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ELEN [110]
3 years ago
8

What are the three fields E⃗ 1E→1, E⃗ 2E→2, and E⃗ 3E→3 created by the three charges? Write your answer for each as a vector in

component form. Express your answer using two significant figures. Enter your answers numerically separated by a comma.

Physics
1 answer:
marishachu [46]3 years ago
8 0

Answer:

E₁ = 8500 N/c i − 2800 N/C j

E₂ = 10,000 N/c i

E₃ = 8500 N/c i + 2800 N/C j

Explanation:

The magnitude of electric field from a point charge Q at a distance r is:

E = kQ / r²

where k is Coulomb's constant (9×10⁹ Nm²/C²).

The direction of the electric field is along the radial line outwards from the point.

Use Pythagorean theorem to find the distance from q₁.

r² = (-0.01 m)² + (0.03 m)²

r² = 0.001 m²

r = 0.0316 m

Plugging in values:

E₁ = (9×10⁹ Nm²/C²) (1×10⁻⁹ C) / (0.001 m²)

E₁ = 9000 N/C

Using ratios to find the components:

E₁ₓ = (x/r) E₁

E₁ₓ = (0.03 m / 0.0316 m) (9000 N/C)

E₁ₓ = 8500 N/C

E₁ᵧ = (y/r) E₁

E₁ᵧ = (-0.01 m / 0.0316 m) (9000 N/C)

E₁ᵧ = -2800 N/C

Therefore:

E₁ = 8500 N/c i − 2800 N/C j

Repeat for E₂:

r = 0.03 m

E₂ = (9×10⁹ Nm²/C²) (1×10⁻⁹ C) / (0.03 m)²

E₂ = 10,000 N/C

E₂ₓ = (x/r) E₂

E₂ₓ = (0.03 m / 0.03 m) (10,000 N/C)

E₂ₓ = 10,000 N/C

E₂ᵧ = (y/r) E₂

E₂ᵧ = (0 m / 0.0 m) (10,000 N/C)

E₂ᵧ = 0 N/C

Therefore:

E₂ = 10,000 N/c i

And of course, E₃ is the same as E₁, except the y component is positive.

E₃ = 8500 N/c i + 2800 N/C j

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How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest? a.3.41 x 10^5 J b.2.73 x 10^5 J
algol13

Answer:

Work done by the frictional force is 3.41\times 10^5\ J

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Initial velocity of car, u = 26.1 m/s

Finally, it comes to rest, v = 0

We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.

W=k_f-k_i

W=\dfrac{1}{2}m(v^2-u^2)

W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)

W = −340605 J

or

W=3.41\times 10^5\ J

Hence, the correct option is (a).

6 0
3 years ago
A quarter is flipped from a height of 1.45 m above the ground. How much time will it take to reach the ground if the person flip
Artemon [7]

It will take the quarter 0.151 seconds to reach the ground.

<u>Given the following data:</u>

  • Height = 1.45 meters
  • Initial velocity = 10.32 m/s

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find how much time it will take the quarter to reach the ground, we  would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

S = ut + \frac{1}{2} at^2

Where:

  • S is the height or distance covered.
  • u is the initial velocity.
  • a is the acceleration.
  • t is the time measured in seconds.

Substituting the values into the formula, we have;

1.45 = 10.32(t) + \frac{1}{2} (9.8)t^2\\\\1.45 = 10.32t + 4.9t^2\\\\4.9t^2 + 10.32t - 1.45 = 0

The standard form of a quadratic equation is:

ax^2 + bx + c = 0

a = 4.9, b = 10.32 and c = 1.45

We would solve the above quadratic equation by using the quadratic equation formula;

x = \frac{-b\; \pm \;\sqrt{b^2 - 4ac}}{2a}

Substituting the values, we have;

t = \frac{-10.32\; \pm \;\sqrt{10.32^2\; - \;4(4.9)(1.45)}}{2(4.9)}\\\\t = \frac{-10.32\; \pm \;\sqrt{106.5024\; - \;28.42}}{9.8}\\\\t = \frac{-10.32\; \pm \;\sqrt{78.0824}}{9.8}\\\\t = \frac{-10.32\; \pm \;8.84}{9.8}\\\\t = \frac{-10.32\; + \;8.84}{9.8}\\\\t = \frac{1.48}{9.8}

<em>Time, t = 0.151 seconds.</em>

Therefore, it will take the quarter 0.151 seconds to reach the ground.

Read more: brainly.com/question/8898885

3 0
2 years ago
A 12 volt battery in a motor vehicle is capable of supplying the starter motor with 150 A. It is noticed that the terminal volta
qwelly [4]

Answer:0.0133 \Omega

Explanation:

Given

Voltage=12 V

Current(I)=150 A

V_{terminal}=10 V

r_{internal}=\frac{\Delta V}{I}

r_{internal}=\frac{12-10}{150}

r_{internal}=\frac{2}{150}=0.0133 \Omega

4 0
3 years ago
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