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In-s [12.5K]
3 years ago
7

An asteroid has acquired a net negative charge of 149 C from being bombarded by the solar wind over the years, and is currently

in equilibrium whereby it expels electrons at the same rate as it acquires them. How many more electrons does it have than protons
Physics
1 answer:
bearhunter [10]3 years ago
5 0

Answer:

93.125 × 10^(19)

Explanation:

We are told the asteroid has acquired a net negative charge of 149 C.

Thus;

Q = -149 C

charge on electron has a value of:

e = -1.6 × 10^(-19) C

Now, for us to determine the excess electrons on the asteroid, we will just divide the net charge in excess on the asteroid by the charge of a single electron.

Thus;

n = Q/e

n = -149/(-1.6 × 10^(-19))

n = 93.125 × 10^(19)

Thus, it has 93.125 × 10^(19) more electrons than protons

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V=12\\I=\text{ ?}\\R=3\\\\\text{Solve for I}:\\12=3I\\3I=12\\I=\frac{12}{3}\\I=4\text{A}

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Two horizontal forces act on a 1.4 kg chopping block that can slide over a friction-less kitchen counter, which lies in an xy pl
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Answer:

Part a)

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Part b)

a = (0.64\hat i + 5.21 \hat j)m/s^2

Part c)

a = (4.92\hat i - 0.5 \hat j)m/s^2

Explanation:

As per Newton's II law we know that

F = ma

so we will have

a = \frac{F}{m}

so we will have

a = \frac{F_1 + F_2}{m}

Part a)

a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i - 4\hat j)}{1.4}

a = \frac{0.9 \hat i - 0.7 \hat j}{1.4}

a = (0.64\hat i - 0.5 \hat j)m/s^2

Part b)

a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i + 4\hat j)}{1.4}

a = \frac{0.9 \hat i + 7.3 \hat j}{1.4}

a = (0.64\hat i + 5.21 \hat j)m/s^2

Part c)

a = \frac{(3.9 \hat i + 3.3 \hat j) + (3\hat i - 4\hat j)}{1.4}

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3 0
3 years ago
A wire carries 3.7 A of current. A second wire is placed parallel to the first 8.0 cm away. What is the current flowing through
IgorC [24]

Answer:

The current in second wire is 5.0 A.

(B) is correct option.

Explanation:

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Current in first wire = 3.7 A

Distance = 8.0 cm

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Using formula of magnetic field

B=\dfrac{\mu_{0}I}{2\pi r}

Where, I = current

r = distance

Put the value into the formula

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B = \dfrac{\mu_{0}\times3.7}{2\pi \times3.4\times10^{-2}}...(I)

For second wire,

The distance is 8-3.7 =  4.3 cm

B' = \dfrac{\mu_{0}\times I'}{2\pi \times(8-3.4)\times10^{-2}}...(II)

The magnetic field in both the wires,

From equation (I) and (II)

\dfrac{\mu_{0}\times3.7}{2\pi \times3.4\times10^{-2}}= \dfrac{\mu_{0}\times I'}{2\pi \times(8-3.4)\times10^{-2}}

I'=\dfrac{3.7\times4.3\times10^{-2}}{3.4\times10^{-2}}

I'=4.68\ A\ approx = 5.0\ A

Hence, The current in second wire is 5.0 A.

8 0
3 years ago
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