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zalisa [80]
3 years ago
10

If oxygen and steel wool are combined is it a physical change or chemical change?

Physics
2 answers:
JulijaS [17]3 years ago
8 0

Answer:

Chemical change

Explanation:

If both combined it will be a <em>chemical</em><em> </em><em>change</em>

<em>Physical</em><em> </em><em>change</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>reversed</em><em> </em><em>easily</em><em>;</em><em> </em><em>chemical</em><em> </em><em>change</em><em> </em><em>cannot</em><em> </em><em>be</em><em> </em><em>changed</em><em> </em><em>back</em><em> </em><em>to</em><em> </em><em>previous</em><em> </em><em>state</em>

Anon25 [30]3 years ago
4 0

Answer:

I think it would be a chemical change.

Explanation:

Because a physical would be like crushing it or something along that line.

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The binding energies of K-shell and L-shell electrons in a certain metal are EK and EL, respectively, If a Kαx ray from this met
Svetach [21]

Answer:

The separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)

Explanation:

The relationship between energy and wavelength is expressed below:

E = hc/λ

λ = hc/EK - EL

Considering the condition of Bragg's law:

2dsinθ = mλ

For the first order Bragg's law of reflection:

2dsinθ = (1)λ

2dsinθ = hc/EK - EL

d = hc/2sinθ(EK - EL)

Where 'd' is the separation distance between the parallel planes of an atom, 'h' is the Planck's constant, 'c' is the velocity of light, θ is the angle of reflection, 'EK' is the energy of the K shell and 'EL' is the energy of the K shell.

Therefore, the separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)

5 0
3 years ago
PLEASEEE HELP ME WITH THIS ALSO. I DONT WANT TO FAIL. You push a merry-go-round on which Kim and Katie are riding. Kim weighs 45
Serjik [45]

Answer:

The body weight

Explanation:

5 0
3 years ago
Consider a roller coaster begins 15m above the ground. If the cart has a mass of 75kg, what is the velocity of the cart halfway
SashulF [63]

Answer:

v = 12.12 m/s

Explanation:

Given that,

The mass of the cart, m = 75 kg

The roller coaster begins 15 m above the ground.

We need to find the velocity of the cart halfway to the ground. Let the velocity be v. Using the conservation of energy at this position, h = 15/2 = 7.5 m

mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 7.5} \\\\v=12.12\ m/s

So, the velocity of the cart is 12.12 m/s.

7 0
3 years ago
The space shuttle fleet was designed with two booster stages. If the second stage provides a thrust of 73 ​kilo-newtons and the
Tresset [83]

Answer:

m = 81281.5 pounds.

Explanation:

Given that,

Force, F = 73 kN

Acceleration of the space shuttle, a = 16000 mi/h²

1 miles/h² = 0.0001241 m/s2

16000 mi/h² = 1.98 m/s²

We need to find the mass of the spacecraft.

According to Newton's second law,

F = ma

m is mass of the spacecraft

m=\dfrac{F}{a}\\\\m=\dfrac{73\times 10^3\ N}{1.98\ m/s^2}\\\\m=36868.68\ kg

Since, 1 kg = 2.20462 pounds

m = 81281.5 pounds

Hence, the mass of the spacecraft is 81281.5 pounds.

8 0
3 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
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