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hram777 [196]
3 years ago
9

Which parts of a compound microscope magnify objects?

Physics
2 answers:
babymother [125]3 years ago
8 0
The parts of a compound microscope that magnify the image are the lenses. A compound microscope is called compound<span> because it has several lenses. The two main lenses are the ocular lens or eyepiece and the objective lenses.</span>
Yuri [45]3 years ago
7 0
The ocular lens or eyepiece lens
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How can meteorologists use the jet stream to predict the weather ?
barxatty [35]
<span>Jet streams act as an invisible director of the atmosphere and are largely responsible for changes in the weather across the globe.
Hope this helps</span>
5 0
3 years ago
Definition of AM and Fm Wave? in your own word
vfiekz [6]

Answer:

In AM broadcasting, the amplitude of the carrier wave is modulated to encode the original sound. In FM broadcasting, the frequency of the carrier wave is modulated to encode the sound. A radio receiver extracts the original program sound from the modulated radio signal and reproduces the sound in a loudspeaker.

8 0
2 years ago
A single-turn circular loop of wire of radius 45 mm lies in a plane perpendicular to a spatially uniform magnetic field. During
Lina20 [59]

Answer:

Magnitude of induced emf is 0.00635 V

Explanation:

Radius of circular loop r = 45 mm = 0.045 m

Area of circular loop A=\pi r^2

A=3.14\times 0.045^2=0.00635m^2

Magnetic field is increases from 250 mT to 350 mT

Therefore change in magnetic field dB=250-350=100mT

Emf induced is given by

e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}

e=-0.00635\times \frac{100\times 10^{-3}}{0.10}=-0.00635V

Magnitude of induced emf is equal to 0.00635 V

7 0
3 years ago
Two wires A and B with circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is
Nesterboy [21]

Answer:

r₁/r₂ = 1/2 = 0.5

Explanation:

The resistance of a wire is given by the following formula:

R = ρL/A

where,

R = Resistance of wire

ρ = resistivity of the material of wire

L = Length of wire

A = Cross-sectional area of wire = πr²

r = radius of wire

Therefore,

R = ρL/πr²

<u>FOR WIRE A</u>:

R₁ = ρ₁L₁/πr₁²   -------- equation 1

<u>FOR WIRE B</u>:

R₂ = ρ₂L₂/πr₂²   -------- equation 2

It is given that resistance of wire A is four times greater than the resistance of wire B.

R₁ = 4 R₂

using values from equation 1 and equation 2:

ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²

since, the material and length of both wires are same.

ρ₁ = ρ₂ = ρ

L₁ = L₂ = L

Therefore,

ρL/πr₁² = 4ρL/πr₂²

1/r₁² = 4/r₂²

r₁²/r₂² = 1/4

taking square root on both sides:

<u>r₁/r₂ = 1/2 = 0.5</u>

6 0
3 years ago
Give 5 examples of physical properties that could be used to describe a solid
ycow [4]
1.) appearance
2.)texture
3.)color
4.)melting point
5.)odor
8 0
3 years ago
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