Answer:
0.60 N, towards the centre of the circle
Explanation:
The tension in the string acts as centripetal force to keep the ball in uniform circular motion. So we can write:
(1)
where
T is the tension
m = 0.015 kg is the mass of the ball
is the angular speed
r = 0.50 m is the radius of the circle
We know that the period of the ball is T = 0.70 s, so we can find the angular speed:
![\omega=\frac{2\pi}{T}=\frac{2\pi}{0.70 s}=8.98 rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D%3D%5Cfrac%7B2%5Cpi%7D%7B0.70%20s%7D%3D8.98%20rad%2Fs)
And by substituting into (1), we find the tension in the string:
![T=(0.015 kg)(8.98 rad/s)^2(0.50 m)=0.60 N](https://tex.z-dn.net/?f=T%3D%280.015%20kg%29%288.98%20rad%2Fs%29%5E2%280.50%20m%29%3D0.60%20N)
And in an uniform circular motion, the centripetal force always points towards the centre of the circle, so in this case the tension points towards the centre of the circle.
Answer:
![\Delta s = 1775.510\,m](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%201775.510%5C%2Cm)
Explanation:
The minimum distance for takeoff is:
![\Delta s = \Delta s_{1} + \Delta s_{2}](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%20%5CDelta%20s_%7B1%7D%20%2B%20%5CDelta%20s_%7B2%7D)
![\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a} + v_{f}\cdot \Delta t](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%20%5Cfrac%7Bv_%7Bf%7D%5E%7B2%7D-v_%7Bo%7D%5E%7B2%7D%7D%7B2%5Ccdot%20a%7D%20%2B%20v_%7Bf%7D%5Ccdot%20%5CDelta%20t)
![\Delta s = \frac{(100\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}}{2\cdot (3.92\,\frac{m}{s^{2}})}+(100\,\frac{m}{s} )\cdot (5\,s)](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%20%5Cfrac%7B%28100%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D-%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D%7D%7B2%5Ccdot%20%283.92%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%29%7D%2B%28100%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5Ccdot%20%285%5C%2Cs%29)
![\Delta s = 1775.510\,m](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%201775.510%5C%2Cm)
![\vec r(t)=bt^2\,\vec\imath+ct^3\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20r%28t%29%3Dbt%5E2%5C%2C%5Cvec%5Cimath%2Bct%5E3%5C%2C%5Cvec%5Cjmath)
The velocity at time
is
![\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=2bt\,\vec\imath+3ct^2\,\vec\jmath](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5Cvec%20r%28t%29%7D%7B%5Cmathrm%20dt%7D%3D2bt%5C%2C%5Cvec%5Cimath%2B3ct%5E2%5C%2C%5Cvec%5Cjmath)
Take two vectors that point in the positive
and positive
directions, such as
and
. The dot products of the velocity vector with
and
are
![\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\imath=2bt=\sqrt{4b^2t^2+9c^2t^4}\cos\theta](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5Cvec%20r%28t%29%7D%7B%5Cmathrm%20dt%7D%5Ccdot%5Cvec%5Cimath%3D2bt%3D%5Csqrt%7B4b%5E2t%5E2%2B9c%5E2t%5E4%7D%5Ccos%5Ctheta)
and
![\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\jmath=3ct^2=\sqrt{4b^2t^2+9c^2t^4}\cos\theta](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5Cvec%20r%28t%29%7D%7B%5Cmathrm%20dt%7D%5Ccdot%5Cvec%5Cjmath%3D3ct%5E2%3D%5Csqrt%7B4b%5E2t%5E2%2B9c%5E2t%5E4%7D%5Ccos%5Ctheta)
We want the angles between these vectors to be 45º, for which we have
. So
![\begin{cases}2\sqrt2\,bt=\sqrt{4b^2t^2+9c^2t^4}\\3\sqrt2\,ct^2=\sqrt{4b^2t^2+9c^2t^4}\end{cases}\implies3\sqrt2\,ct^2-2\sqrt2\,bt=0](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D2%5Csqrt2%5C%2Cbt%3D%5Csqrt%7B4b%5E2t%5E2%2B9c%5E2t%5E4%7D%5C%5C3%5Csqrt2%5C%2Cct%5E2%3D%5Csqrt%7B4b%5E2t%5E2%2B9c%5E2t%5E4%7D%5Cend%7Bcases%7D%5Cimplies3%5Csqrt2%5C%2Cct%5E2-2%5Csqrt2%5C%2Cbt%3D0)
![\implies t(3ct-2b)=0](https://tex.z-dn.net/?f=%5Cimplies%20t%283ct-2b%29%3D0)
![\implies t=0\text{ or }t=\dfrac{2b}{3c}](https://tex.z-dn.net/?f=%5Cimplies%20t%3D0%5Ctext%7B%20or%20%7Dt%3D%5Cdfrac%7B2b%7D%7B3c%7D)
When
, the velocity vector is equal to the zero vector, which technically has no direction/doesn't make an angle with any other vector. So the only time this happens is for
![\boxed{t=\dfrac{2b}{3c}}](https://tex.z-dn.net/?f=%5Cboxed%7Bt%3D%5Cdfrac%7B2b%7D%7B3c%7D%7D)
Answer:
1.6 ns
Explanation:
Given data :
delay in 10-bit adder = 1ns
setup time = 0.3 ns
hold time = 0.1 ns
propagation delay = 0.2 ns
<u>calculate the minimum cycle time </u>
∑ delay + setup time + hold time + propagation delay
= 1 + 0.3 + 0.1 + 0.2
= 1.6 ns
D is your answer.
Just like with a magent, opposites attract.
-Seth