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andreev551 [17]
4 years ago
11

Which example illustrates a chemical change?

Physics
2 answers:
ad-work [718]4 years ago
8 0
<span>Chemical changes are those that yield to a different substances or compounds. This happens when compounds change the bonding ot the atoms which meanns that the composition of the substance has changed. Rust, fire, combustion, food cooking are examples of chemical changes. Freezing and vaporizations are not chemical changes, they are examples of physical changes.</span>
IRINA_888 [86]4 years ago
5 0

Answer:

Burning wood

Explanation: The wood is still wood, even if it's burnt. But it has a different identity.

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A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
ki77a [65]

Answer:

a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.  Pin = 216 w

d. PL = Pin = 216 w

e. Pg = 478.4 w , Pzg = 262.4 w

Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

βl = 2π / λ * 0.15 λ = 54 °

Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

4 0
4 years ago
Narcotics are highly addictive and affect the driver's
adelina 88 [10]

Answer:

D

that's should be the answer

6 0
3 years ago
Read 2 more answers
g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr
Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

a = \frac{GM}{r^2}

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

\frac{da}{dr} = -2\frac{GM}{r^3}

da = -2\frac{GM}{r^3}dr

At the same time since Newton's second law we know that:

F_w = ma

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

F_W = m (\frac{GM}{r^2} ) = 600N

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

dF_W = mda

dF_W = m(-2\frac{GM}{r^3}dr)

dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})

dF_W = -2F_W(\frac{dr}{r})

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})

dF_W = -0.3N

Therefore there is a weight loss of 0.3N every kilometer.

4 0
4 years ago
If the magnitude of the magnetic force on a proton is F when it is moving at 14.0 o with respect to the field, what is the magni
sergeinik [125]

Answer:

The magnitude of the magnetic force at 32.5⁰ to the field, is 2.22 F

Explanation:

Given;

magnitude of the magnetic force is F at 14.0⁰ with respect to the field.

To determine the magnitude of the magnetic force at 32.5⁰ to the field, we apply the following formula and solve with proportion;

f*Sin(14.0⁰) = F

f*Sin(32.5⁰) = ?

= \frac{f*Sin(32.5^o)}{f*Sin(14.0^o)}.F\\\\ =2.22 \ F

Therefore, the magnitude of the magnetic force at 32.5⁰ to the field, is 2.22 F

6 0
4 years ago
A frictionless piston–cylinder device contains 5 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly accordin
Arte-miy333 [17]

Answer:

The work input during this process is -742 kJ

Explanation:

Given;

Initial temperature of nitrogen T₁ = 250 K

final temperature of nitrogen T₂ = 450 K

mass of nitrogen, m = 5 kg

PV^{1.4} = constant

The work input during the process is calculated as;

W = \frac{m*R(T_2-T_1)}{1-n}

where;

R is gas constant = 0.2968 kJ/kgK

substitute given values in above equation.

W = \frac{m*R(T_2-T_1)}{1-n} = \frac{5*0.2968(450-250)}{1-1.4} = -742 \ kJ

Therefore, the work input during this process is -742 kJ

8 0
4 years ago
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