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sertanlavr [38]
2 years ago
13

Which statement about images is correct? a) A virtual image cannot be formed on a screen. b) A virtual image cannot be viewed by

the unaided eye. c) A virtual image cannot be photographed. d) A real image must be erect. e) Mirrors always produce real images because they reflect light.
Physics
1 answer:
ankoles [38]2 years ago
5 0

Answer:

A). A virtual image cannot be formed on a screen.

Explanation:

A virtual image can not be formed on a screen.

For image:

1.A virtual image can be viewed by the unaided eye.

2. A real image must be erect or maybe inverted.

3.Mirrors can produce virtual as well as real image ,it depends on which type of mirror is.

4.A virtual image can be photographed.

So the option A is correct.

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Now that we have a feel for the state of the circuit in its steady state, let us obtain the expression for the current in the ci
vesna_86 [32]

Answer:

i(t) = (E/R)[1 - exp(-Rt/L)]

Explanation:

E−vR−vL=0

E− iR− Ldi/dt = 0

E− iR = Ldi/dt

Separating te variables,

dt/L = di/(E - iR)

Let x = E - iR, so dx = -Rdi and di = -dx/R substituting for x and di we have

dt/L = -dx/Rx

-Rdt/L = dx/x

interating both sides, we have

∫-Rdt/L = ∫dx/x

-Rt/L + C = ㏑x

x = exp(-Rt/L + C)

x = exp(-Rt/L)exp(C)     A = exp(C) we have

x = Aexp(-Rt/L) Substituting x = E - iR we have

E - iR = Aexp(-Rt/L) when t = 0, i(0) = 0. So

E - i(0)R = Aexp(-R×0/L)

E - 0 = Aexp(0) = A × 1

E = A

So,

E - i(t)R = Eexp(-Rt/L)

i(t)R = E - Eexp(-Rt/L)

i(t)R = E(1 - exp(-Rt/L))

i(t) = (E/R)(1 - exp(-Rt/L))

5 0
2 years ago
Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac
alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = 47\times 10^{-9}\ C/m^2

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

E=\dfrac{\sigma}{2\epsilon_0}

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Now, plug in  47\times 10^{-9}\ C/m^2 for 'σ' and 8.85\times 10^{-12}\ F/m for \epsilon_0 and solve for the electric field. This gives,

E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0
2 years ago
The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan
PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

Explanation:

Magnetic field around a long current carrying wire is given by

B=\frac{\mu _o I}{2\pi r}

where B= magnetic field

           \mu _o= permeability of free space

           I= current in the long wire and

           r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  

Now if I'=3I and r'=2r then magnetic field B' is given by

B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

   

7 0
2 years ago
Read 2 more answers
What keeps an inflated balloon from falling down if you rub it against your hair and place it against a wall? 1. rubbing distort
Keith_Richards [23]
2 because when you are doing this it causes friction Which then cause the balloon to stick
~dany-ley
3 0
2 years ago
The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the e
Lapatulllka [165]

Answer:

0.2cm towards the retina.

Explanation:

the focal length of the frog eye is

(1/f) = (1/10) + (1/0.8)

f = 0.74cm

Since the distance of the object is 15cm Hence

(1/0.74) = (1/15) + (1/V)

V = 0.78cm

Therefore the distance the retina is to move is

0.78cm - 0.8cm = 0.02cm towards the retina.

3 0
2 years ago
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