All categories

3 years ago

Which one of these is a chemical change?

Physics

2 answers:

Fofino [41]3 years ago

4 0

D burning a log I think

satela [25.4K]3 years ago

3 0

Burning a log because you are turning the log into ash from wood.

You might be interested in

Air pressure is 1.0 · 105 N/m2, air density is 1.3 kg/m3, and the density of soft drinks is 1.0 · 103 kg/m3. If one blows carefu

natita [175]

Answer:

v = 27.456 m/s

Explanation:

The support pressure needed of the water in the straw can be calculated by the formula

Given that,

P = r*g*h

= 1000*9.8*0.05 Pa.= 490 Pa

This pressure is compensated by 0.5*r*v^2 of the air,

Hence,

0.5*1.3*v^2 = 490

velocity of air blown into the straw =

v = 27.456 m/s

8 0

3 years ago

What are the benefits of the cool-down period following exercise?

kvv77 [185]

The benefits of the cool down period are quite important, it allows your body to slow your heart rate at a nice healthy safe pace, if you stop right away it can cause breathing, heart, and muscle problems.

5 0

3 years ago

Which statement best describes the direction of the buoyant force on any object?

poizon [28]

The buoyant force on any object acts in the direction opposite to the force of gravity. <em>(A)</em>

5 0

3 years ago

Read 2 more answers

What happens when you put a metallic object on a charger while it is chstging?

Ganezh [65]

It gets attracted due to electro magnetizing

3 0

3 years ago

A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow directi

Sidana [21]

Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, $V_{\infty}$ = 200 m/s

Given that the flow is laminar.

$Re_L=\frac{\rho V L}{\mu _{\infty}}$

$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$

$= 4.10 \times 10^7$

So boundary layer thickness,

$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$

$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

= 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

$=\frac{1}{2} \times 1.225 \times 200^2$

$=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

= 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

= 270 N

Therefore the net drag = 270 x 2

= 540 N

7 0

3 years ago