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Arte-miy333 [17]

3 years ago

8

If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force betw

een them?

1 answer:

LiRa [457]3 years ago

4 0

Answer:

5.619×10⁶ N

Explanation:

Applying,

F = kqq'/r²................... Equation 1

Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges

From the questiion,

Given: q = 2.5 C, q' = 2.5 C, r = 100 m

Constant: 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.5×2.5×8.99×10⁹)/100²

F = 56.19×10⁵

F = 5.619×10⁶ N

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igomit [66]

Answer:

$1.06\times 10^6 J$

Explanation:

We are given that

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$V_2=-5 V$

Charge on 1 electron, 1 e= $-1.6\times 10^{-19} C$

Avogadro's number, $N_A=6.02\times 10^{23}$

$q=ne$

$q=6.02\times 10^{23}\times 1.6\times 10^{-19}=-9.632\times 10^4 C$

$W=q\Delta V$

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8 0

3 years ago

Copper Pot A copper pot with a mass of 2 kg is sitting at room temperature (20°C). If 200 g of boiling water (100°C) are put in

mr Goodwill [35]

Answer:

<em>The final temperature is 61.65 °C</em>

Explanation:

mass of copper pot $m_{c}$ = 2 kg

temperature of copper pot $T_{c}$ = 20 °C (the pot will be in thermal equilibrium with the room)

specific heat capacity of copper $C_{c}$ = 385 J/kg-°C

The heat content of the copper pot = $m_{c}$ $C_{c}$ $T_{c}$ = 2 x 385 x 20 = 15400 J

mass of boiling water $m_{w}$ = 200 g = 0.2 kg

temperature of boiling water $T_{w}$ = 100 °C

specific heat capacity of water $C_{w}$ = 4182 J/kg-°C

The heat content of the water = $m_{w}$ $C_{w}$ $T_{w}$ = 0.2 x 4182 x 100 = 83640 J

The total heat content of the water and copper mix $H_{T}$ = 15400 + 83640 = 99040 J

This same heat is evenly distributed between the water and copper mass to achieve thermal equilibrium, therefore we use the equation

$H_{T}$ = $m_{c}$ $C_{c}$ $T_{f}$ + $m_{w}$ $C_{w}$

where $T_{f}$ is the final temperature of the water and the copper

substituting values, we have

99040 = (2 x 385 x $T_{f}$ ) + (0.2 x 4182 x

99040 = 770 $T_{f}$ + 836.4

99040 = 1606.4 $T_{f}$

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3 years ago

A disk (radius = 2.00: mm) is attached to a high-speed drill at a dentist's office and is turning at 7.85 times 10^4 rad/s. Dete

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Answer:

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Explanation:

Tangential speed = Angular speed x Radius

Angular speed = 7.85 x 10⁴ rad/s

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3 0

3 years ago

The filament of a lightbulb has a resistance of R0=12Ω at 20 ∘C and 140 Ω when hot. Part APart complete Part B In this temperatu

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Answer:

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Given data:

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where $R_O$ is resistance at $_o,$ and $\alpha$ = coefficient of resistivity

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SOLVING FOR $\Delta T$

$\Delta T = 2370$ degree C

WE KNOW

$\Delta T = T - T_o$

T = 2370 + 20

T = 2390 degree C

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3 years ago

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4 years ago