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2 years ago

How does food affects carbon footprint?

2 answers:

7 0

Answer: Food accounts for 10-30% of a household's carbon footprint, typically a higher portion in lower-income households.

Explanation:

eduard2 years ago

6 0

Answer:

Food accounts for 10-30% of a household's carbon footprint, typically a higher portion in lower-income households. Production accounts for 68% of food emissions, while transportation accounts for 5%.

Explanation:

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Which of these is a mixture?

kicyunya [14]

It is salt water because they are two things combined together.

4 0

3 years ago

Read 2 more answers

The density of an object is dependent upon the object’s mass and ---

kotegsom [21]

Answer:Volume

Explanation:

Density = mass/ Volume

7 0

3 years ago

Two blocks, stacked one on top of the other, slide on a frictionless horizontal surface. The surface between the two blocks is r

devlian [24]

Answer:

$F_{net} = 31.88 N$

Explanation:

When top block is just or about to slide on the lower block then we can say that the frictional force on it will be maximum static friction

So we will have

$F_{net} = ma$

$F_s = ma$

$\mu_s mg = ma$

$a = \mu_s g$

$a = (0.50)(9.81)$

$a = 4.905 m/s^2$

now for the Net force on two blocks to move together

$F_{net} = (m_1 + m_2) a$

$F_{net} = (2.3 + 4.2)(4.905)$

$F_{net} = 31.88 N$

4 0

3 years ago

A 75 N box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each

shusha [124]

Answer:

μk = 0.26885

Explanation:

Conceptual analysis

We apply Newton's second law:

∑Fx = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

a= -0.9 m/s²,

g = 9.81 m/s² : acceleration due to gravity

W= 75 N : Block weight

W= m*g

m = W/g = 75/9.8= 7.65 kg : Block mass

Friction force : Ff

Ff= μk*N

μk: coefficient of kinetic friction

N : Normal force (N)

Problem development

We apply the formula (1)

∑Fy = m*ay , ay=0

N-W-25 = 0

N = 75 +25

N= 100N

∑Fx = m*ax

20-Ff= m*ax

20-μk*100 = 7.65*(-0.90 )

20+7.65*(0.90) = μk*100

μk = ( 20+7.65*(0.90)) / (100)

μk = 0.26885

4 0

3 years ago

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00

xenn [34]

Answer:

$0.496 kg m^2$

Explanation:

The torque exerted is given by

$\tau = Fd$

where

$F=2.00 \cdot 10^3 N$ is the force applied

d = 3.10 cm = 0.031 m is the length of the lever arm

Substituting,

$\tau=(2.00\cdot 10^3 N)(0.031 m)=62 Nm$

The equivalent of Newton's second law for rotational motion is:

$\tau = I \alpha$

where

$\tau = 62 Nm$ is the net torque

I is the moment of inertia

$\alpha = 125 rad/s^2$ is the angular acceleration

Solving the equation for I, we find

$I=\frac{\tau}{\alpha}=\frac{62 Nm}{125 rad/s^2}=0.496 kg m^2$

6 0

3 years ago