Answer:
The object measures 6 m in distance and 2 m in height.
It creates a virtual image that is upright.
Explanation:
Provided data includes:
Focal length = 0.25 m
Image height = 0.080 m
Image distance = 0.24 m
We are to determine the object's distance.
Using the lens formula:
Substituting values into the formula:
We also need to calculate magnification:
Applying the magnification formula:
Substituting values into this formula:
Next, we need to find the height of the object:
Using the magnification formula once more:
A convex mirror generates a virtual and upright image on its backside.
Consequently, the object is at a distance of 6 m and has a height of 2 m.
The image formed is virtual and upright.
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
A. a<span> = 1.3 m/s^2</span><span>; </span>FN<span> = 63.1 N</span>
Answer:
C = 4,174 10³ V / m^{3/4}, E = 7.19 10² / ∛x, E = 1.5 10³ N/C
Explanation:
In this problem, we are tasked with determining the constant value and the generated electric field.
We will begin with computing the constant C:
V = C 
C = V / x^{4/3}
C = 220 / (11 10⁻²)^{4/3}
C = 4,174 10³ V / m^{3/4}
Next, we will find the electric field by utilizing the formula:
V = E dx
E = dx / V
E = ∫ dx / C x^{4/3}
E = 1 / C x^{-1/3} / (- 1/3)
E = 1 / C (-3 / x^{1/3})
We consider the evaluation from the lower limit x = 0 where E = E₀ = 0 to the upper limit x = x, resulting in E = E:
E = 3 / C (0- (-1 / x^{1/3}))
E = 3 / 4,174 10³ (1 / x^{1/3})
E = 7.19 10² / ∛x
Substituting x = 0.110 cm:
E = 7.19 10² /∛0.11
E = 1.5 10³ N/C
