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

1 answer:

monitta2 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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A mass of 0.34 kg is fixed to the end of a 1.4 m long string that is fixed at the other end. Initially at rest, he mass is made

frozen [14]

At time $t$ seconds, the mass has angular speed

$\omega = \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) t$

and hence linear speed

$v = (1.4\,\mathrm m) \omega = (1.4\,\mathrm m) \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) t$

After 8 s, its linear speed is

$v = (1.4\,\mathrm m) \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) (8\,\mathrm s) = 37.072 \dfrac{\rm m}{\rm s} \approx 37 \dfrac{\rm m}{\rm s}$

and has centripetal acceleration with magnitude

$a = \dfrac{v^2}{1.4\,\rm m} \approx 981.667\dfrac{\rm m}{\mathrm s^2} \approx 980 \dfrac{\rm m}{\mathrm s^2}$

To maintain this linear speed, by Newton's second law the required centripetal force should have magnitude

$F = (0.34\,\mathrm{kg}) a \approx 333.767\,\mathrm N \approx \boxed{330 \,\mathrm N}$

5 0

1 year ago

A ball is thrown vertically upward with a speed of 27.9 m/s from a height of 2.0 m. How long does it take to reach its highest p

Bas_tet [7]

Answer:

2.84403 seconds

2.91483 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-27.9}{-9.81}\\\Rightarrow t=2.84403\ s$

It takes 2.84403 seconds to reach the highest point

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-27.9^2}{2\times -9.81}\\\Rightarrow s=39.67431 m$

The ball will travel 39.67431+2 = 41.67431 m while going down to the ground

$s=ut+\frac{1}{2}at^2\\\Rightarrow 41.71479=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{41.67431\times 2}{9.81}}\\\Rightarrow t=2.91483\ s$