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Sonbull [250]
4 years ago
13

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Physics
2 answers:
igor_vitrenko [27]4 years ago
7 0

Answer:

svbsshsjjsjs Hello :)

Ede4ka [16]4 years ago
3 0

Answer:

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Suppose you sketch a model of an atom using the ones here as a guide. How would you build a model that is ionized? How would you
kap26 [50]

Answer:

Explanation:

An atom is constructed of three different particles known as electrons, protons and neutrons.

These particles have different mass and charges and are responsible for various characters than an atom posses.

An electron has a negative charge, a proton has positive charge and charge of neutron is neutral. Equal number of electrons and protons are present in an atom that make it electrically neutral but different conditions can occur if we remove these particles from an atom.

1 : Model of an ionized atom - an ionized atom is one which has some net charge on this. It can be either a positive charge or a negative charge.

If we need to sketch the model of an ionized atom then one should either keep the number of electrons less or proton.

2: Model of radioactive atom : A radioactive atom is one an unstable atom and has access of energy in its center. It can be caused by adding either neutrons or protons.

3 0
3 years ago
A steel wire of length 1.5 m and diameter 1 mm is joined to an aluminum wire of identical dimensions to make a composite wire of
Anarel [89]

Answer:

1.805 mm

Explanation:

Extension in the steel wire = WL_{steel}/AE_{steel}

Extension in the aluminium wire = WL_{Al}/AE_{Al}

Total extension = W/A * (L_{steel}/E_{steel} + L_{Al}/E_{Al})

we have:

W = mg

W = 5 × 9.8

W = 49 N

Area A = π/4 × (0.001)²

= 7.85398 × 10 ⁻⁷ m²

Total extension = W/A * (L_{steel}/E_{steel} + L_{Al}/E_{Al})

Total extension = 49/ 7.85398 × 10 ⁻⁷ ( (1.5/ 200×10⁹) + 1.5/ 70×10⁹))

Total extension =  0.0018048

Total extension = 1.805 mm

Thus, the total extension = the resulting change in the length of this composite wire = 1.805 mm

4 0
3 years ago
On an icy road, a 1100 kg car moving at 55 km/h strikes a 480 kg truck moving in the same direction at 37 km/h . The pair is soo
statuscvo [17]

Answer:

Speed of the wreckage = 49.29 km/hr

Explanation:

This question is solved simply by using the conservation of momentum law.

The momentum of the three moving bodies are calculated below:

Momentum of Car 1 : 1100 * 55 = 60500 kg.km/hr

Momentum of Truck : 480 * 37 = 17760 kg.km/hr

Momentum of Car 2 : 1300 * 49 = 63700 kg.km/hr

Total mass of all three vehicles: 1100 + 480 + 1300 = 2880 kg

The final momentum equals the initial momentum if it is conserved. Thus we have the following equation:

Final Momentum = Initial Momentum

Final Velocity * Total mass = Momentum of all three vehicles combined

Final Velocity * 2880 = 60500 + 17760 + 63700

Final Velocity = 49.29 km/hr

7 0
4 years ago
Read 2 more answers
Please help<br> How would you explain 2 m/s/s [1<br> (1 Point)<br> Enter your answer
AlexFokin [52]
2 m/s/s means the velocity increases by 2 m/s every second.
3 0
4 years ago
An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
irinina [24]

Answer:

a) \sigma_{\rm in} = -2.18~{\rm \mu C/m^2}

b) \sigma_{\rm out}= 1.12~{\rm \mu C/m^2}

c) E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}

Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}

Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}

c) The electric field outside the cylinder can be found by Gauss' Law:

\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}

4 0
3 years ago
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