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Lena [83]
3 years ago
13

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Calculus proof of second equation of motion ~ ​
Physics
1 answer:
kozerog [31]3 years ago
5 0

Newton's second equation of motion :-

S=ut+1/2at^2 [where, u is the initial velocity, a is the acceleration and t is the time interval]

This Equation simply finds a relation between distance travelled by a particle (classically) under uniform acceleration.

So let's see what pieces of information (bundles of equations) do we have with us, initially.

We have, a very primary equation with us,

dS/dt = v… (I)

(Considering motion in a straight line only)

And we also have the equation

dv/dt = a…(II)

Simply replacing the v in eqn (II) by eqn (I), we find

d2S/dt^2 = a…(III)

This is what we need to solve. It's easy.

You know,

d2S/dt^2 = d/dt(dS/dt) = a

⟹ dS/dt = ∫adt = at+c1

Since, dS/dt is the velocity of the particle,

Therefore, at t = 0, dS/dt|t = 0 = u

⟹ u = a∗0 + c1 = c1

⟹ c1 = u

Therefore, dS/dt = u + at

Thus, S = ∫(udt + atdt)

⟹ S = ut + 1/2at^2 +c^2

If say, the particle is already having a displacement S0 the moment you start measuring it's motion. Then, at t = 0, S = S0

This makes S = S0 +ut + 1/2at^2

Since, in most of the practical cases, we start measuring a motion when the particle starts displacing (i.e., when S0=0 ),

We get

S = ut + 1/2at^2

Hope it helps :)

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Answer:Thus, if each of the charges were reduced by one-half, the repulsion would be reduced to one-quarter of its former value. Also In electrostatics, the electrical force between two charged objects is inversely related to the distance of separation between the two objects. ... And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects.

Hope this helps have a awesome night/day❤️✨

Explanation:

Nothing, until those two objects physically touch each other; contact aligns polarity among the now single shared mass.

(Your question never states if both objects are unique or similar polar charges, so I just assumed they were both neutral objects existing within an electric field.)

So a better question would then be, what is gravity’s relationship with an electric field?

You could solve this with the following: confine the electric field’s volume to a set variable (never increasing nor decreasing in size or shape); density is variable and easily definable. This creates the limit to build upon. This density has to be fluid and has electron mass (full of electrons at rest mass, so with substance but no movement). Within, create a closed system (the hard part in this equation; outside interference like ambient light will eschew results) where each variable of kinetic energy then is accounted for or measurable (including heat and light, and the physical movement of the two objects)

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4 0
3 years ago
To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the d
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The question is incomplete. Here is the complete question.

To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity when the intensity as measured in W/m² changes by a multiplicative factor. The number of decibels increase by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is

\beta=10log(\frac{I}{I_{0}} )dB,

where I_{0} is a reference intensity. for sound waves, I_{0} is taken to be 10^{-12} W/m^{2}. Note that log refers to the logarithm to the base 10.

Part A: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity, i.e. I=10I_{0}? Express the sound intensity numerically to the nearest integer.

Part B: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity, i.e. I=100I_{0}? Express the sound intensity numerically to the nearest integer.

Part C: Calculate the change in decibels (\Delta \beta_{2},\Delta \beta_{4} and \Delta \beta_{8}) corresponding to f = 2, f = 4 and f = 8. Give your answer, separated by commas, to the nearest integer -- this will give an accuracy of 20%, which is good enough for sound.

Answer and Explanation: Using the formula for sound intensity level:

A) I=10I_{0}

\beta=10log(\frac{10I_{0}}{I_{0}} )

\beta=10log(10 )

β = 10

<u>The sound Intensity level with intensity 10x is </u><u>10dB</u>.

B) I=100I_{0}

\beta=10log(\frac{100I_{0}}{I_{0}} )

\beta=10log(100)

β = 20

<u>With intensity 100x, level is </u><u>20dB</u>.

C) To calculate the change, take the f to be the factor of increase:

For \Delta \beta_{2}:

I=2I_{0}

\beta=10log(\frac{2I_{0}}{I_{0}} )

\beta=10log(2)

β = 3

For \Delta \beta_{4}:

I=4I_{0}

\beta=10log(\frac{4I_{0}}{I_{0}} )

\beta=10log(4)

β = 6

For \Delta \beta_{8}:

I=8I_{0}

\beta=10log(\frac{8I_{0}}{I_{0}} )

β = 9

Change is

\Delta \beta_{2},\Delta \beta_{4}, \Delta \beta_{8} = 3,6,9 dB

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Answer:

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Explanation:

We know that weight of water displaced= buoyant force=weight of man

Now volume of water displaced v=3.38\times 10^{-2}+6.42\times 10^{-2}=9.8\times 10^{-2}m^3

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So weight of water displaced =9.8\times 10^{-2}\times 1000=98kg

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So density of jacket =\frac{mass}{volume}=\frac{23}{3.38\times 10^{-2}}=680.4733kg/m^3

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Snezhnost [94]

Answer:

The current in the motor in this case is 13.2 A

Explanation:

Given:

Resistance R = 5 Ω

Emf E = 120 V

Induced emf E _{induced} = 108 V

When motor run at half speed due to load increased then induced emf is also reduced to half of its value

So new induced emf in our case is given by,

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Where V = E - E_{Induced }

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