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galben [10]
3 years ago
11

What is the ultimate energy source for most wind?

Physics
2 answers:
Varvara68 [4.7K]3 years ago
6 0

Answer:

solar radiation

Explanation:

Vinvika [58]3 years ago
5 0

The ultimate energy source for wind comes from Earth's uneven heating on Earth's surface which can be caused mainly by the sun.

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A mother pushes a baby stroller 50 meters by applying 10 newtons of force. How much work was done?
Alex
A mother pushes a baby stroller 50 meters by applying 10 newtons of force. How much work was done?

Answer.- 500 joules
5 0
3 years ago
Read 2 more answers
When an object is located 32 cm to the left of the lens, the image is formed 17 cm to the right of the lens. What is the focal l
Harman [31]

Answer: 11.1cm

Explanation:

Object distance (u) = 32cm

Image distance(v) = 17cm

Focal length(f) =?

Lens formular:

1/f = 1/u + 1/v

1/f = 1/32 + 1/17

Taking the L. C. M of 17 and 32

1/f = (17 + 32) / 544

1/f = 49/544

Taking the reciprocal

f = 544/49

f = 11.1 cm

Therefore, Focal length of the lens is 11.1cm

7 0
3 years ago
Small robots that can move around on the surface of a planet are called space shuttles.
Luden [163]

No.  "Shuttle" has been pretty much used to indicate shuttling
between Earth and Earth-orbit. 

The devices that have tooled around on the surface of Mars
have been referred to as "rovers".  That was also the nickname
of the buggy on which the Apollo astronauts moved around on
the Moon in the early 1970s ... the "lunar rover".

5 0
3 years ago
The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p
Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

4.85x10^{-6}pc

(b) How many meters are in a parsec?

3.081x10^{16}m

(c) How many meters in a light-year?

9.46x10^{15}m

(d) How many astronomical units in a light-year?

63325AU

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun (1.496x10^{11} m) is defined as 1 astronomical unit:

\tan{p} = \frac{1AU}{d}

Where d is the distance to the star.

Since p is small it can be represent as:

p(rad) = \frac{1AU}{d}  (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}

p('') = 2.06x10^5 p(rad)

p(rad) = \frac{p('')}{2.06x10^5} (2)

Then, equation 2 can be replace in equation 1:

\frac{p('')}{2.06x10^5} = \frac{1AU}{d}  

\frac{d}{1AU} = \frac{2.06x10^5}{p('')}  (3)

From equation 3 it can be see that 1pc = 2.06x10^5 AU

<em>a) How many parsecs are there in one astronomical unit? </em>

1AU . \frac{1pc}{2.06x10^5AU} ⇒ 4.85x10^{-6}pc

<em>(b) How many meters are in a parsec? </em>

2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU} ⇒ 3.081x10^{16}m

<em>(c) How many meters in a light-year? </em>

To determine the number of meters in a light-year it is necessary to use the next equation:

x = c.t

Where c is the speed of light (c = 3x10^{8}m/s) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

x = (3x10^{8}m/s)(31536000s)

x = 9.46x10^{15}m

<em>(d) How many astronomical units in a light-year?</em>

9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m} ⇒ 63325AU

<em>(e) How many light-years in a parsec?</em>

2.06x10^{5}AU . \frac{1ly}{63235AU} ⇒ 3.26ly

5 0
4 years ago
An electric field of intensity 3.80 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.
lakkis [162]

Answer: 1.55 x 10⁴ Nm²c^-1

Explanation: The electric flux, electric field intensity and area are related by the formulae below.

Φ= EAcosθ,

Where Φ= electric flux (Nm²c^-1)

E =electric field intensity (N/m²)

A = Area (m²)

θ= this is angle between the planar area and the magnetic flux

For our question E=3.80KN/c= 3800 N/c

A= 0.700 x 0.350= 0.245m²

θ= 0° ( this is because the electric field was applied along the x axis, thus the electric flux will be parallel to the area).

Hence Φ= 3800 x 0.245 x cos(0)

= 3800 x 0.245 x 1 (value of cos 0° =1)

= 1.55 x 10⁴ Nm²c^-1

Thus the electric field is 1.55 x 10⁴ Nm²c^-1

4 0
3 years ago
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