All categories

3 years ago

Can i eat unhealthy while i’m drinking green tea for weight loss ?

2 answers:

6 0

Answer:I don't think it's a better idea to eat unhealthy food while having green tea because it will act as a strumbling rock in your attempt of getting weight loss.

Explanation:I don't say you have to mark my ans as brainliest but if ypu think it has really helped you plz don't forget to thank me...

storchak [24]3 years ago

6 0

Weight loss treatments vary among all people. Any health improvements could help, but I don’t think this plan is meant for fast action results.

You might be interested in

SMAW and GMAW use constant current arc welding machines.<br><br> True<br><br> False

Oksi-84 [34.3K]

It is true. Good luck

4 0

3 years ago

Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2

motikmotik

Answer:

The answer is " $112.97 \ \frac{ft}{s}$ "

Explanation:

Air flowing into the $p_1 = 20 \ \frac{lbf}{in^2}$

Flow rate of the mass $m = 230.556 \frac{lbm}{s}$

inlet temperature $T_1 = 700^{\circ} F$

Pipeline $A= 5 \times 4 \ ft$

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

$\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}$

$= \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}$

$V= \frac{mv}{A}$

$= \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}$

8 0

3 years ago

The diameter of a cylinder is called the a.) stroke b.)bore c.) principal length d.) radius

Amiraneli [1.4K]

Answer:

Explanation:

7 0

4 years ago

Two employees were installing aluminum siding on a farmhouse when it became necessary to remove a 36-foot high metal pole CB ant

Evgesh-ka [11]

Answer:using rubber gloves and choping the antena in 2

Explanation:

7 0

3 years ago

A forklift raises a 90.5 kg crate 1.80 m. (a) Showing all your work and using unity conversion ratios, calculate the work done b

klasskru [66]

Answer:

(a) Work done is 1.59642 kJ

(b) Useful power supplied = 0.1298 kW

Explanation:

(a) Work done = mass of crate × acceleration due to gravity × distance = 90.5 kg × 9.8 m/s^2 × 1.8 m = 1596.42 kgm^2/s^2 × 1 J/1 kgm^2/s^2 × 1 kJ/1000 J = 1.59642 kJ

(b) Useful power supplied = work done ÷ time = 1.59642 kJ ÷ 12.3 s = 0.1298 kJ/s × 1 kW/1 kJ/s = 0.1298 kW

5 0

3 years ago

Read 2 more answers