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Fed [463]
2 years ago
10

Rahul sees a flock of birds. He watches as the flying birds land in neat little rows on several power lines. Which change of sta

te does the motion of the birds most closely model?
melting
vaporization
deposition
sublimation
Physics
2 answers:
weqwewe [10]2 years ago
8 0

Answer:

Number 3, deposition. Does this help?

Explanation:

Lapatulllka [165]2 years ago
5 0
Think it deposition
You might be interested in
A lamp uses a 230 V mains supply and transfers 96 J of energy every second. Work out the current through the lamp. Give your ans
sertanlavr [38]

Answer:

0.4 A

Explanation:

From the question,

Electric power = Voltage×current

P = VI.......................... Equation 1

Make I the subject of the equation

I = P/V..................... Equation 2

Given: P = 96 J/s, V = 230 V.

Substitute into equation 2

I = 96/230

I = 0.4 A.

Hence the current is 0.4 A

8 0
2 years ago
the distance between the sun and earth is about 1.5X10^11 m. express this distance with an SI prefix and kilometers
Angelina_Jolie [31]
First, we write the SI prefixed. The SI unit for distance is meters.

Kilo = 10³
Mega = 10⁶
Giga = 10⁹
Terra = 10¹²

Because our value has ten to the power of 11, we will use the closest and lowest power prefix, which is giga. 

1.5 x 10¹¹ /  10⁹
= 1.5 x 10² Gm or 150 Gm

Writing in kilometers, we simply repeat the procedure except we divide by 10³ this time.

1.5 x 10¹¹ / 10³
= 1.5 x 10⁸ km
5 0
2 years ago
The intensity at a distance of 6.0 m from a source that is radiating equally in all directions is 6.0 × 10-10 w/m2 . what is the
satela [25.4K]
The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:
I= \frac{P}{A} (1)
where
P is the power
A is the area

In our problem, the intensity is I=6.0 \cdot 10^{-10} W/m^2. At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:
A=4 \pi r^2 = 4 \pi (6.0 m)^2 = 452.2 m^2

And so if we re-arrange (1) we find the power emitted by the source:
P=IA = (6.0 \cdot 10^{-10}W/m^2)(452.2 m^2)=2.7 \cdot 10^{-7} W
3 0
2 years ago
The Palo Verde nuclear power generator of Arizona has three reactors that have a combined generating capacity of 3.937×109 W . H
den301095 [7]

Answer:

t = 2.68 x 10¹⁴ years

Explanation:

First we need to find the amount of energy that Sun produce in one day.

Energy = Power * Time

Energy of Sun in 1 day = (3.839 x 10²⁶ W)(1 day)(24 hr/1 day)(3600 s/ 1 hr)

Energy of Sun in 1 day = 3.32 x 10³¹ J

Now, the time required by the nuclear power generator, in years, will be:

Energy of power generator = Energy Sun in 1 day = 3.32 x 10³¹ J

3.32 x 10³¹ J = Power * Time

3.32 x 10³¹ J = (3.937 x 10⁹ W)(t years)(365 days/1 year)(24 hr/1 day)(3600 s/ 1 hr)

t = 3.32 x 10³¹ /1.24 x 10¹⁷

<u>t = 2.68 x 10¹⁴ years</u>

8 0
2 years ago
The critical angle for a substance is measured at 53.7 degrees. light enters from air at 45.0 degrees. at what angle it will con
faust18 [17]
When light travels from a medium with greater refractive index n_1 to a medium with smaller refractive index n_2, there exists an angle (called critical angle) above which the light is totally reflected, and the value of this angle is given by
\theta_c = \arcsin ( \frac{n_2}{n_1} )
In this problem, we know that the critical angle is\theta_c = 53.7^{\circ}, so we can find the ratio between the refractive indices of the two mediums:
\frac{n_2}{n_1} = \sin \theta_c = \sin 53.7^{\circ} =0.81
and since the second medium is air (n=1.00), the refractive index of the first medium is
n_1=  \frac{n_2}{0.81}= \frac{1.00}{0.81}=1.23

In the second part of the problem, we have light entering from air (n_i = 1.00) at angle of incidence of \theta_i = 45.0 ^{\circ}, into the second medium with n_r = 1.23. By using Snell's law, we can find the angle of refraction of the light inside the medium:
n_i \sin \theta_i = n_r \sin \theta_r
\sin \theta_r =  \frac{n_i}{n_r}  \sin \theta_i = \frac{1.00}{1.23} \sin 45^{\circ}=0.574
\theta_r = \arcsin(0.574)=35.1^{\circ}
3 0
2 years ago
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