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3 years ago

The magnification of a microscope is increased when_________.

Physics

1 answer:

azamat3 years ago

6 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,

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

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

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12. Which of the following is the property of a system?

Sati [7]

Answer:

Explanation:

Pressure, temperature are measurable properties and they are also known as physical properties.

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2 years ago

Which method of popcorn popping transfers heat into the kernels without any direct

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3 years ago

A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

skelet666 [1.2K]

Answer:

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

$A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}$

Differentiate both sides with respect to d.

$\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)$

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

$\frac{d^{2}A}{dt^{2}}= + ve$

(a) So, the area is maximum when the side of square is 29 m

(b) so, the area is minimum when the side of square is 8.76 m

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3 years ago

A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magni

irinina [24]

The magnitude of the magnetic field inside the solenoid is $6.46 \times 10^{-3} \ T$ .

The given parameters;

length of the solenoid, L = 91 cm = 0.91 m
radius of the solenoid, r = 1.5 cm = 0.015 m
number of turns of the solenoid, N = 1300
current in the solenoid, I = 3.6 A

The magnitude of the magnetic field inside the solenoid is calculated as;

$B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\$

where;

$\mu_o$ is the permeability of frees space = 4π x 10⁻⁷ T.m/A

$B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T$

Thus, the magnitude of the magnetic field inside the solenoid is $6.46 \times 10^{-3} \ T$ .

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2 years ago

If the magnitude of the magnetic force on a proton is F when it is moving at 18.0 ∘ with respect to the field, what is the magni

Lelu [443]

Answer:

The force when θ = 33° is 1.7625 times of the force when θ = 18°

Explanation:

The force on a moving charge through a magnetic field is given by

F = qvB sin θ

q = charge of the moving particle

v = Velocity of the moving charge

B = Magnetic field strength

θ = angle between the magnetic field and the velocity (direction of the motion) of the moving charge

Because qvB are all constant, we can call the expression K.

F = K sinθ

when θ = 18°,

F = K sin 18° = 0.309K

when θ = 33°, let the force be F₁

F₁ = K sin 33° = 0.5446K

(F₁/F) = (0.5446K/0.309K) = 1.7625

F₁ = 1.7625 F

Hope this Helps!!!

5 0

3 years ago