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3 years ago

How does knowledge of the energy content of food help diabetics make smart choices

1 answer:

monitta3 years ago

7 0

The knowledge of the energy content of food can help diabetics make smart diet choices because diabetic patients need the foods that have high carbohydrate and foods with rich fiber content in them. If the diabetic patient is not aware about the energy contents of a food, they can make a wrong choice and ultimately fall ill. Most diabetic patients require foods that have high energy content and low sugar content. Diabetic patients need full care about the food they intake. Taking the right kind of foods rich in protein and carbohydrates can immensely help a diabetic patient control the amount of sugar in the blood.

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For the following statements, choose the word or words inside the parentheses that serve to make a correct statement. Each state

AnnyKZ [126]

Answer:

a) Temperatura, b) Temperature, c) Constant , d) None of these , e) Gibbs enthalpy and free energy (G)

Explanation:

a) the expression for ideal gases is PV = nRT

Temperature

b) The internal energy is E = K T

Temperature

c) S = ΔQ/T

In an isolated system ΔQ is zero, entropy is constant

Constant

d) all parameters change when changing status

None of these

e) Gibbs enthalpy and free energy

7 0

3 years ago

Which would you use with a circuit to determine if a magnet was moving in close proximity to the circuit?

stealth61 [152]

Answer:

i think it would be Galvanometer

Explanation:

because it would have to be a type of coil or wire

7 0

3 years ago

Read 2 more answers

If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m

AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

$mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}$

We can cancel the common factor of $m$ which leaves us with

$gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}$

Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

We are given $g,v_{i},h_{i},h_{f}$ .

We can now solve for the final velocity.

${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

Anything multiplied by 0 is 0.

${v_{f}=\sqrt{2*9.81*2.41$

${v_{f}=\sqrt{47.2842$

$v_f=6.88$

7 0

1 year ago

Two examples of hydraulic brake

rjkz [21]

Answer:

1). Brake pedal or lever

2). A pushrod also called an actuating rod

3 0

2 years ago

A bowling ball weighing 71.2 N (16.0 lb) is attached to the ceiling by a 3.80-m rope. The ball is pulled to one side and release

Elina [12.6K]

Answer:

a) For this case we want to find the acceleration of the bowling ball, and we now that the only acceleration for this case would be the centrifugal acceleration becuase since the pendulum is in phase with the equilibrium point the tangential acceleration is 0, and if we find the centrifugal acceleration we got:

$a= \frac{v^2}{R}= \frac{(4.2 m/s)^2}{3.8 m}=4.64 m/s^2$

b) For this case the tendsion is an opposite force against the weight and the centrifugal force acting in the ball so then we can find the tension like this:

$T= mg +ma$

In order to find the mass we know that $W=mg$ and solving for m we got:

$m =\frac{W}{g}=\frac{71.2 N}{9.8 m/s^2}=7.265 kg$

And replacing into the tension we got:

$T= m(g+a) =7.265 kg *(9.8 m/s^2 +4.64 m/s^2)=104.907 N$

Explanation:

For this case we have the following data given:

$W= 71.2 N$ represent the weigth for the object

$R=3.8 m$ the length of the rope or the radius

$v= 4.2 m/s$ represent the velocity of the bowling ball

Part a

For this case we want to find the acceleration of the bowling ball, and we now that the only acceleration for this case would be the centrifugal acceleration becuase since the pendulum is in phase with the equilibrium point the tangential acceleration is 0, and if we find the centrifugal acceleration we got:

$a= \frac{v^2}{R}= \frac{(4.2 m/s)^2}{3.8 m}=4.64 m/s^2$

Part b

For this case the tendsion is an opposite force against the weight and the centrifugal force acting in the ball so then we can find the tension like this:

$T= mg +ma$

In order to find the mass we know that $W=mg$ and solving for m we got:

$m =\frac{W}{g}=\frac{71.2 N}{9.8 m/s^2}=7.265 kg$

And replacing into the tension we got:

$T= m(g+a) =7.265 kg *(9.8 m/s^2 +4.64 m/s^2)=104.907 N$

7 0

3 years ago