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2 years ago

D. 'g' vanishes at centre of

Physics

1 answer:

guapka [62]2 years ago

6 0

Answer:

a. 0.8 cm

Explanation:

The distance of the object from the lens, u = 1 cm

The magnification of the lens, m = 5

The focus of a lens formula is given as follows;

$f = \dfrac{1}{\dfrac{1}{v} + \dfrac{1}{v} }$

The magnification of the lens, m = -v/u

Where;

v = The distance of the image from the lens

Therefore, we have;

v = m × u

∴ v = 5 × 1 cm = 5 cm (on the other side of the lens)

From which we get;

$f = \dfrac{1}{\dfrac{1}{1} + \dfrac{1}{5} } = \dfrac{5}{6} \approx 0.8$

The focal length ≈ 0.8 cm

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A 30-m-long rocket train car is traveling from Los Angeles to New York at 0.5c when a light at the center of the car flashes. Wh

Novosadov [1.4K]

Given that,

Distance =30 m

speed = 0.5c

(A). We need to find the bell and siren simultaneous events for a passenger seated in the car

According to given data

The distance travelled by the light to reach either side of the rocket train car is same.

So, The two events are simultaneous and the bell and siren are the simultaneous events for a passenger seated in the car.

(B). We need to calculate time interval between the events

Using formula of time dilation

$\Delta t=\dfrac{\Delta t'}{\sqrt{1-\dfrac{v^2}{c^2}}}$ .....(I)

Where, $\delta t'$ = proper time

$\delta t$ = time interval between the events

The time interval between the events measured in a reference frame

The proper time in this case is

$\Delta t'=\Delta t_{1}-\dfrac{v\Delta x}{c^2}$

For the second interval,

Put the value of $\Delta t'$ in the equation (I)

$\Delta t_{2}=\dfrac{\Delta t_{1}-\dfrac{v\Delta x}{c^2}}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value in the equation

$\Delta t_{2} = \dfrac{0-\dfrac{0.5c\times30}{c^2}}{\sqrt{1-\dfrac{0.5^2c^2}{c^2}}}$

$\Delta t_{2}=\dfrac{-15}{3\times10^{8}\sqrt{1-0.25}}$

$\Delta t_{2}=-5.77\times10^{8}\ s$

Negative sign shows the siren rings before the bell ring.

Hence, (A). Yes, the bell and siren are simultaneous events.

(B). The siren sounds before the bell rings.

8 0

2 years ago

Three parallel wires of length l each carry current Iin the same direction. They’re positioned at the vertices of an equilateral

cluponka [151]

Answer:

F = μi²l/πa

Explanation:

The magnetic force F on a length of wire, l carrying a current i in a magnetic field B is given by

F = Bilsinθ

The magnetic field due to one wire of length, l carrying a current, i at a distance a from it is given by B = μi/2πa

So, the force on the first wire due to the second wire is F₁ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

the force on the first wire due to the third wire is F₂ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

Since the magnetic field due to the one wire is perpendicular to the length of the other wire its field acts upon, θ = 90

So, F₁ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa and

F₂ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa

Since the angle between F₁ and F₂ is 60° (since it is an equilateral triangle)

The resultant force F is thus

F = √(F₁² + F₂² + 2F₁F₂cos60°)

F = √(F₁² + F₂² + 2F₁F₂ × 0.5)

F = √(F₁² + F₂² + 2F₁F₂) (since F₁ = F₂)

F = √(2F₁² + 2F₁²) = √(4F₁²)

F = 2F₁

F = 2μi²l/2πa

F = μi²l/πa

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3 years ago

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kumpel [21]

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3 years ago

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An automobile starter motor has an equivalent resistance of 0.055 Ï and is supplied by a 12.0 v battery which has a 0.0305 Ï int

LiRa [457]

12 V is the f.e.m. $\epsilon$ of the battery. The potential difference that is applied to the motor is actually the fem minus the voltage drop on the internal resistance r:
$\epsilon - Ir$
this is equal to the voltage drop on the resistance of the motor R:
$RI$
so we can write:
$\epsilon - Ir = RI$
and using $r=0.0305~\Omega$ and $R=0.055~\Omega$ we can find the current I:
$I= \frac{\epsilon}{R+r}=140~A$

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2 years ago

A wind-tunnel experimentis performed on a 1/25scale model of a supersonic aircraft. The prototype aircraft flies at 450 m/s in c

KengaRu [80]

Answer: V = 504m/a

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2 years ago