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OverLord2011 [107]
2 years ago
7

How can you return a negatively or positively charged object back to its neutral state?

Physics
1 answer:
Dmitriy789 [7]2 years ago
3 0

\huge\fbox\orange{A} \huge\fbox\red{N}\huge\fbox\blue{S}\huge\fbox\green{W}\huge\fbox\gray{E}\huge\fbox\purple{R}

\huge\underline\mathtt\colorbox{cyan}{For Positively Charged:}

By gaining electrons from the ground, the object will have a balance of charge and therefore be neutral. Grounding is the grounding of a positively charged object and involves the transfer of electrons from the ground into the object.

\huge\underline\mathtt\colorbox{red}{For Negatively Charged:}

If it is to have its charge removed, then it will have to lose its excess electrons. Once the excess electrons are removed from the object, there will be equal numbers of protons and electrons within the object and it will have a balance of charge.

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If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star
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1) In the first case, the correct answer is
<span>A.Wavelengths measured would match the actual wavelengths emitted.
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2) In this second case, the correct answer is
</span><span>A.Wavelengths measured would be shorter than the actual wavelengths emitted.
</span>in fact, since the stars in this case are moving towards the Earth, then apparent frequency of their emitted light will be larger than the actual frequency, because of the Doppler effect, according to the formula:
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The average kinetic energy of the particles in an object is directly proportional to its A) heat. B) volume. C) temperature. D)
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Two blocks of masses M 1 and M 2 are connected by a massless string that passes over a massless pulley as shown in the figure. M
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The mass M1 is 7.8 kg

Explanation:

Block M1 is hanging on the string while block M2 is on the frictionless ramp.

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M_1 g - T = M_1 a

where

M_1 is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

a is the acceleration of the system

- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, M_2 g sin \theta. So the equation of motion is

T-M_2 g sin \theta = M_2 a

where

M_2 = 13.5 kg is the mass of the 2nd block

\theta=35.5^{\circ} is the angle of the ramp

In order for the two blocks to be in equilibrium, the acceleration must be zero:

a=0

So the two equations become:

M_1 g - T=0\\T-M_2 g sin \theta = 0

Isolating T from the 1st equation,

T=M_1 g

And substituting into the 2nd equation, we can find the value of the mass M_1:

M_1 g - M_2 g sin \theta = 0\\M_1 = M_2 sin \theta = (13.5)(sin 35.5^{\circ})=7.8 kg

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