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poizon [28]
3 years ago
6

A lunar eclipse can only happen during a(1) new moon.(2) solstice.(3) first quarter moon.(4) full moon.(5) perihelion passage of

the Sun.
Physics
1 answer:
Ray Of Light [21]3 years ago
7 0

Answer:

(4) full moon.

Explanation:

Lunar eclipse can only occur on a full moon night when the sun the earth and the moon are very much in a straight line.

During this period the the light of the sun that incidents on the moon is blocked by the earth and so we have the phases of the moon due to the relative motion of the three bodies which partially enables the light of the sun to reach the moon.

The moon appears orange-red during this time because the light that reaches the moon is after the refraction through the earth's atmosphere from which the other wavelengths have been absorbed by the earth's atmosphere.

Lunar eclipse can only occur at night and hence it can only be observed from about half of the earth.

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A certain AM radio wave has a frequency of 1.12 x 100 Hz. Given that radio waves travel at
riadik2000 [5.3K]

Answer: 267 m

Explanation:

2.99x10^8 m/s

———————-

1.12 x 10^6 Hz

3 0
3 years ago
A certain geothermal power station has a power output of 200,000W.
Ne4ueva [31]

Answer:4800 kWh

Explanation:

Given

Power output of station =2,00,000 W or 200 kW

because 1 kW=1000 W

For 24 hours, it will generate 200\times 24=4800\ kWh

5 0
3 years ago
A rocket moves through outer space at 11,000 m/s. At this rate, how much time would be required to travel the distance from Eart
VladimirAG [237]
Speed = 11000 m/s = 11km/s

D = 380000 km,

t = D/s = 380000 km/ 11km/s

t = 34 545.45 seconds.
4 0
3 years ago
Hunter aims directly at a target (on the same level) 75.0 m away. (a) If the bullet leaves the gun at a speed of 180 m/sby how m
tankabanditka [31]

Answer:

Part a)

\delta y = 0.85 m

Part b)

\theta = 0.65 degree

Explanation:

Part a)

As we know that the target is at distance 75 m from the hunter position

so here we will have

x = v_x t

here we know that

v_x = 180 m/s

so we have

75.0 = 180 (t)

t = 0.42 s

now in the same time bullet will go vertically downwards by distance

\delta y = \frac{1}{2}gt^2

\delta y = \frac{1}{2}(9.81)(0.42^2)

\delta y = 0.85 m

Part b)

In order to hit the target at same level we need to shot at such angle that the range will be 75 m

so here we have

R = \frac{v^2 sin(2\theta)}{g}

75 = \frac{180^2 sin(2\theta)}{9.81}

\theta = 0.65 degree

4 0
4 years ago
A
True [87]

Answer:

\Delta T=3.615^{\circ}C is the drop in the water temperature.

Explanation:

Given:

  • mass of ice, m_i=14.7\ g=0.0147\ kg
  • mass of water, m_w=324\ g=0.324\ kg

Assuming the initial temperature of the ice to be 0° C.

<u>Apply the conservation of energy:</u>

  • Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.

<u>Now from the heat equation:</u>

Q_i=Q_w

m_i.L=m_w.c_w.\Delta T ......................(1)

where:

L= latent heat of fusion of ice =333.55\ J.g^{-1}

c_w= specific heat of water =4.186\ J.g^{-1}.^{\circ}C^{-1}

\Delta T= change in temperature

Putting values in eq. (1):

14.7 \times 333.55=324\times 4.186\times \Delta T

\Delta T=3.615^{\circ}C is the drop in the water temperature.

8 0
3 years ago
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