Total distance = 36500 m
The average velocity = 19.73 m/s
<h3>Further explanation</h3>
Given
vo=initial velocity=0(from rest)
a=acceleration= 1 m/s²
t₁ = 20 s
t₂ = 0.5 hr = 1800 s
t₃= 30 s
Required
Total distance
Solution
State 1 : acceleration
![\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m](https://tex.z-dn.net/?f=%5Ctt%20d%3Dvo.t%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5C%5Cd%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%201%5Ctimes%2020%5E2%5Crightarrow%20vo%3D0%5C%5C%5C%5Cd%3D200~m)
![\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s](https://tex.z-dn.net/?f=%5Ctt%20vt%3Dvo%2Bat%5C%5C%5C%5Cvt%3Dat%5Crightarrow%20vo%3D0%5C%5C%5C%5Cvt%3D1%5Ctimes%2020%5C%5C%5C%5Cvt%3D20~m%2Fs)
State 2 : constant speed
![\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m](https://tex.z-dn.net/?f=%5Ctt%20d%3Dv%5Ctimes%20t%5C%5C%5C%5Cd%3D20%5Ctimes%201800%5C%5C%5C%5Cd%3D36000~m)
State 3 : deceleration
![\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)](https://tex.z-dn.net/?f=%5Ctt%20vt%3Dvo%2Bat%5Crightarrow%20vt%3D0%28stop%29%5C%5C%5C%5Cvo%3D-at%5C%5C%5C%5C20%3D-a.30~s%5C%5C%5C%5Ca%3D-%5Cdfrac%7B2%7D%7B3%7Dm%2Fs%5E2%28negative%3Ddeceleration%29)
![\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m](https://tex.z-dn.net/?f=%5Ctt%20d%3Dvot%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5C%5Cd%3D20.30-%5Cdfrac%7B1%7D%7B2%7D.%5Cdfrac%7B2%7D%7B3%7D.30%5E2%5C%5C%5C%5Cd%3D300~m)
Total distance : state 1+ state 2+state 3
![\tt 200 + 36000 + 300=36500~m](https://tex.z-dn.net/?f=%5Ctt%20200%20%2B%2036000%20%2B%20300%3D36500~m)
the average velocity = total distance : total time
![\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s](https://tex.z-dn.net/?f=%5Ctt%20avg~velocity%3D%5Cdfrac%7B36500%7D%7B20~s%2B1800~s%2B30~s%7D%3D19.73~m%2Fs)
<span>The radio frequency characteristic that best determines the range of a 2.4 GHz ism signal is the wavelength.
This frequency can be used in WiFi and can reach up to 46 meters when indoors and about 92 meters when outdoors.
</span><span>
</span>
The concept of power is given by the relationship between intensity and area, that is to say that power is defined as
![P = A*I](https://tex.z-dn.net/?f=P%20%3D%20A%2AI)
Our values are given under the condition of,
![r_1 = 18m](https://tex.z-dn.net/?f=r_1%20%3D%2018m)
![r_2 = m](https://tex.z-dn.net/?f=r_2%20%3D%20m)
The power is proportional to the Area, and in turn, we know that the Area of a circle is the product between
times the radius squared, therefore the power is proportional to the radius squared.
![\text{Power} \propto r^2](https://tex.z-dn.net/?f=%5Ctext%7BPower%7D%20%5Cpropto%20r%5E2)
For both panels we would have to
![\frac{\text{Power by panel 1}}{\text{Power by panel 2}} = \frac{r_1^2}{r_2^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BPower%20by%20panel%201%7D%7D%7B%5Ctext%7BPower%20by%20panel%202%7D%7D%20%3D%20%5Cfrac%7Br_1%5E2%7D%7Br_2%5E2%7D)
![\frac{P_1}{P_2} = (\frac{18}{6})^2](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7BP_2%7D%20%3D%20%28%5Cfrac%7B18%7D%7B6%7D%29%5E2)
![\frac{P_1}{P_2} = 9](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7BP_2%7D%20%3D%209)
Therefore the correct option is option C.9
Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg