Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?
1 answer:
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Response:
Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.
Details:
The locomotive's acceleration is 1.6
The duration taken to pass the crossing is 2.4 seconds.
We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.
When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 .
32 = 0 + 1.6 * t
t = 20 seconds.
Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.
Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.
1) For x = 6.6 cm,
2) For x = 6.6 cm,
3) For x = 1.45 cm,
4) For x = 1.45 cm,
5) Surface charge density at b = 4 cm:
6) At x = 3.34 cm, the x-component of the electric field equals zero
7) Surface charge density at a = 2.9 cm:
8) None of these regions
Explanation:
1)
The electric field from an infinite charge sheet is perpendicular to it:
where
is the surface charge density
represents vacuum permittivity
Outside the slab, the electric field behaves like that of an infinite sheet.
Consequently, the electric field at x = 6.6 cm (situated to the right of both the slab and sheet) results from the combination of the fields from both:
where
The field from the sheet points left (negative, inward), and the slab’s field points right (positive, outward).
Thus,
and the negative sign indicates a rightward direction.
2)
Both the sheet’s and slab’s fields are perpendicular to their surfaces, directing along the x-axis, hence there's no y-component for the total field.
<pThus, the y-component totals zero.This happens because both the sheet and slab stretch infinitely along the y-axis. Choosing any x-axis point reveals that the y-component of the field, generated by a surface element dS of either the sheet or slab, , will be equal and opposite to the corresponding component from the opposite side,
. Thus, the combined y-direction field is always zero.
3)
This scenario resembles part 1), but the point here is
x = 1.45 cm
which lies between the sheet and the slab. The fields from both contribute leftward as the slab has a negative charge (resulting in an outward field). Thus, the total field computes to
Replacing with expressions from part 1), we get
where the negative illustrates a leftward direction.
4)
This portion parallels part 2). Since both fields remain perpendicular to the slab and sheet, no component exists along the y-axis, thus the electric field's y-component is zero.
5)
Notably, the slab behaves as a conductor, signifying charge mobility within it.
The net charge on the slab is positive, indicating a surplus of positive charge. With the negatively charged sheet on the left of the slab, positive charges shift towards the left slab edge (at a = 2.9 cm), while negative charges move to the right edge (at b = 4 cm).
The surface charge density per unit area of the slab is
This average denotes the surface charge density on both slab sides at points a and b:
(1)
Additionally, the infinite sheet at x = 0 negatively charged , induces an opposite net charge on the slab's left surface, thus
(2)
Having equations (1) and (2) allows for solving the surface charge densities at a and b, yielding:
6)
We aim to compute the x-component of the electric field at
x = 3.34 cm
This point lies inside the slab, bounded at
a = 2.9 cm
b = 4.0 cm
In a conducting slab, the electric field remains at zero owing to charge equilibrium; thus, the x-component thereof in the slab is zero
7)
From part 5), we determined the surface charge density at x = a = 2.9 cm is
8)
As mentioned in part 6), conductors have zero electric fields internally. Since the slab is conductive, the electric field inside remains zero; therefore, the regions where the electric field is null are
2.9 cm < x < 4 cm
Thus, the suitable answer is
"none of these regions"
Learn more about electric fields:
Objects in vertical motion are an illustration of non-uniform motion. At the peak of the circle, centripetal force is balanced by the object's weight. Therefore, the minimum speed required at this top point is given by v = 
= 
= 5.23 m/s. As the sphere descends from the top to the bottom of the circle, according to the law of conservation of energy, potential energy can be expressed as 
where u is the velocity at the lowest point. Consequently, the modified equation is
=
=
= 11.71 m/s. The collision of the dart with the bullet is an inelastic one. According to the conservation of momentum: v =
=
=
= 58.55 m/s. Thus, the dart's minimum initial speed for the combined system to complete a circular loop post-collision is 58.55 m/s.