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2 years ago

In a tropic level, energy flows from the producer to the primary consumer and to a secondary consumer. True False

Physics

1 answer:

Ganezh [65]2 years ago

4 0

Answer:

True

Explanation:

This is true because in an energy flow, the primary producers such as plants produce their own foods by using energy from the sun. These primary producers are consumed by primary consumers such as rabbits for food, while secondary consumers such as snakes would consume the primary consumers.

By so doing, energy flows from one trophic level to the other.

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Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowing this,

insens350 [35]

Answer:

The distance travel before stopping is 1.84 m

Explanation:

Given :

coefficient of kinetic friction $\mu_{k} = 0.250$

Zak's speed $v = 3 \frac{m}{s}$

Gravitational acceleration $g = 9.8$ $\frac{m}{s^{2} }$

Work done by frictional force is given by,

$W = \Delta K$

$\mu _{k} mg d = \frac{1}{2} m v^{2}$

$d = \frac{v^{2} }{2 g \mu _{k} }$

$d = \frac{9}{2 \times 9.8 \times 0.250}$

$d = 1.84$ m

Therefore, the distance travel before stopping is 1.84 m

3 0

3 years ago

During combustion reactions, explain why the energy of the reactants must exceed the total energy of the products

maxonik [38]

Answer:

In these reactions the products are higher in energy than the reactants. ... This barrier is due to the fact that to make CO2 and H2O we have to break 4 carbon-hydrogen bonds and some ...

Explanation:

8 0

2 years ago

If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is

viktelen [127]

To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

5 0

2 years ago

Which of the following is a zone found in the open ocean

Svetlanka [38]

What are your answer choices?

5 0

2 years ago

Read 2 more answers

9. A 50 kg physics i student trudges from the 1st floor to the 3rd floor, going up a total height of 13 m.

SVETLANKA909090 [29]

Answer:

6370 J

Explanation:

By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m

$W = E_p = mgh$

where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference

$W = 50*9.8*13 = 6370 J$

3 0

2 years ago