Pls help it’s easy ik but I want to make sure of them both I don’t have much time

Which of the following sequences describes the path by which electrons travel downhill energetically in aerobic respiration

Vlad1618 [11]

Answer:

glucose → NADH → electron transport chain → oxygen.

Explanation:

Aerobic respiration is also called as cellular respiration, in which electrons are picked up by the FADH, and NADH from the food. Then through a proton pump electrons are transferred to the electron transport chain, in this, the activity of the proton pump generates an electrochemical gradient, then this gradient used by the ATP synthase enzyme to produce ATP.

In aerobic respiration, the last acceptor of electron is oxygen. An electron is donated to oxygen for the formation of the water.

5 0

3 years ago

Match the organisms to the descriptions.

makvit [3.9K]

Answer:

Lacks colored blood - Scorpion

Soft, unsegmented body - Octopus

A vertebrate - Bird

Lacks antennae - Starfish

Explanation:

Scorpion belongs to the Phylum Arthropoda of the Kingdom Animalia. All the organisms of this phylum lack coloured blood.

Octopus comes under the Phylum Molusca of the Kingdom Animalia. Soft, unsegmented body is the property of the organisms of this phylum.

Bird belongs to the Vertebrata class of the Phylum Chordata of Kingdom Animalia.

Starfish belongs to Echinodermata Phylum of the Kingdom Animalia. The organisms of this category do not posses antennae.

6 0

3 years ago

Read 2 more answers

At what time of day would you be most likely to find that the air over water is significantly warmer than the air over land near

Brut [27]

This would happen later at night or early in the morning.

The reason being land becomes warm and cold quicker than the water because of the heat capacity. So during the day water warms up because of sunlight but at night the land becomes a lot cooler as compared to the water which is still war. So the air over water is significantly warmer than the air over land.

4 0

3 years ago

Read 2 more answers

Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc

lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

$E = \frac{V}{L}$

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

$J = \frac{E}{p}$

Substitute (V/L) for E in the above equation of current density.

$J = \frac{V}{pL} ------(1)$

Substitute iR for V in equation (1)

$J = \frac{iR}{pL} ------(2)$

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

$J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2$

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

$p = \frac{RA}{L}$

$A = \frac{pL}{R}$

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A and λ is use for (m/L) in the eqn above

$\lambda = d\frac{p}{\frac{R}{L} } ------(3)$

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m

$\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1$

c) Using the equation (2) current density for aluminum cable is,

$J = \frac{iR}{pL}$

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

$J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2$

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

$\lambda = d\frac{p}{\frac{R}{L} }$

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

$\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m$

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0

3 years ago

Read 2 more answers

Dimension of density <br>

Rom4ik [11]

Answer:

See the explanation below.

Explanation:

We know that density is defined as the relationship between mass and volume.

$Ro = m/V$

where:

m = mass [kg]

V = volume [m³]

Therefore Ro is given in:

$[kg/m^{3} ]$

8 0

2 years ago