be best suitable to make a sound proof room.

4 0
2 years ago
Read 2 more answers
A balloon drifts 140m toward the west in 45s ; then the wind suddenly changes and the balloon flies 90m toward the east in the n
Bogdan [553]

Answer: 140 m

Explanation:

Let's begin by stating clear that motiont is the change of position of a body at a certain time. So, during this motion, the balloon will have a trajectory and a displacement, being both different:

The<u> trajectory</u> is <u>the path followed by the body, the distance it travelled</u> (is a scalar quantity).  

The displacement is <u>the distance in a straight line between the initial and final position</u> (is a vector quantity).  

So, according to this, the distance the balloon traveled during the first 45 s (its trajectory) is 140 m.

But, if we talk about displacement, we have to draw a straight line between the initial position of the balloon (point 0) to its final position (point 90 m).  Being its displacement 95 m.

8 0
2 years ago
A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by
Airida [17]

Answer:

Mobility of the minority carriers, \mu_{n} =1184.21 cm^{2} /V-sec

Diffusion coefficient for minority carriers,D_{n} = 29.20 cm^2 /s

Verified from Einstein relation as  \frac{D_{n} }{\mu_{n} }  = 25 mV

Explanation:

Length of sample, l_{s} = 2 cm

Separation between the two probes, L = 1.8 cm

Drift time, t_{d} = 0.608 ms

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), \mu_{n} = \frac{V_{d} }{E}

Where the drift velocity, V_{d} = \frac{L}{t_{d} }

V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s

and the Electric field strength, E = \frac{V}{l_{s} }

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec

The electron diffusion coefficient, D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }

\triangle x = (\triangle t )V_{d}, where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3}  }\\\triangle x =0.533 cm

D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s

For the Einstein equation to be satisfied, \frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V

\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV

Verified.

4 0
2 years ago
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