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fenix001 [56]
3 years ago
7

When it is brought near a neutral object, a charged object can _________ a charge in a neutral object.A) induce, B) cancel,C) ne

utralize, D) add or subtract
Physics
1 answer:
Bess [88]3 years ago
3 0
The answer is a induce.
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The density of a substance equals its mass divided by its volume. Talia listed the density of some common materials at 20 °C.
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mercury

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Three long, straight wires are carrying currents that have the same magnitude. In C the current is opposite to that in A and B.
Nadusha1986 [10]

Answer:

(b) B

Explanation:

The direction of force on a current carrying wire in a magnetic field can be found using the right hand rule, which states that-"stretch the thumb in the direction of the current, and point the fingers in the direction of magnetic field. The direction of palm will then give the direction of force on the wire

On wire B the forces due to A and C act in the same direction and so strengthen each other. they get added up because the forces act in the same direction.

on wires A and C the forces (due to B and C and A and B

respectively) act in opposite directions and therefore tend to cancel out.

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3 years ago
Please help :<
Artemon [7]

Answer: methyl

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7 0
2 years ago
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta l
butalik [34]

Answer:

a)   f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} } , b)   Δf = 2 f₀ \frac{v}{c}

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

                    f ’= fo\frac{c+v}{c}

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

                   f ’’ = f’ \frac{c}{ c-v}

where c represents the sound velocity in stationary blood

therefore the received frequency is

                 f ’’ = f₀   \frac{c}{c-v}

let's simplify the expression

                f ’’ = f₀ \frac{c+v}{c-v}

                f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} }

         

b) At the low speed limit v <c, we can expand the quantity

                 (1 -x)ⁿ = 1 - x + n (n-1) x² + ...

                 ( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}

 

                f ’’ = fo ( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )

                f ’’ = fo ( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )

leave the linear term

               f ’’ = f₀ + f₀ 2\frac{v}{c}

the sound difference

               f ’’ -f₀ = 2f₀ v/c

               Δf = 2 f₀ \frac{v}{c}

4 0
2 years ago
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