Maggots feed on dead and decaying organisms for energy. What are maggots?autotrophs
producers
decomposers
heterotrophs

Answers

Answer 1
Answer:

Answer:

Explanation:

Decomposers is the correct answer

Answer 2
Answer:

Answer:

Decomposers is the right answer

Explanation:

Maggots are decomposers because they eat the dead bodys for energy

I don't know if the thing I wrote it truse so ya


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A 8.00-μF capacitor that is initially uncharged is connected in series with a 3.00-Ω resistor and an emf source with E = 70.0 V and negligible internal resistance. At the instant when the resistor is dissipating electrical energy at a rate of 300 W, how much energy has been stored in the capacitor?

Answers

Answer:

The energy stored is E_s = 0.0064 \  J

Explanation:

From the question we are told that  

   The capacitance  is  C =  8 \ \mu F = 8*10^(-6) \  F

    The resistance is  R = 3.00-Ω

    The emf is  E_t = 70.0 V

      The power  is  P = 300 W

Generally the total  emf is mathematically represented as

   E_t = E_c + E_r

Here  E_c is the emf across that capacitor which is mathematically represented as

      E_c  =  (q)/(C)

and E_r is the emf across the resistor which is mathematically represented as

       E_r  = √(P R)

So  

       E_t  =  √(PR)  + (q)/(C)

=>    q = C[E_t - √(PR) ]

Generally the energy stored in a capacitor is mathematically represented as

          E_s = (q^2)/(2C)

=>      E_s = ([C [ E_t - √(PR) ]]^2)/(2C)

=>      E_s = ([8.0*10^(-6) [ 70 - √(300 * 3))/(2 *(8.0*10^(-6)))

=>      E_s = 0.0064 \  J

Final answer:

The energy stored in the capacitor is 0 J.

Explanation:

When a 8.00-μF capacitor that is initially uncharged is connected in series with a 3.00-Ω resistor and an emf source with E = 70.0 V

At the instant when the resistor is dissipating electrical energy at a rate of 300 W, we can calculate the current flowing through the circuit using Ohm's law: I = V/R = 70.0 V / 3.00 Ω = 23.33 A.

The energy stored in a capacitor can be calculated using the formula: E = 1/2 * C * V^2, where C is the capacitance and V is the voltage across the capacitor.

Since the capacitor is initially uncharged, the voltage across it is also zero. So the energy stored in the capacitor is 0.5 * 8.00 x 10^-6 F * (0 V)^2 = 0 J.

Learn more about Energy stored in a capacitor here:

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when a ball is dropped off a cliff in free fall, it has an acceleration of 9.8 m/s^2. what is its acceleration as it gets closer to the ground

Answers

acceleration due to gravity is contract for the purposes of this question, so the acceleration would remain at 9.8 m/s^2

If the ball, the cliff, and the ground are all on the Earth, and everything is bathed in an ocean of air, then the ball's acceleration will decrease as it falls, because of the friction of air resistance.  If it has far enough to fall, it's possible that its acceleration may even become zero, and the ball settle on a constant speed (called "terminal velocity") before it hits the ground.

But until we get to College-level Physics and Engineering, we ALWAYS ignore that stuff, and assume NO AIR RESISTANCE.  The ball is in FREE FALL, and the ONLY force acting on it is the force of gravity.   We also assume that the distance of the fall is small enough so that the value of gravity is constant over the entire fall.

Under those assumptions, there's nothing present to change the acceleration of the falling ball.  It's 9.81 m/s² when it rolls off the edge of the cliff, and it's 9.81 m/s² when it hits the ground.

If a barometer reads 772 mm hg, what is the atmospheric pressure expressed in pounds per square inch?

Answers

15.23.....................
I think it would be 15.23 not so sure but
hope this helps! (:

Consider a bird that flies at an average speed of 10.7 m/s and releases energy from its body fat reserves at an average rate of 3.70 W (this rate represents the power consumption of the bird). Assume that the bird consumes 4.00g of fat to fly over a distance db without stopping for feeding. How far will the bird fly before feeding again?Fat is a good form of energy storage because it provides the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories, compared to 4.20 (food) Calories per 1.00 grams of carbohydrate. Remember that Calories associated with food, which are always capitalized, are not exactly the same as calories used in physics or chemistry, even though they have the same name. More specifically, one food Calorie is equal to 1000 calories of mechanical work or 4186 joules. Therefore, in this problem use the conversion factor 1Cal=4186J.

Answers

Answer:

The distance covered by the bird before feeding is 4.55 * 10^(5)m.

Explanation:

As the bird consumes 4 g of fat before flying, the amount of initial food energy (E_(F)) stored by it is given by

E_(F) = 4 g * 9.4 (food) cal = 37.6 (food) cal

So the mechanical energy stored by the bird (E_(M)) is given by

E_(M) = E_(F) * 4186 J = 1.57 * 10^(5) J

Given, the power consumed by the bird P = 3.7 W

So, the time (t) required to consume this power by the bird is

t = (E_(M))/(P) = (1.57 * 10^(5) J)/(3.7 W) = 4.24 * 10 ^(4) s

As the bird flies at an average speed (v) of 10.7 ms^(-1), so the distance (d) covered by the bird before feeding again is given by

d = v * t = 10.7 ms^(-1) * 4.25 * 10 ^(4) s = 4.55 * * 10^(5) m

The distance of  the bird'sflight before him/her feeds again is mathematically given as

d = 4.55* 10^{5} m

What is the distance of the bird's flight before him/her feeds again?

Question Parameter(s):

a bird that flies at an averagespeed of 10.7 m/s

its body fat reserves at an average rate of 3.70 W

the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories,

Generally, the initial food energy  is mathematically given as

Ex= 4 g*9.4

Ex= 37.6cal

Therefore, the mechanical energy

Em = Ex * 4186

Em = 1.57*10^{5} J

In conclusion, time of flight

t = (E_(M))/(P) \n\n t=(1.57 *10^(5) J)/(3.7 W) \n

t= 4.24*10 ^{4} s

Th distance hence is

d = v* t

d= 10.7 *4.25*10 ^{4}

d = 4.55* 10^{5} m

Read more about distance

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When an external magnetic field is applied, what happens to the protons in a sample?A) All protons align with the field.
B) All protons align opposite to the field.
C) Some protons align with the field and some align opposite to it.
D) All protons assume a random orientation.

Answers

On account of external magnetic field, the protons will align with the magnetic field. Hence, option (a) is correct.

The given problem is based on the concept of magnetic field. The region where the magnetic force is experienced is known as magnetic field. Generally, the protons are the charged entities carrying the positive polarity and are one of the major constituents of modern atomic structure.

  • The origin of magnetic field occurs due to charged particles present in a specific space. And the magnetic field is due to the flowing of liquid metal in the outer core of the planet generates electric currents.
  • In the condition when an external field is applied, the majority of protons align to the field because these protons tend to act like small magnets under the effect of this external field.

Thus, we can conclude that on account of external magnetic field, the protons will align with the field.

Learn more about the magnetic field here:

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Answer:

Some protons align with the field and some align opposite to it.

Explanation:

Majority align to the field because these protons tend to act like small magnets under the effect of this external field

The planet uranus is tilted nearly on its side so that its axis or rotation is only 8 degress abway from its orbit plane. if you lived at latitude 45 degrees on uranus for what fraction of the uranian year would answer?

Answers

The rotation of Uranus, like that of Venus, is retrograde and its axis of rotation is inclined almost ninety degrees above the plane of its orbit. During its orbital period of 84 years one of the poles is permanently illuminated by the Sun while the other remains in the shade. Exactly its rotation period is equivalent to 17 hours and 14 Earth minutes and its translation period is equivalent to 84 years, 7 days and 9 Earth hours.

Only a narrow band around the equator experiences a rapid cycle of day and night, but with the Sun very low on the horizon as in the polar regions of the Earth. On the other side of the orbit of Uranus, the orientation of the poles in the direction of the Sun is inverse. Each pole receives about 42 years of uninterrupted sunlight, followed by 42 years of darkness. Therefore an observer at latitude of 45 degrees in Uranus will probably experience a long winter night that is equivalent to one third of the year uranium.