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lozanna
1 day ago
7

An uncharged spherical conducting shell surrounds a charge –q at the center of the shell. Then charge +3q is placed on the outsi

de of the shell.
When static equilibrium is reached, the charges on the inner and outer surfaces of the shell are respectively:

a) +q, -q
b) -q, +q
c) +q, +2q
d) +2q, +q
Physics
1 answer:
Maru [2.3K]1 day ago
8 0

Answer:The charges on the inner and outer surfaces of the shell are +q and -q, respectively

Explanation:

In static equilibrium within a conductor, the total electric field, denoted as E, equals zero

This indicates that no charge can be in motion, maintaining the static state of equilibrium within the conductor.

Since the Electric field, E remains zero, the flux through the shell’s surface is also zero.

According to Gauss' law, the total enclosed charge must be zero.

Given that the center of the shell has a charge of -q, the positive charge on the inner surface must be +q in order for the total charge enclosed to balance out to zero.

As the charge resides in static equilibrium, there will be a corresponding negative charge on the outer surface, which totals to -q.

This leads us to the conclusion that the charges on the inner and outer surfaces of the shell are +q and -q, respectively

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An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte
Ostrovityanka [2208]

Answer:

35.79 meters

Explanation:

We have an archer, and there is a target. Denote the distance between them as d.

The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:

v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time t_{sound}. Thus, we can derive:

v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d.

Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:

t_{arrow} + t _{sound} = 1 s.

Using this relationship in the distance formula for sound allows us to write:

340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d.

Substituting the value of d from the first equation yields:

40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow}).

Now, after some calculations, we can proceed further:

40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s.

Finally, the value is inserted into the initial equation:

40 \frac{m}{s} * t_{arrow} = d

40 \frac{m}{s} * 340/380 s = 35,79 s = d

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A hard rubber rod with an electric potential energy of 5.2 × 10–3 J has a charge of 4.0 µC at the tip. What is the electric pote
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1) The electric potential energy can be defined as the product of the electric potential and the associated charge:
U=q V
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V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3}J}{4.0 \cdot 10^{-6} C}=1300 V=1.3 \cdot 10^3 V

2) Using this same formula, if the charge changes to q=2.0 \mu C = 2.0 \cdot 10^{-6} C, the resulting electric potential will be:
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8 0
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the minute hand on a clock is 9 cm long and travels through an arc of 252 degrees every 42 minutes. To the nearest tenth of a ce
Ostrovityanka [2208]

Answer:

The distance covered by the minutes hand is 39.60 cm.

Explanation:

Note: A clock has a circular shape, where the minutes hand acts as the radius, and its motion creates an arc.

Length of an arc is calculated as ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Here, L represents the arc’s length, ∅ is the angle made by the arc, and r is the arc’s radius.

Given: ∅ = 252°, r = 9 cm, π = 3.143.

Upon substituting these values into equation 1,

L = 252/360(2×3.143×9)

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L = 39.60 cm.

Thus, the distance traversed by the minutes hand is 39.60 cm.

4 0
1 month ago
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