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nekit [7.7K]
3 years ago
5

If the vertical axis and time on the horizontal axis if the speed is steadily increasing could the speef line ever become perfec

tly vertical
Physics
1 answer:
Phantasy [73]3 years ago
8 0

A graph of real speed can have a section that's as steep as you want,
but it can never be a perfectly vertical section.

Any vertical line on a graph, even it it's only a tiny tiny section, means
that at that moment in time, the speed had many different values.

It also means that the speed took no time to change from one value to
another, and THAT would mean infinite acceleration.

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Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13
lyudmila [28]

Answer:

y = 0.14 Cos\left ( 2.512x-34.66t \right )

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

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The general equation of the transverse harmonic wave which is travelling right is given by

y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right  )

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

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Ф = π/2

So, the equation is

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If we put a convex lens made by crown glass <br>in Carbondia sulphide what will happen?​
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a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe
Nikitich [7]

Answer:

(A) The spring constant is <u>98 N/m.</u>

(B) The period of oscillation is <u>0.635 s.</u>

Explanation:

Given:

Mass of the body is, m=1\ kg

Extension length of the spring is, x=10\ cm=0.1\ m

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

T=2\pi\sqrt{\frac{m}{k}}

Plug in the given values and solve for period 'T'. This gives,

T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s

Therefore, the period of oscillation is 0.635 s.

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