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3 years ago

PLEASE HELP ASAP!!!!!!!!!!WILL GIVE BRAINLIEST...

Physics

2 answers:

Delicious77 [7]3 years ago

7 0

The answer to this is D. Green.

joja [24]3 years ago

6 0

The answer is A because of the pigment chlorophyll that absorbs blue and red light

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Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon'

Ne4ueva [31]

Explanation:

Below is an attachment containing the solution

4 0

4 years ago

Small birds can migrate over long distances without feeding, storing energy mostly as fat rather than carbohydrate. (Figure 1) F

Katen [24]

Answer: 3.1107grams of fat.

Explanation:The speed to cover 800 km in 20 hours will be 40km/hr, Because

Speed =Distance/Time.

This is equal to 72,000 seconds. At a power consumption of 1.7 W, the bird will consume (1.7 X 72,000) =122,400 J of energy.

A gram of fat contains about (9.4 Calories X 4186 J/Calorie) ,=39348 joules of energy, the bird will need (122,400 / 39,348) = 3.1107 grams of fat.

4 0

3 years ago

Define reflection and give to examples define refraction and give two examples

pickupchik [31]

Reflection- looking at yourself in a mirror and seeing the moon in the sky
Refraction- putting on your glasses in order to see more clearly and the pencil looks like it’s bent when you stick one end of it in water.

6 0

4 years ago

Un reloj de péndulo de largo L y período T, aumenta su largo en ΔL (ΔL << L). Demuestre que su período aumenta en: ΔT = π

Kruka [31]

Answer:

ΔT = $\pi \ \frac{\Delta L}{\sqrt{Lg} }$

Explanation:

In a simple harmonic motion, specifically in the simple pendulum, the angular velocity

w = $\sqrt{\frac{g}{L} }$

angular velocity and period are related

w = 2π / T

we substitute

2π / T = \sqrt{\frac{g}{L} }

T = $2\pi \ \sqrt{\frac{L}{g} }$

In this exercise indicate that for a long Lo the period is To, then and increase the long

L = L₀ + ΔL

we substitute

T = $2\pi \ \sqrt{\frac{L + \Delta L}{g} }$

T = $2\pi \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }$

in general the length increments are small ΔL/L «1, let's use a series expansion

$\sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...$

we keep the linear term, let's substitute

T = $2\pi \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L} )$

if we do

T = T₀ + ΔT

T₀ + ΔT = $2\pi \sqrt{\frac{\Delta L}{g} } + \pi \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}$

T₀ + ΔT = T₀ + $\pi \sqrt{\frac{1}{Lg} } \ \Delta L$

ΔT = $\pi \ \frac{\Delta L}{\sqrt{Lg} }$

4 0

3 years ago

A scientist must only base his or her conclusions on observable evidence from investigation

Nezavi [6.7K]

is this a true of false cuz if so this it true

8 0

3 years ago