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Dafna1 [17]
3 years ago
13

A 2.00-gram air inflated balloon is given an excess negative charge, q1 = -3.25 × 10-8 C, by rubbing it with a blanket. It is fo

und that a charged rod can be held above the balloon at a distance of d = 5.00 cm to make the balloon float. In order for this to occur, what polarity of charge must the rod possess? How much charge, q2, does the rod have? Assume the balloon and rod to be point charges. The Coulomb force constant is 1/(4π ε0) = 8.99 × 109 N·m2/C2 and the acceleration due to gravity g = 9.81 m/s2.
Physics
1 answer:
Llana [10]3 years ago
3 0

Answer:

The rod`s charge must be positive,  because the gravity force is pointing downwards and the electrostatic force must be pointing upwards (in order to balance the gravity force)

The charge is q_2 = 1.667 times 10^(-7) C

Explanation:

F_e = F_g

where F_g = m g and F_e= (1/4 pi e_0)*(q_1*q_2)/d^2,

please see the file attached for more details.

Download pdf
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duplication being slightly more expensive.

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Find the current flowing out of the battery.​
klemol [59]

Answer:

0.36 A.

Explanation:

We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:

Resistor 1 (R₁) = 35 Ω

Resistor 2 (R₂) = 20 Ω

Equivalent Resistance (Rₑq) =?

Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:

Rₑq = (R₁ × R₂) / (R₁ + R₂)

Rₑq = (35 × 20) / (35 + 20)

Rₑq = 700 / 55

Rₑq = 12.73 Ω

Next, we shall determine the total resistance in the circuit. This can be obtained as follow:

Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω

Resistor 3 (R₃) = 15 Ω

Total resistance (R) in the circuit =?

R = Rₑq + R₃ (they are in series connection)

R = 12.73 + 15

R = 27.73 Ω

Finally, we shall determine the current. This can be obtained as follow:

Total resistance (R) = 27.73 Ω

Voltage (V) = 10 V

Current (I) =?

V = IR

10 = I × 27.73

Divide both side by 27.73

I = 10 / 27.73

I = 0.36 A

Therefore, the current is 0.36 A.

6 0
3 years ago
A 55.4 g sample of water at 99.61 °C is placed in a constant pressure calorimeter. Then, 23.4 g of zinc metal at 21.6 °C is adde
Zolol [24]

Answer:

The specific heat capacity of the zinc metal measured in this experiment is 0.427 J/g.°C

Explanation:

From the experimental data, the water loses heat because its initial temperature is greater than the final temperature of the mixture. On the other hand, the zinc metal gains heat because its initial temperature is less than the final temperature of the mixture

Heat loss by water = Heat gain by zinc metal

m1C1(T1 - T3) = m2C2(T3 - T2)

m1 is mass of water = 55.4 g

C1 is specific heat capacity of water = 4.2 J/g.°C

m2 is mass of zinc metal = 23.4 g

C2 is specific heat capacity of zinc metal

T1 is the initial temperature of water = 99.61 °C

T2 is the initial temperature of zinc metal = 21.6 °C

T3 is the final temperature of the mixture = 96.4 °C

55.4×4.2(99.61 - 96.4) = 23.4×C2(96.4 - 21.6)

746.9028 = 1750.32C2

C2 = 746.9028/1750.32 = 0.427 J/g.°C

3 0
3 years ago
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