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3 years ago

5

Define: New Moon, Full Moon, Waxing phases and Waning phases.

1 answer:

Nutka1998 [239]3 years ago

5 0

Full moon: All of the moon or sun appears covers by the shadow.
When the moon is waxing, it means that the lit part is getting larger (going towards a full moon)
When the moon is waning , it means that the lit part is getting smaller (going towards a new moon)
The new moon represents the start of a new lunar cycle and occurs approximately every 29 days.

Hope that help:)

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Seafloor spreading is a process that occurs along ___________1____________. At ___________2____________ boundary, two plates mov

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Answer:

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Explanation:

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3 years ago

Light is incident on an air-water interface at an angle of 30 degree to the normal. What angle does the refracted ray make with

Vikentia [17]

Answer:

22 degree

Explanation:

Angle of incidence, i = 30 degree

the refractive index of water with respect to air is 4/3.

As the ray of light travels from rarer medium to denser medium, that mean air to water, the refraction takes place.

According to Snell's law,

Refractive index of water with respect to air = Sin i / Sin r

Where, r be the angle of refraction

4 / 3 = Sin 30 / Sin r

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Sin r = 0.375

r = 22 degree

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6 0

3 years ago

A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a

Nimfa-mama [501]

Answer:

$F_0 = 393 N$

Explanation:

As we know that amplitude of forced oscillation is given as

$A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}$

here we know that natural frequency of the oscillation is given as

$\omega_0 = \sqrt{\frac{k}{m}}$

here mass of the object is given as

$m = \frac{W}{g}$

$\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}$

$\omega_0 = 8.48 rad/s$

angular frequency of applied force is given as

$\omega = 2\pi f$

$\omega = 2\pi(10.5) = 65.97 rad/s$

now we have

$0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}$

$F_0 = 393 N$

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3 years ago

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

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Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

7 0

3 years ago