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expeople1 [14]
3 years ago
8

A point charge of -3.0 x 10-C is placed at the origin of coordinates. Find the clectric field at the point 13. X= 5.0 m on the x

-axis.​
Physics
1 answer:
mezya [45]3 years ago
3 0

Answer:

-1.0778×10⁻¹⁰ N/C

Explanation:

Applying,

E = kq/r²................ equation 1

Where E = elctric field, q = charge, r = distance, k = coulomb's law

From the question,

Given: q = -3.0×10 C, r = 5.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values in equation 1

E = (-3.0×10)(8.98×10⁹)/5²

E = -1.0778×10⁻¹⁰ N/C

Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C

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Question 4 What would be the UCS for Tab the dog? O The shock O The flowers and fence O Fear O The boundary Question 5​
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2 years ago
You place a cup of 210 degrees F coffee on a table in a room that is 68 degrees F, and 10 minutes later, it is 200 degrees F. Ap
Inga [223]
The answer is 35 minutes

The Newton's law of cooling is:
T(x) = Ta + (To - Ta)e⁻ⁿˣ

T(x) - the temperature of the coffee at time x
Ta - the ambient temperature
To - the initial temperature
n - constant

step 1. Calculate constant k:

We have:
T(x) = 200°F
x = 10 min
Ta = 68°F
To = 210°F
n = ?

T(x) = Ta + (To - Ta)e⁻ⁿˣ
200 = 68 + (210 - 68)e⁻ⁿ*¹⁰
200 = 68 + 142 * e⁻¹⁰ⁿ
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ln(e⁻¹⁰ⁿ) = ln(0.93)
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Step 2. Calculate time x when T(x) = 180°F:
We have:
T(x) = 180°F
x = ?
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T(x) = Ta + (To - Ta)e⁻ⁿˣ
180 = 68 + (210 - 68)e⁻⁰.⁰⁰⁷*ˣ
180 - 68 = 142 * e⁻⁰.⁰⁰⁷*ˣ
112 = 142 * e⁻⁰.⁰⁰⁷⁾*ˣ
e⁻⁰.⁰⁰⁷*ˣ = 112/142
e⁻⁰.⁰⁰⁷*ˣ = 0.79

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-0.007x * 1 = -0.24
-0.007x = -0.24
x = -0.24 / -0.007
x ≈ 35
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