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2 years ago

Does increase in metabolism accelerates weight loss?.

Physics

1 answer:

erma4kov [3.2K]2 years ago

8 0

Answer: yes

Explanation: Having a higher metabolism can help you lose weight and keep it off, while also giving you more energy.

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A spherical shell of radius 9.0 cm carries a uniform surface charge density σ= 9.0 nC/m2. The electric field at r= 9.1 cm is app

maria [59]

Answer:

995.12 N/C

Explanation:

R = 9 cm = 0.09 m

σ = 9 nC/m^2 = 9 x 10^-9 C/m^2

r = 9.1 cm = 0.091 m

q = σ x 4π R² = 9 x 10^-9 x 4 x 3.14 x 0.09 x 0.09 = 9.156 x 10^-10 C

E = kq / r^2

E = ( 9 x 10^9 x 9.156 x 10^-10) / (0.091 x 0.091)

E = 995.12 N/C

8 0

3 years ago

What force could be applied to a box to make the net force in the horizontal direction of zero

Kazeer [188]

If there is no friction, the force that moves the box forward horizontally must be matched by the same force.

If there is friction, then the force moving it forward = frictional force + the additional force you need to add.

6 0

3 years ago

If the truck has a mass of 2,000 kilograms , what's its momentum?(v=35 m/s)

katen-ka-za [31]

Answer:

$\boxed {\boxed {\sf 70,000 \ kg*m/s}}$

Explanation:

Momentum is the product of mass and velocity.

$p=m*v$

The mass of the truck is 2,000 kilograms and the velocity is 35 meters per second.

$m= 2000 \ kg \\v= 35 \ m/s$

Substitute the values into the formula and multiply.

$p= 2000 \ kg * 35 \ m/s \\p= 70,000 \ kg*m/s$

The truck's momentum is <u>70,000 kilograms meters per second.</u>

8 0

3 years ago

The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of t

EastWind [94]

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

$Ay = By = \frac{w * 6}{2} = 3w$

$P_c_r = 3w * F.S = 3w * 2.0 = 6w$

$I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7$

To find the maximum intensity, w, let's take the Pcr formula, we have:

$P_c_r = \frac{\pi^2 E I}{(KL)^2}$

Let's take k = 1

$E = 200*10^9$

Substituting figures, we have:

$6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}$

Solving for w, we have:

$w = \frac{67266.84}{6}$

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

$\sigma _c_r = \frac{w}{A}$

$\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y$ . This means it is safe

The maximum intensity w = 11.211KN/m

3 0

3 years ago

A syringe of volume 16 cm3 is filled with air to a pressure of 1.03 atm. If the piston of the syringe is pushed to change the vo

denpristay [2]

<h3>Answer:</h3>

189.07 kPa

<h3>Explanation:</h3>

Concept tested: Boyle's law

<u>We are given;</u>

Initial volume of the syringe, V1 is 16 cm³
Initial pressure of the syringe, P1 is 1.03 atm
New volume of the syringe, V2 is 8.83 cm³

We are required to calculate the new pressure of the syringe;

We are going to use the concept on Boyle's law of gases.
According to the Boyle's law, for a fixed mass of a gas, the pressure is inversely proportional to its volume at constant temperature.

That is; P α 1/V

At varying pressure and volume, k(constant) = PV and P1V1=P2V2

Therefore, to get the new pressure, P2, we rearrange the formula;

P2 = P1V1 ÷ V2

= ( 16 cm³ × 1.03 atm) ÷ 8.83 cm³

= 1.866 atm.

Thus, the new pressure is 1.866 atm
But, we need to convert pressure to Kpa
Conversion factor is 101.325 kPa/atm

Thus;

Pressure = 1.866 atm × 101.325 kPa/atm

= 189.07 kPa

Hence, the new pressure of the air in the syringe is 189.07 kPa

3 0

3 years ago