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Travka [436]
3 years ago
5

What is a warming sign that suggest a weight loss plan is a fad diet rather than a healthy weight loss plan?

Physics
1 answer:
Serga [27]3 years ago
5 0

Answer:

Warming signs before fad diet includes:

1. When a special type of diet is recommended for weight loss.

2. When quick weight loss is promised instead of advice on the need to exercise, eat healthy foods including fruits and vegetables, and adopt a good feeding behavior.

3. Promises rapid weight loss in a short period of time.

After taking the fad diet the following might be experienced:

1. Constipation.

2. Weakness and fatigue.

3. More accumulation of fats in the body in the long run after a short term normal  BMI (Body mass index).

4. Lower metabolic rates over time.

Explanation:

weight loss is a practice adopted by people for various reasons ranging from the desire to stay healthy to, ability to pursue a career that requires a good Body Mass Index. For instance, A BMI (Body mass index) which is equal to a person's weight (kg) / (height(m))^2 of 30 and higher is considered obese.

Fad diet which may give a temporary result of weight loss is not considered totally healthy as it doesn't capture the need to eat a healthy balanced diet instead focuses on more routine consumption of a particular kind of diet.

A healthy weight loss plan captures the need to exercise, reduce body fats while eating healthy. Eating fruits, vegetables are also encouraged.

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The volume of 3200 kg of petrol is 4 m^3.

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3 years ago
A power plant uses Uranium<br> to produce energy
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Answer:

how is that a question?

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3 years ago
a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet
ANTONII [103]

Answer:

Explanation:

Given

radius of circular path r=4\ in.

Position is given by

\theta =\cos 2t---1

Differentiate 1  to angular velocity we get

\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2

Differentiate 2 to get angular acceleration

\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3

Net acceleration is the vector summation of tangential and centripetal force

a_t=\alpha \times r

a_t=-4\cos 2t\times 4=-16\cos 2t

a_r=\omega ^2\cdot r

a_r=(-2\sin 2t)^2\cdot 4

a_r=16\sin^2(2t)

a_{net}=\sqrt{a_r^2+a_t^2}

a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}

a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}                                                    

6 0
3 years ago
(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us
konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m

(b) The maximum height of the projectile in meters;

H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m

(c) The speed with which the projectile hits the ground is;

v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s

5 0
3 years ago
A nuclear reactor is:
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A device used to initiate and control a sustained nuclear chain reaction. 
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