Answer:
The carbon atom has unique properties that allow it to form covalent bonds to as many as four different atoms, making this versatile element ideal to serve as the basic structural component, or “backbone,” of the macromolecules.
Answer: -33.3 * 10^9 C/m^2( nC/m^2)
Explanation: In order to solve this problem we have to use the gaussian law, the we have:
Eoutside =0 so teh Q inside==
the Q inside= 4.6 nC/m*L + σ *2*π*b*L where L is the large of the Gaussian surface and b the radius of the shell.
Then we simplify and get
σ= -4.6/(2*π*b)= -33.3 nC/m^2
Answer:
Explanation:
Given
Initially Reading on the odometer is
Final reading on the odometer is
Time taken is
average velocity
Thus the average velocity of mail truck is
Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr