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
Dafna11 [192]
2 years ago
14

Which trophic level has the least available energy in kilojoules in this food web?

Physics
1 answer:
insens350 [35]2 years ago
7 0

The highest trophic level has the least available energy in kilojoules.

Even though the food web is not shown in the question, but we know that energy decreases steadily as it is passed on from one trophic level to the next according to the second law of thermodynamics.

Energy enters into the system from the sun. The primary producers utilize this energy to produce food. As plants are eaten by animals, this energy is transferred along the food web an diminishes at each higher trophic level.

At the highest trophic level, the the least available energy in kilojoules in this food web is found.

Learn more: brainly.com/question/2233704

You might be interested in
Henry slides across an icy pond. The coefficient of kinetic friction betweenhis shoes and the ice is 0.09. If his mass is 115 kg
Kruka [31]

Answer:

101 N

Explanation:

4 0
2 years ago
A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls
Mars2501 [29]

Answer:

Therefore % increase in velocity is 18.23 %

Explanation:

we use the equality of mass flow rate and the areas

m_1 = m_2\\p_1v_1 = p_2v_2\\p_1A_1v_1 = p_2A_2v_2\\v_2 = \frac{p_1}{p_2} v_1

The percentage increase in velocity is

Δ v% = \frac{v_2 - v_1}{v_1} \\100%

= \frac{p_1}{p_2} v_1 - v_1.100%

= \frac{\frac{1.2}{1.015} - 1}{1} . 100%

= Therefore % increase in velocity is 18.23 %

5 0
2 years ago
~~~NEED HELP ASAP~~~<br>Please solve each section and show all work for each section.
anastassius [24]

Explanation:

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass m_A as

x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)

y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)

Substituting (2) into (1), we get

\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)

where f_N= \mu_kN, the frictional force on m_A. Set this aside for now and let's look at the forces on m_B

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on m_B as

x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)

y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)

From (5), we can solve for <em>N</em> as

N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)

Substituting (7) into (4) we get

m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba

Collecting similar terms together, we get

(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag

or

a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)

Putting in the numbers, we find that a = 1.4\:\text{m/s}. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get T = 21.3\:\text{N}. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get N = 50.9\:\text{N}

8 0
3 years ago
A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are
igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

T = I\alpha = 0.303*0.6454 = 0.196 Nm

So the force acting on the other end to generate this torque mush be:

F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N

4 0
3 years ago
A compact car has a maximum acceleration of 2.0 m/s2 when it carries only the driver and has a total mass of 1100 kg . you may w
nadya68 [22]

According to Newton`s  law. Force exerted by car,

F = m a = 1100 kg \times 2 m/s^2 = 2200 \ N

After adding an additional 400 kg of mass, the force will be same therefore the acceleration

F = 2200 \ N = (1100 \ kg + 400 \ kg)  a \\\\ a = \frac{2200 \ N}{1500 \ kg} = 1.47 \ m/s^2

Thus, the acceleration after adding the masses is 1.47 \ m/s^2.

4 0
3 years ago
Other questions:
  • Which statement best describes perigee?
    8·2 answers
  • You need to prepare a 0.199-mm-diameter tungsten wire with a resistance of 3.09 kÏ. how long must the wire be? the resistivity o
    9·1 answer
  • State one way of reducing the risks to a doctor who uses gamma radiation.
    11·2 answers
  • Please Help!!!
    11·2 answers
  • How is calculating amplitude​
    10·1 answer
  • Hai vận động viên chạy trên cùng 1 đoạn đường, vận động
    10·1 answer
  • This is the given equation of vibration of
    15·1 answer
  • How much force is required to accelerate a 0.10 gram mosquito at 20m/?
    7·2 answers
  • If the slowest instruction in the SCA executes in 12.5 ns, then what is maximum system clock frequency in MHz
    5·1 answer
  • Review. An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT . The angular momen
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!