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2 years ago

To ensure the greatest nutritional benefits, dietary guidelines suggest which of the following?

Physics

2 answers:

Flura [38]2 years ago

4 0

D. Eating nutrient-dense foods.

oee [108]2 years ago

3 0

The answer would be D. Eating nutrient-dense foods

You might be interested in

Draw the vector C⃗ =1.5A⃗ −3B⃗ . The length and orientation of the vector will be graded. The location of the vector is not impo

Nutka1998 [239]

I made the drawing in the attached file.

I included two figures.

The upper figure shows the effect of:

- multiplying vector A times 1.5.
It is drawn in red with dotted line.

- multiplying vector B times - 3 .
It is drawn in purple with dotted line.

In the lower figure you have the resultant vector: C = 1.5A - 3B.

The method is that you translate the tail of the vector -3B unitl the point of the vector 1,5A, preserving the angles.

Then you draw the arrow that joins the tail of 1,5A with the point of -3B after translation.

The resultant arrow is the vector C and it is drawn in black dotted line.

Download pdf

7 0

2 years ago

Read 2 more answers

The chemical formula for water, H2O, means that each water molecule contains

Ainat [17]

The chemical formula for water, H2o means that each water molecule contains one oxygen atom and two hydrogen atoms. This is the formula for water which has a liquid form, a solid form as ice, and also a gaseous form as water vapor.

5 0

2 years ago

Read 2 more answers

The force F required to compress a spring a distance x is given by F 2 F0 5 kx where k is the spring constant and F0 is the prel

IrinaVladis [17]

Answer:

a)W=8.333lbf.ft

b)W=0.0107 Btu.

Explanation:

<u>Complete question</u>

The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.

Solution

Preload = F₀=0 lbf

Spring constant k= 200 lbf/in

Initial length of spring x₁=0

Final length of spring x₂= 1 in

At any point, the force during deflection of a spring is given by;

F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

$W=\int\limits^2_1 {} \, Fds \\\\\\W=\int\limits^2_1( {F_0+kx} \,) dx \\\\\\W=\int\limits^a_b {kx} \, dx ; F_0=0\\\\\\W=k\int\limits^2_1 {x} \, dx \\\\\\W=k*\frac{1}{2} (x_2^{2}-x_1^{2} )\\\\\\W=200*\frac{1}{2} (1^2-0)\\\\\\W=100.lbf.in\\\\$

Change to lbf.ft by dividing the value by 12 because 1ft=12 in

100/12 = 8.333 lbf.ft

work required to compress the spring, W=8.333lbf.ft

The work required to compress the spring in Btu will be;

1 Btu= 778 lbf.ft

?= 8.333 lbf.ft----------------cross multiply

(8.333*1)/ 778 =0.0107 Btu.

6 0

2 years ago

When boiling water, a hot plate takes an average of 8 minutes and 55 seconds to boil 100 milliliters of water. Assume the temper

alexandr1967 [171]

Answer:

90.9 seconds

Explanation:

m = Mass of liquid = Volume×Density

c = Specific heat

$\Delta T$ = Change in temperature

t = Time taken

Room temperature = 75 °F

Converting to Celsius

$(75-32)\times \frac{5}{9}=23.889\ ^{\circ}C$

Heat required to raise the temperature of water

$Q=mc\Delta T\\\Rightarrow Q=100\times 10^{-6}\times 1000\times 4186\times (100-23.889)\\\Rightarrow Q=31860.0646\ J$

Power

$P=\frac{Q}{t}\\\Rightarrow P=\frac{31860.0646}{8\times 60+55}\\\Rightarrow P=59.55152\ W$

Efficiency of the plate

$\frac{59.5512}{283}\times 100=21.04282\%$

Heat required to raise the temperature of water

$Q=mc\Delta T\\\Rightarrow Q=100\times 10^{-6}\times 784\times 2150\times (56-23.889)\\\Rightarrow Q=5412.63016\ J$

$P=\frac{Q}{t}\\\Rightarrow t=\frac{Q}{P}\\\Rightarrow t=\frac{5412.63016}{0.2104282\times 283}\\\Rightarrow t=90.9\ s$

Time taken to heat the aceton is 90.9 seconds

4 0

2 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

2 years ago