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MrRissso [65]
3 years ago
14

Think of an animal and list two resources that it might compete for in its community . Then describe what adaptions the animals

has to compete for these resources
Physics
1 answer:
Viefleur [7K]3 years ago
6 0
An animal is a lion, it may compete for food and it’s territory. The animal may have to fight other animals to get these things
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List six forms of electromagnetic radiation from the shortest waves(highest energy) to the longest waves (lowest energy)
romanna [79]
That's the reverse of RIVUXG
so your answer is
gamma rays
x rays
ultraviolet light
visible light
infrared
radio waves
8 0
3 years ago
(a) State and explain which of the arrangements would have the greater extension of spring(s). (b) Explain if there are any chan
swat32

Answer:
arrangement 2

Explanation:
arrangement 1's spring would broke idek

3 0
1 year ago
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When a single source of light shines through an extremely thin rectangular slit and projects on a far away viewing screen, a sin
saveliy_v [14]
True
If it helps you plz brainlest me
7 0
3 years ago
If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave
kotegsom [21]

Answer:

0.82 mm

Explanation:

The formula for calculation an n^{th} bright fringe from the central maxima is given as:

y_n=\frac{n \lambda D}{d}

so for the distance of the second-order fringe when wavelength \lambda_1 = 745-nm can be calculated as:

y_2 = \frac{n \lambda_1 D}{d}

where;

n = 2

\lambda_1 = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y_2 = 0.00276 m

y_2 = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength \lambda_2 = 660-nm is as follows:

y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y^'}_2 = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

\delta y = y_2-y^{'}_2

\delta y = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

\delta y = 10 ⁻³ (2.76 - 1.94)

\delta y = 10 ⁻³ (0.82)

\delta y = 0.82 × 10 ⁻³ m

\delta y =  0.82 × 10 ⁻³ m (\frac{1.0mm}{10^{-3}m} )

\delta y = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

6 0
3 years ago
What is the potential energy of a puppy that weighs 18 N istting in a high chair 2 m high?
kykrilka [37]

Answer:

Potential energy =mass* acceleration due to gravity * height

mass*acceleration due to gravity =weight

hence potential energy of the puppy= weight * height

=18*2

=36 joule

6 0
3 years ago
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