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3 years ago

Dont skip need help ASAP and pls hurry will give 10 pts + brainliest

Physics

2 answers:

forsale [732]3 years ago

8 0

C, believe you need to test others, can’t just test one thing with eachother several times

Gemiola [76]3 years ago

3 0

Answer:

Explanation:

In order for an experiment to be valid need to be conducted with several tested variables

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Einstein's theory of relativity is currently the best explanation of gravity. What is likely to happen to this theory in the fut

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I believe the answer is C

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1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.

NikAS [45]

Answer:

109.6 cm , - 1.74 , real

1.5

Explanation:

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{40} = \frac{1}{63} + \frac{1}{d}$

d = 109.6 cm

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{-109.6}{63}$

m = - 1.74

The image is real

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}$

a = 10

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{- (- (a +5))}{a}$

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3 years ago

Eukaryotic cells can be unicellular and multicellular true or false

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Yes, it can be unicellular and multicellular

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3 years ago

If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is t

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Answer:

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Explanation:

Given;

first magnetic field at first distance, B₁ = 2.50 mT

first distance, r₁ = 12.6 cm = 0.126 m

Second magnetic field at Second distance, B₂ = ?

Second distance, r₂ = ?

Magnetic field for a straight wire is given as;

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

Where:

μ is permeability

B is magnetic field

I is current flowing in the wire

r distance to the wire

$Let \ \frac{\mu I}{2\pi} \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT$

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

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