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2 years ago

An engineer can increase the magnitude of the magnification of a compound microscope by

Physics

1 answer:

lakkis [162]2 years ago

5 0

Answer:

Option B

Explanation:

Magnification of Microscope is

$M = M_o \times M_e$

Mo= Magnification of objective lens and

Me= magnification of the eyepiece.

Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.

Magnification,

$M\ \alpha\ \dfrac{1}{f}$

when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.

Thus. Magnification will increase by decreasing the focal length.

The correct answer is Option B i.e. using shorter focal length

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Why does frost bite occur in mountain climbers ?

san4es73 [151]

Answer:

The higher the altitude the colder

Explanation:

Mountains are very high up obviously higher then the ground. It is colder so it makes it easier for mountain climbers to get frost bite because of the high altitude.

Hope this helps:)

7 0

2 years ago

Read 2 more answers

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

2 years ago

A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 245 Hz. A person on the pl

Luba_88 [7]

Answer:

Explanation:

by Doppler effect

F apparent = F real x (Vair +- Vobserver) / (Vair +- Vsource)

case1 :if observer is approaching a source then put a + in numerator and - in denominator

case 2 : if observer is going away from source then do opposite.

it is case1 cos observer is moving toward wall . actually wall is acting as a source for observer

F apparent = 240 x (344 + vp ) / (344 - vp )

now F apparent - Freal = beat frequency of 6.00 Hz

so {240 x (344 + vp ) / (344 - vp )} - 230 = 6

240 x (344 + vp -344 + vp) / (344 - vp ) = 6

240 x 2 x vp = 6 x (344 - vp )

480 vp = 2064 - 6 x vp

486 x vp = 2064

vp = 4.24 m/s

4 0

2 years ago

What is the energy of the electron in a helium ion with a charge of +1 in an orbit with a value of n=4?

goldenfox [79]

Answer:

The energy of the electron in a helium ion is $-5.44\times10^{-19}\ J$

Explanation:

Given that,

Charge = +1

Value of n = 4

Suppose $k=2.179\times10^{-18}\ J$

We need to calculate the energy of the electron in a helium ion

Using formula of energy

$E_{n}=-k\dfrac{z^2}{n^2}$

Put the value into the formula

$E_{n}=-2.179\times10^{-18}\times\dfrac{2^2}{4^2}$

$E_{n}=-5.44\times10^{-19}\ J$

Hence, The energy of the electron in a helium ion is $-5.44\times10^{-19}\ J$

8 0

2 years ago

Janice's mother often lets her 6-month-old baby sit in front of the television, watching episodes of Sesame Street. What is Jani

brilliants [131]

B.

The child is too old to be gaining something from the screen time.

6 0

2 years ago