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grin007 [14]
3 years ago
6

An infinite line charge of linear density λ = 0.30 μC/m lies along the z axis and a point charge q = 6.0 µC lies on the y axis a

t y = 2.0 m. The x component of the electric field at the point P on the x axis at x = 3.0 m is approximately . A) 1.8 kN/C B) 4.2 kN/C C) 0.96 kN/C D) 5.2 kN/c E) 0.64 mN/C
Physics
2 answers:
ira [324]3 years ago
6 0
The line charge E-field Ec = λ/(2πr*e0),
where λ = charge/length and e0 is the permittivity constant = 8.8542E-12 F/m. 
<span>The point charge E-field Ep = kq/r^2 where electrostatic constant k = 1/(4π*e0) = 8.99E9 N-m^2/C^2.</span>
Ivahew [28]3 years ago
6 0

Answer:

D) 5.2 kN/c

Explanation:

Electric field due to infinity line charge

λ /2πε₀r = .3 x 10⁻⁶ x 2 x 9 x 10⁹ / 3 = 1.8 x 10³. It will be towards x axis.

Electric field due to point charge = k.q / r² = 9 x 10⁹ x 6 x 10⁻⁶ / 13 = 4.15 x 10³

x component of this field = 4.15 x 10³ x cos Ф = 4.15 x 10³ x 3 / √13

3.458 x 10³

Total electric field = 5.2 kN / C.

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Streams have a detectable current, while rivers do not.
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8 0
2 years ago
An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo
timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

Where:

v_{a}, v_{b} - Initial and final velocities, measured in meters per second.

t_{a}, t_{b} - Initial and final times, measured in seconds.

a(t) - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

v_{4} = 2\,\frac{m}{s}  +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt

v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)

v_{4} = -6\,\frac{m}{s}

Region II (t = 4 s to t = 6 s)

v_{6} = -6\,\frac{m}{s}  +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt

v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)

v_{6} = -4\,\frac{m}{s}

Region III (t = 6 s to t = 10 s)

v_{10} = -4\,\frac{m}{s}  +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt

v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)

v_{10} = 4\,\frac{m}{s}

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0
3 years ago
Calculate the force of a particle with a net charge of 170 coulombs traveling at a speed of 135 meters/second perpendicular to t
amm1812

Answer:

F=1.14N j

Explanation:

The magnitude of the magnetic force over a charge in a constant magnetic field is given by the formula:

|\vec{F}|=|q\vec{v} \ X\ \vec{B}|=qvsin\theta  (|)

In this case v and B vectors are perpendicular between them. Furthermore the direction of the magnetic force is:

-i X k = +j

Finally, by replacing in (1) we obtain:

\vec{F}=(170C)(135\frac{m}{s})(5.0*10^{-5}T)=1.14N\ \hat{j}

hope this helps!

6 0
3 years ago
Read 2 more answers
Which statement correctly defines displacement
Andreyy89

Displacement is B) the shortest distance between the starting point and the ending point of a motion

Explanation:

Displacement is a vector quantity; it is a vector connecting the initial position to the final position of motion of an object.

Since it is a vector, it has both a magnitude and a direction:

  • The magnitude of the displacement is the length of the vector, therefore it corresponds to the shortest distance in a  straight line between the starting point and the ending point of the motion
  • The direction goes from the starting point to the ending point

Therefore, the correct answer is

B) the shortest distance between the starting point and the ending point of a motion

Note that displacement is very different  from distance. Consider for example an object moving in a circle, returning to its initial position: in this case, the distance covered by the object is not zero (it is the length of the circle), however the  displacement is zero, because the initial position corresponds to the ending position.

Learn more about distance and displacement:

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#LearnwithBrainly

6 0
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