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3 years ago

Malia has developed Type II Diabetes. When she eats, her body

Physics

1 answer:

AlladinOne [14]3 years ago

3 0

Answer:

The correct option is c.

Explanation:

Metabolism is a sum of anabolic and catabolic reactions. The body's inability to produce/synthesize enough insulin is the cause of type II diabetes. Generally, metabolism is the process in which most compounds (proteins, carbohydrates and lipids) are produced (anabolism) or broken down (catabolism) in the body. Insulin is a protein that can be produced in less amount due to metabolic disorder in the body.

Maria's disease means she already has an exponentially high amount of blood sugar against the required insulin to balance it out, hence the disease already slowed down her rate of metabolism (catabolism) of blood sugar EXCEPT she decides to increase of metabolism by medication and exercise.

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3 years ago

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3 years ago

For installation with a 25-kVA, 3-phase transformer, a 440-volt primary, and a 120-volt secondary. Calculate the maximum overcur

andre [41]

Answer:

41.053 A

Explanation:

given,

three phase kVA = 25-kVA

voltage = 440 Volt

current = ?

To determine three phase kVA when volts and amperes are know

three phase kVA = 1.73 x V x I

$I = \dfrac{three\ phase\ kVA}{1.73 \times V}$

$I = \dfrac{25000}{1.73 \times 440}$

I = 32.84 A

the maximum overcurrent protection value is equal to

= 125 % of I

= 1.25 x 32.84

= 41.053 A

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3 years ago

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Explanation:

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3 years ago

Read 2 more answers

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$

The box is initially at rest, so $v_i=0$ . Solving for $v_f$ :

$v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}$

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3 years ago