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kvasek [131]
4 years ago
5

Melting, boiling, anf bending are physical changes but why is burning a chemical change

Physics
2 answers:
bija089 [108]4 years ago
7 0
Because burning is a non-reversible chemical change.
Nady [450]4 years ago
6 0
Burning is a chemical change because if it flames it could not back to the way it was it will be ash. Plus to start a fire with a lighter that has a chemical liquid that makes it flammable.

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In what direction must a force be applied so that the forces on the 1 kg object are balanced
choli [55]

Answer:

towards the object

Explanation:

7 0
3 years ago
Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = \sqrt{2I/ce_{0} }

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9}  = <em>1.55 x 10^-8 v/m</em>

6 0
4 years ago
Question 30
stellarik [79]

Answer: 0.69\°

Explanation:

The angular diameter \delta of a spherical object is given by the following formula:

\delta=2 sin^{-1}(\frac{d}{2D})

Where:

d=16 m is the actual diameter

D=1338 m is the distance to the spherical object

Hence:

\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})

\delta=0.685\° \approx 0.69\° This is the angular diameter

3 0
3 years ago
Jvhftfcticgvigybvhbjnjndjanvjisn
nirvana33 [79]

Answer:

dnndmssnxnxdnskkaamzmzma nzz.

4 0
3 years ago
Read 2 more answers
A 50-kg block is at rest on a 15o slope. A force of 250 N is acting on the block up the slope parallel to it. If the block does
Iteru [2.4K]

The minimum value of the coefficient of static friction between the block and the slope is 0.53.

<h3>Minimum coefficient of static friction</h3>

Apply Newton's second law of motion;

F - μFs = 0

μFs = F

where;

  • μ is coefficient of static friction
  • Fs is frictional force
  • F is applied force

μ = F/Fs

μ = F/(mgcosθ)

μ = (250)/(50 x 9.8 x cos15)

μ = 0.53

Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.

Learn more about coefficient of friction here: brainly.com/question/20241845

#SPJ1

6 0
2 years ago
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