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2 years ago

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]2 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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A car is traveling in a uniform circular motion on a section of road whose radius is r. The road is slippery, and the car is jus

Basile [38]

Answer:

a)r' = 4 r

b) $V=\sqrt{\mu rg}$

Explanation:

Given that car is moving with uniform speed.

At the verge of sliding

Friction force = Force due to circular motion

$\mu mg=\dfrac{mV^2}{r}$

So we can say that

$V=\sqrt{\mu rg}$ -----1

Where is the coefficient of friction

r is the radius of circular path

V is the velocity of car

if the car speed become double

It means that new speed of car =2V

lets tale new radius of circular path is r'

$2V=\sqrt{\mu r'g}$ ---------2

From equation 12 and 2 we cay that

r' = 4 r

It means that we have to increase radius 4 times to avoid sliding

$V=\sqrt{\mu rg}$

In the above expression there is no any terms of mass ,it means that sliding speed does not depends on the weight of car.

So the sliding speed will remain same.

$V=\sqrt{\mu rg}$

4 0

2 years ago

At distances greater than 2 femtometers, the nuclear strong force is stronger than (select all that apply):

grigory [225]

The correct answer is
<span>nuclear weak force
</span><span>
In fact, the nuclear strong force has a very limited range of action (approximately below 1-2 femtometers), and its magnitude decays exponentially for larger distances. Only the nuclear weak force is weaker than the nuclear strong force at large distance, while the gravity and the electromagnetic force become dominant for distances > 2 femtometers.</span>

3 0

2 years ago

Read 2 more answers

A projectile is fired vertically from earth's surface with an initial speed of 5.3 km/s. neglecting air drag, how far above the

MaRussiya [10]

It will rise, and then fall back to the Earth’s surface.

6 0

2 years ago

a 1.5 kg ball is thrown vertically upward with an initial speed of 15 m/s. if the initial potential energy is taken as zero, fin

trapecia [35]

Answer:

a) $E_{p} = 0$

$E_{k} = 168.7 J$

$E_{m} = 168.7 J$

b) $E_{p} = 73.6 J$

$E_{k} = 95.8 J$

$E_{m} = 169.4 J$

c) $E_{p} = 169.2 J$

$E_{k} = 0$

$E_{m} = 169.2 J$

Explanation:

We have:

m: is the ball's mass = 1.5 kg

v₀: is the initial speed = 15 m/s

g: is the gravity acceleration = 9.81 m/s²

a) In the initial position we have:

h: is the height = 0

The potential energy is given by:

$E_{p} = mgh = 0$

The kinetic energy is:

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(15)^{2} = 168.7 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 0 + 168.7 J = 168.7 J$

b) At 5 m above the initial position we have:

h = 5 m

The potential energy is:

$E_{p} = mgh = 1.5*9.81*5 = 73.6 J$

Now, to find the kinetic energy we need to calculate the speed at 5 m:

$v_{f}^{2} = v_{0}^{2} - 2gh = (15)^{2} - 2*9.81*5 = 126.9$

$v_{f} = \sqrt{126.9} = 11.3 m/s$

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(11.3)^{2} = 95.8 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 73.6 + 95.8 J = 169.4 J$

c) At its maximum height:

$v_{f}$ : is the final speed = 0

$h = \frac{v_{0}^{2}}{2g} = \frac{(15)^{2}}{2*9.81} = 11.5 m$

Now, the potential, kinetic and mechanical energies are:

$E_{p} = mgh = 1.5*9.81*11.5 = 169.2 J$

$E_{k} = \frac{1}{2}mv^{2} = 0$

$E_{m} = 169.2 J + 0 = 169.2 J$

I hope it helps you!

7 0

2 years ago

Marissaâs car accelerates uniformly at a rate of +2.60 m/sÂ². How long does it take for Marissaâs car to accelerate from a speed

DIA [1.3K]

Answer:

The time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds.

Explanation:

Given that,

Acceleration of the car, $a=+2.6\ m/s^2$

Initial speed of the car, u = 24.6 m/s

Final speed of the car, v = 26.8 m/s

We need to find the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s. The acceleration of an object is given by :

$t=\dfrac{v-u}{a}$

$t=\dfrac{(26.8-24.6)\ m/s}{2.6\ s}$

t = 0.84 seconds

So, the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds. Hence, this is the required solution.

4 0

2 years ago