Accuracy in scientific investigation is important because

When you go "down" the food chain by continuing to ask "what does it eat?" at what category of living things do you always end u

Klio2033 [76]

Answer:

Autotrophs

Explanation:

When you go down a food chain continuing to ask "what does it eat?" the last living thing that you will land upon is an autotroph.

Autotrophs are the primary producers as they (photoautotrophs) use the energy either from the sun to prepare there food by the process of photosynthesis or, more rarely, obtain chemical energy through oxidation (chemoautotrophs) to make organic substances from inorganic ones.

Autotrophs get consumed by the primary consumers in the food chain.

7 0

3 years ago

An owl weighing 40N is sitting in a tree waiting to dive down to catch a mouse. If the owl's potential energy is 800 J with resp

Natali5045456 [20]

Answer:

Explanation:

<u>Gravitational Potential Energy</u>

Gravitational potential energy (GPE) is the energy stored in an object due to its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or $9.8 m/s^2$ .

The weight of an object of mass m is:

W = m.g

Thus, the GPE is:

U=W.h

Solving for h:

$\displaystyle h=\frac{U}{W}$

The weight of the owl is W=40 N and its GPE is U=800 J.

$\displaystyle h=\frac{800}{40}=20$

h = 20 m

3 0

3 years ago

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu

Alekssandra [29.7K]

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰, force = 4.07 N

(b) When the angle, θ = 90 ⁰, force = 4.7 N

(c) When the angle, θ = 120 ⁰, force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

$F = BIl \ sin(\theta)$

(a) When the angle, θ = 60 ⁰

$F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N$

(b) When the angle, θ = 90 ⁰

$F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N$

(c) When the angle, θ = 120 ⁰

$F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N$

6 0

3 years ago

Atom in the ground state is said to be

koban [17]

Answer:

Ground-state atom

Explanation:

When an atom is not excited, it is in its ground-state, which we refer as "standard" or "normal" state.

(Hopefully that helped you!)

GOOD LUCK

Astrophysicist Dr. D

3 0

3 years ago

A produce distributor uses 800 packing crates a month, which it purchases at a cost of $10 each. The manager has assigned an ann

strojnjashka [21]

Answer:

$364.29

Explanation:

given,

Packing of crates per month (u)= 800

annual carrying cost of 35 percent of the purchase price per crate.

Ordering cost(S) = $ 28

D = 800 x 12 = 9600 crates/year

H = 0.35 P

H = 0.35 x $10

H = $3.50/crate per yr.

Present Total cost

= $\dfrac{800}{2}\times 3.50 + \dfrac{9600}{800}\times 28$

= 1400 + 336

= $ 1,736

$Q_0 = \sqrt{\dfrac{2DS}{H}}$

$Q_0 = \sqrt{\dfrac{2\times 9600 \times 28}{3.50}}$

$Q_0 =\$ 391.92$

Total cost at EOQ

= $\dfrac{391.92}{2}\times 3.50 + \dfrac{9600}{391.92}\times 28$

= 685.86 + 685.85

= $ 1,371.71

the firm save annually in ordering and carrying costs by using the EOQ

= $ 1,736 - $ 1,371.71

= $364.29

4 0

3 years ago