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2 years ago

Why does eating less meat conserve more water than just eating plants?

2 answers:

8 0

You turn off the faucet when you’re brushing your teeth and you definitely don’t take twenty-minute long showers. While these acts, along with the use of appliances that save water like toilets and showerheads, are fantastic ways to conserve water … if you really want to make a difference with water conservation, then there is a bigger piece of the puzzle to consider.
Every food and beverage product we consume: from soda to pineapples to cheese crackers and pork chops, etc. requires water for its production. Obviously, fruits and veggies need water. That’s a given, but surprisingly, there is another group of foods that is the biggest water guzzler of them all and that is animal products. Meat, dairy, and eggs require more water to “produce” than any other food out there.

Afina-wow [57]2 years ago

3 0

Answer:

Studies show that eating fewer animal-based products could reduce water use since animal production uses more water than crops do. In addition, reducing the amount of food that's lost or wasted at various points in the food supply chain could feed about 1 billion extra people while simultaneously reducing water use.

You might be interested in

If the spring constant is doubled , what value does the period have for a mass on a spring?

zhuklara [117]

Answer:

D. The period would decrease by sqrt (2)

Explanation:

The period of a mass-spring system is given by:

$T=2\pi \sqrt{\frac{m}{k}}$

where

m is the mass

k is the spring constant of the spring

If the spring constant is doubled,

k' = 2k

So the new period will be

$T'=2\pi \sqrt{\frac{m}{(2k)}}=\frac{1}{\sqrt{2}}(2\pi \sqrt{\frac{m}{k}})=\frac{T}{\sqrt{2}}$

So the correct answer is

D. The period would decrease by sqrt (2)

8 0

3 years ago

Read 2 more answers

5th Grade. Please help! SC standards

lesantik [10]

The answer would be D) The tropical convection region because the states below SC are all tropical and the current would go up to SC. I hope this helps.

6 0

3 years ago

The bell rings and a physics student heads to class. They stop to talk to a few friends. They slow

grin007 [14]

Answer:

420m

Explanation:

Given parameters:

Time = 5minutes

Average speed = 1.4m/s

Unknown:

Distance covered = ?

Solution:

Speed is the rate of change of distance with time.

Mathematically;

Speed = $\frac{distance}{time}$

Distance = speed x time

Insert the parameters and solve;

Convert the time to seconds;

1 minute = 60s

5 minute = 5 x 60 = 300s

So,

Insert the parameters and find the distance;

Distance = 300 x 1.4 = 420m

3 0

2 years ago

What are the rules for setting up an integral of rotation?

Angelina_Jolie [31]

Setting up an integral of rotation is used as a method of of calculating the volume of a 3D object formed by a rotated area of a 2D space. Finding the volume is similar to finding the area, but there is one additional component of rotating the area around a line of symmetry.

First the solid of revolution should be defined. The general function is y=f(x), on an interval [a,b].

Then the curve is rotated about a given axis to get the surface of the solid of revolution. That is the integral of the function.

It all depends of the function f(x), which must be known in order to calculate the integral.

3 0

2 years ago

A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is

BlackZzzverrR [31]

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by

∑F = ma (1)

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:

a_max = -w^2A (2)

3- The angular frequency w of a wave is related to the frequency f by:

w = 2π f (3)

Given Data

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)= 0.04 m.

- The frequency of oscillation of the piston is steadily increased.

Required Data

In part (a), we are asked to determine the point at which the penny first loses contact with the piston.

In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.

Solution

(a)

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:

∑F_y -mg= ma

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So

0 = m(g + a)

a = —g

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):

a = —w^2A

where a = —g at the highest point. So

g = w^2A

Solve for w:

w =√g/A

Substitute for w from Equation (3):

2πf = √g/A

Solve for f :

f = 1/2π√g/A

Substitute numerical values:

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

6 0

3 years ago