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cricket20 [7]
2 years ago
15

Guy can someone help me with this

Physics
1 answer:
Mrac [35]2 years ago
3 0

I believe it should be B.

Here’s one desctiption:

When the balloon has been rubbed enough times to gain a sufficient negative charge, it will be attracted to the wall. Although the wall should normally have a neutral charge, the charges within it can rearrange so that a positively charged area attracts the negatively charged balloon

Hopefully it’s right!

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What would happen to the apparent change in mass if the direction of the current is reversed?
pantera1 [17]
When you reverse the direction of the current, the current loop generated by the magentic field is revered.
8 0
3 years ago
A current of 3.6-A flows through a conductor. Calculate how much charge passed through and cross-section of the conductor in 25s
Anastaziya [24]

Answer:

90 C

Explanation:

Electric current: This can be defined as the rate of flow of electric charge in a circuit. This can be expressed mathematically as,

I = dQ/dt

dQ = Idt

∫dQ = ∫Idt

Q = It................................ Equation 1

Where Q = amount of charge, I = current, t = time.

Given: I = 3.6 A, t = 25 s.

Substituting into equation 1,

Q = 3.6(25)

Q = 90 C.

Hence the amount of charge passing through the cross section of the conductor = 90 C

6 0
3 years ago
A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a
DENIUS [597]

Answer:

   C = 4,174 10³ V / m^{3/4} ,  E = 7.19 10² / ∛x,    E = 1.5  10³ N/C

Explanation:

For this exercise we can calculate the value of the constant and the electric field produced,

Let's start by calculating the value of the constant C

           V = C x^{4/3}

           C = V / x^{4/3}

            C = 220 / (11 10⁻²)^{4/3}

            C = 4,174 10³ V / m^{3/4}

To calculate the electric field we use the expression

            V = E dx

             E = dx / V

             E = ∫ dx / C x^{4/3}

            E = 1 / C  x^{-1/3} / (- 1/3)

            E = 1 / C (-3 / x^{1/3})

We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E

            E = 3 / C     (0- (-1 / x^{1/3}))

            E = 3 / 4,174 10³   (1 / x^{1/3})

           E = 7.19 10² / ∛x

for x = 0.110 cm

          E = 7.19 10² /∛0.11

          E = 1.5  10³ N/C

6 0
3 years ago
A ball having a mass of 0.20 kilograms is placed at a height of 3.25 meters. If it is dropped from this height, what will be the
asambeis [7]
EC_1 + EP_1 = EC2 + EP_2

EC_2 = 0

EC_2 = EP_1 - EP_2

EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J
7 0
2 years ago
Read 2 more answers
The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
kotykmax [81]

Answer:

2.1\times 10^{-12} c

Explanation:

We are given that

Surface area of membrane=5.3\times 10^{-9} m^2

Thickness of membrane=1.1\times 10^{-8} m

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

Substitute the values then we get

Capacitance between parallel plate capacitor=\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}

C=0.25\times 10^{-12}F

V=85.9 mV=85.9\times 10^{-3}

Q=CV

Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c

Hence, the charge resides on the outer surface=2.1\times 10^{-12} c

5 0
3 years ago
Read 2 more answers
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