Answer:
v = 3.08 m/s.
Explanation:
Given,
- Initial velocity of the block = u = 43 cm/s = 0.43 m/s
- mass of the block = m = 1.40 kg
- spring constant = k = 15.5 N/m.
- x = 0.650 m
At x = 0.650 m
Let 'a' be the acceleration of the block at x.
Total force due to the spring force on the blcok at x= kx
![\therefore F_s\ =\ kx\\\Rightarrow ma\ =\ kx\\\Rightarrow a\ =\ \dfrac{kx}{m}\\\Rightarrow a\ =\ \dfrac{15.5\times 0.650}{1.40}\\\Rightarrow a\ =\ 7.19\ m/s^2](https://tex.z-dn.net/?f=%5Ctherefore%20F_s%5C%20%3D%5C%20kx%5C%5C%5CRightarrow%20ma%5C%20%3D%5C%20kx%5C%5C%5CRightarrow%20a%5C%20%3D%5C%20%5Cdfrac%7Bkx%7D%7Bm%7D%5C%5C%5CRightarrow%20a%5C%20%3D%5C%20%5Cdfrac%7B15.5%5Ctimes%200.650%7D%7B1.40%7D%5C%5C%5CRightarrow%20a%5C%20%3D%5C%207.19%5C%20m%2Fs%5E2)
Hence the acceleration of the block at x is 7.19\ m/s^2.
Let 'v' be the velocity of the block at x.
From the kinematics,
![v^2\ =\ u^2\ +\ 2ax\\\Rightarrow v\ =\ \sqrt{u^2\ +\ 2ax}\\\Rightarrow v\ =\ \sqrt{0.43^2\ +\ 2\times 7.19\times 0.650}\\\Rightarrow v\ =\ 3.08\ m/s.](https://tex.z-dn.net/?f=v%5E2%5C%20%3D%5C%20u%5E2%5C%20%2B%5C%202ax%5C%5C%5CRightarrow%20v%5C%20%3D%5C%20%5Csqrt%7Bu%5E2%5C%20%2B%5C%202ax%7D%5C%5C%5CRightarrow%20v%5C%20%3D%5C%20%5Csqrt%7B0.43%5E2%5C%20%2B%5C%202%5Ctimes%207.19%5Ctimes%200.650%7D%5C%5C%5CRightarrow%20v%5C%20%3D%5C%203.08%5C%20m%2Fs.)
Hence at the compression x = 0.650 m the velocity of the block is 3.08 m/s.
Answer:
it is very large and invisible
Answer:
When light is reflected, the "outgoing" ray is also called the reflected ray.
Explanation:
when an light ray is incident on a boundary of two mediums it an reflect or refract.
reflected ray is the part of the ray that gets back with same mediu that it was coming from. angle of reflection and incident are equal . angle of reflection and incident are the angle made by the reflected and incident ray with the normal.
total internal reflection occurs when the incident angle is greater than the critical angle of pair of given medium and when light travels from denser to rarer medium . in total internal reflection ray gets reflected with no loss at all.
in refracted ray the the light ray passes to the other medium angle of incidence and refraction are not same.
Answer:
The constant value is ![k\approx 0.17\,m/s^4](https://tex.z-dn.net/?f=k%5Capprox%200.17%5C%2Cm%2Fs%5E4)
The net displacement is ![D=53.64\, m](https://tex.z-dn.net/?f=D%3D53.64%5C%2C%20m)
Explanation:
If the acceleration as a function of time is given
then, first of all, knowing that the units of acceleration should be
we should have
where
stands for The dimension of k (these are just the units of k in a less formal way of saying it.
On the other hand we have only information about the velocity, but we only have the acceleration function, it turns out we can integrate the expression of acceleration in order to obtain the velocity as a function of time:
where
as a constant of integration which should have units of
in order to be consistent with the fact that it is a velocity function, it is therefore natural to think of
as the initial velocity of the the particle.
Let's now get our hands dirty by integrating ![a(t)](https://tex.z-dn.net/?f=a%28t%29)
.
By having the velocity as a function of time we can now use the conditions given at t=0 and t=6.
At t=0 we have:
![v(0)=-k\frac{0^3}{3}+v_0=12\implies v_0=12\, m/s](https://tex.z-dn.net/?f=v%280%29%3D-k%5Cfrac%7B0%5E3%7D%7B3%7D%2Bv_0%3D12%5Cimplies%20v_0%3D12%5C%2C%20m%2Fs)
At t=6 the particle start reversing direction, that means at that very instant it velocity should be zero in order to start traveling the other way. This can only mean the following
.
We have a full description now of the acceleration and the velocity function. In order to get the net displacement we need to integrate the velocity function
![x(t)=\int v(t)=-\frac{k}{3}\int 3 dt+\int v_0 dt=-\frac{k}{12}t^4+v_0\cdot t+x_0](https://tex.z-dn.net/?f=x%28t%29%3D%5Cint%20v%28t%29%3D-%5Cfrac%7Bk%7D%7B3%7D%5Cint%203%20dt%2B%5Cint%20v_0%20dt%3D-%5Cfrac%7Bk%7D%7B12%7Dt%5E4%2Bv_0%5Ccdot%20t%2Bx_0%20)
Where
is the initial displacement. If we subtract
on both sides we get the net displacement or distance traveled
![x(t)-x_0=D=-\frac{kt^4}{12}+12t](https://tex.z-dn.net/?f=x%28t%29-x_0%3DD%3D-%5Cfrac%7Bkt%5E4%7D%7B12%7D%2B12t)
Plugging the value of 6 above gives us the net displacement
.