1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
enot [183]
3 years ago
7

What is the life cycle of an invertebrate?

Physics
2 answers:
Softa [21]3 years ago
7 0

Invertebrates vary tremendously in their body composition. It's common for invertebrate species to have four distinct stages to their life cycle: egg, larva, pupa, and adult, although this life cycle can vary. For example, octopus and squid hatch and grow into adults without a larva or pupa stage.

erica [24]3 years ago
7 0

The circle of life means that everything go back to where it started from

You might be interested in
What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh
Goshia [24]

Answer:

a) R_s = 0.092

b) R_p = 0.085

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2

using snell's law to calculate θ t

sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}

cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}

a) R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2

R_s = 0.092

b) R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2

R_p = 0.085

3 0
3 years ago
How does a toaster oven use micro waves
vekshin1
It uses microwaves as little heatwaves and uses them to heat the food
5 0
3 years ago
the half-life of carbon-14 is 5,730 years. After 11,460 year, how much of original carbon-14 remains?
Inessa [10]

Via the half-life equation:

A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}

Where the time elapse is 11,460 year and the half-life is 5,730 years.

A_{final}=A_{initial}(\frac{1}{2})^\frac{11460}{5730} \\\\A_{final}=A_{initial}\frac{1}{4} \\\\A_{final}=\frac{1}{4}A_{initial}

Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.

6 0
3 years ago
A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t
Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=\sqrt{\frac{Gmm}{R+h} }         [(R+h) is the distance of the satellite   from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = h_{i} = 98 km = 98000 m

∴Initial Energy (E_{i})  = \frac{1}{2}mv^{2} + \frac{-GMm}{(R+h_{i} )}

Substituting v=\sqrt{\frac{GMm}{R+h_{i} } } in the above equation and simplifying we get,

E_{i} = \frac{-GMm}{2(R+h_{i}) }

Similarly for final condition,

h=h_{f} = 198km = 198000 m

∴Final Energy(E_{f}) = \frac{-GMm}{2(R+h_{f}) }

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = E_{f} - E_{i}

        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

Substituting ,

M = 6 × 10^{24} kg

m = 1036 kg

G = 6.67 × 10^{-11}

R = 6400000 m

h_{i} = 98000 m

h_{f} = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = \frac{1}{2}m[v_{f} ^{2} - v_{i} ^{2}]

                                                          = \frac{GMm}{2}[\frac{1} {R+h_{f} } - \frac{1} {R+h_{i} }]

                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} }]

                                                             = 2ΔE

                                                             = 968.907 MJ

3 0
3 years ago
The engine of a jet airplane pushes exhaust gases from burning fuel backward
olga nikolaevna [1]
For more boost and to stop chases of fire
6 0
3 years ago
Other questions:
  • How long does it take a person to skate the width of a hockey rink (85 feet) at a constant speed of 15 feet per second?
    8·1 answer
  • Why is it warmer near the equator and cooler near the poles?
    9·2 answers
  • A sky diver jumps from a reasonable height above the ground. The air resistance she experiences is proportional to her velocity,
    15·1 answer
  • Is F=mxa expressed in equation in newtons second law of motion?
    12·1 answer
  • For an object moving with constant negative acceleration, draw the following:
    13·1 answer
  • At the instant the traffic light turns green, a car starts with a const acceleration of 7.00 ft/sA2. At the same instant a truck
    6·1 answer
  • how can you find the mechanical advantage of the six different simple machines? (simples mechinals are pulleys, inclined plane,
    9·1 answer
  • What force causes something to stop and slow down
    13·1 answer
  • True or false
    13·1 answer
  • How is the centripetal force related to the tangential velocity of the mass?.
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!