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lions [1.4K]
2 years ago
15

Which of the following statements about lunar phases is true? (A) Only one quarter of the first-quarter moon is illuminated by t

he Sun.(B) It is possible to have two full moons during January, but not during February.(C) It is possible to have two full moons during November, but not during December.(D) The time between new moons is two weeks.(E) The full moon sometimes rises around midnight.
Physics
1 answer:
saul85 [17]2 years ago
7 0

Answer:

A) False B) True C) False D) False E) False

Explanation:

A) At any point of time half of the moon is always illuminated by the sun. We are able to see only the part of the moon which is facing us.

B)  Typically a Lunar cycle phases take 29.5 days to complete one cycle i.e. it will take 29.5 days for one event say full moon to reoccur. So except for February all the other month can have two full moons.

C) As explained in the above answer any month except February can have two full moons.

D) The tame taken between new moon is 29.5 days.

E) Full moon always rises at the sunset and sets at the sunrise

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A star produces 2x10^26 watts. how much energy does it lose every minutes
Step2247 [10]

Answer:

Energy loss per minute will be 120\times 10^{26}j

Explanation:

We have given the star produces power of 2\times 10^{26}W

We know that 1 W = 1 J/sec

So 2\times 10^{26}W=2\times 10^{26}J/sec

Given time = 1 minute = 60 sec

So the energy loss per minute =2\times 10^{26}\times 60=120\times 10^{26}j

We multiply with 60 we have to calculate energy loss per minute

7 0
3 years ago
A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

7 0
3 years ago
Difference between male and female gamete​
Inga [223]

Answer:

The male gamete is smaller in comparison to that of the female gamete.

The male gamete is conical from the front while the female gamete is spherical.

The cytoplasm in the male gamete is less in comparison to that of the female gamete.

The male gamete can move with the help of the tail while the female gametes are immobile and do not have any tail or flagella present.

The number of mitochondria present in the sperm is less than the number of mitochondria present in the ovum.

The male gamete has acrosome present in the head region that contains enzymes for dissolving the membranes present around the ovum. The ovum does not contain such digestive enzymes.

Explanation:

5 0
3 years ago
Read 2 more answers
Calculate the Latent Heat of Vaporization. (Please see picture attached)
Hunter-Best [27]

Answer:

20 J/g

Explanation:

In this question, we are required to determine the latent heat of vaporization

  • To answer the question, we need to ask ourselves the questions:

What is latent heat of vaporization?

  • It is the amount of heat required to change a substance from its liquid state to gaseous state without change in temperature.
  • It is the amount of heat absorbed by a substance as it boils.

How do we calculate the latent heat of vaporization?

  • Latent heat is calculated by dividing the amount of heat absorbed by the mass of the substance.

In this case;

  • Mass of the substance = 20 g
  • Heat absorbed as the substance boils is 400 J (1000 J - 600 J)

Thus,

Latent heat of vaporization = Quantity of Heat ÷ Mass

                                             = 400 Joules ÷ 20 g

                                             = 20 J/g

Thus, the latent heat of vaporization is 20 J/g

7 0
3 years ago
An object weighing 4 newtons swings on the end of a string as a simple pendulum. At the bottom of the swing, the tension in the
Anit [1.1K]

Answer:

(B) 0.5 g

Explanation:

Newton's second law says ∑  F i = m a .

the rate of change in momentum of a body is proportional to the force applied on the body.

f∝ma

f=kma

were k is constant and equal to 1

The centripetal acceleration is an acceleration.

the tension on the swing and object weight goes to the left hand side while the centripetal acceleration goes to the right handside

At the bottom of the swing, ΣF = FT – mg = mac;

notice that the tension in the swing is 1.5 times the weight of the object

we can write

1.5mg – mg = mac,

0.5mg = mac

0.5 g=ac

8 0
3 years ago
Read 2 more answers
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