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3 years ago

In The United States how much more sugar is the average person consume each year that in 1970?

1 answer:

7 0

In the 1970, the average American ate only 2 pounds of sugar a year. In 1970, we ate 123 pounds of sugar per year. Today, the average American consumes almost 152 pounds of sugar in one year. This is equal to 3 pounds (or 6 cups) of sugar consumed in one week!

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Pls help this is due today.

bixtya [17]

Answer:

11 moments docx has the answer

Explanation:

6 0

2 years ago

Why does sodium form an ion +1 charge?

Rashid [163]

First of all we know about the ionic bond that the attraction b/w two atoms the atom which lose electrons is become cation and have positive sing which one get electrons it become ionic. so in such sitivation the sodium lose one electons "that's why sodium has 1+ charge in Nacl chemacal bond."
hope it will help you and please get my answer in brainlist
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4 0

3 years ago

A 4 kg block is moving at 12 m/s on a horizontal frictionless surface. a constant force is applied such that the block slows wit

weeeeeb [17]

Answer:

Is there a picture?

Explanation:

8 0

3 years ago

CP Bang! A student sits atop a platform a distance h above the ground. He throws a large firecracker horizontally with a speed.

Firlakuza [10]

Answer:

h = v₀ g / a

Explanation:

We can solve this problem using the kinematic equations. As they indicate that the air does not influence the vertical movement, we can find the time it takes for the body to reach the floor

y = $v_{oy}$ t - ½ g t²

The vertical start speed is zero

t² = 2t / g

The horizontal document has an acceleration, with direction opposite to the speed therefore it is negative, the expression is

x = v₀ₓ t - ½ a t²

Indicates that it reaches the same exit point x = 0

v₀ₓ t = ½ a t2

v₀ₓ = ½ a (2h / g)

v₀ₓ = v₀

h = v₀ g / a

7 0

3 years ago

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top

Anton [14]

Answer:

$y_{max} = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}$

$v = 2.913\,\frac{m}{s}$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s$

$1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_{1} \approx 0.128\,m$

$\Delta s_{2} \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}$

$y_{max} = 0.829\,m$

4 0

3 years ago

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