ANSWER
![6.84\times 10^9](https://tex.z-dn.net/?f=6.84%5Ctimes%2010%5E9)
EXPLANATION.
To write in scientific notation means , we write in form,
![k\times 10^n](https://tex.z-dn.net/?f=k%5Ctimes%2010%5En)
Where n is an integer and
![1\le k\:](https://tex.z-dn.net/?f=1%5Cle%20k%5C%3A%3C%5C%3A10)
The given number is,
![6,840,000,000](https://tex.z-dn.net/?f=6%2C840%2C000%2C000)
We move the decimal point backward to obtain our k, the n here is 9, because we moved backward 9 times.
This implies that,
![6,840,000,000=6.84\times 10^9](https://tex.z-dn.net/?f=6%2C840%2C000%2C000%3D6.84%5Ctimes%2010%5E9%20)
The correct answer is D
Answer:
9.81 × 10 = 98.1 meters
vertical displacement is s=1/2 at^2 + vt
initial vertical velocity is 0 so s=1/2 at^2
a in this instance is gravitational acceleration so 60m= 1/2 (9.81)t^2
solve for t, t = 3.497s. //I corrected this answer as just now I misread horizontal as vertical.
Answer:
gravitational potential energy
Answer:
Dry-strength additives, or dry-strengthening agents, are chemicals that improve paper strength normal conditions. These improve the paper's compression strength, bursting strength, tensile breaking strength, and delamination resistance. Typical chemicals used include cationic starch and polyacrylamide derivatives.
Explanation:
It is known that relation between torque and angular acceleration is as follows.
![\tau = I \times \alpha](https://tex.z-dn.net/?f=%5Ctau%20%3D%20I%20%5Ctimes%20%5Calpha)
and, I = ![\sum mr^{2}](https://tex.z-dn.net/?f=%5Csum%20mr%5E%7B2%7D)
So, ![I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}](https://tex.z-dn.net/?f=I_%7B1%7D%20%3D%202%20kg%20%5Ctimes%20%281%20m%29%5E%7B2%7D%20%2B%202%20kg%20%5Ctimes%20%281%20m%29%5E%7B2%7D)
= 4 ![kg m^{2}](https://tex.z-dn.net/?f=kg%20m%5E%7B2%7D)
![\tau_{1} = 4 kg m^{2} \times \alpha_{1}](https://tex.z-dn.net/?f=%5Ctau_%7B1%7D%20%3D%204%20kg%20m%5E%7B2%7D%20%5Ctimes%20%5Calpha_%7B1%7D)
![\tau_{2} = I_{2} \alpha_{2}](https://tex.z-dn.net/?f=%5Ctau_%7B2%7D%20%3D%20I_%7B2%7D%20%5Calpha_%7B2%7D)
So, ![I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}](https://tex.z-dn.net/?f=I_%7B2%7D%20%3D%202%20kg%20%5Ctimes%20%280.5%20m%29%5E%7B2%7D%20%2B%202%20kg%20%5Ctimes%20%280.5%20m%29%5E%7B2%7D)
= 1 ![kg m^{2}](https://tex.z-dn.net/?f=kg%20m%5E%7B2%7D)
as ![\tau_{2} = I_{2} \alpha_{2}](https://tex.z-dn.net/?f=%5Ctau_%7B2%7D%20%3D%20I_%7B2%7D%20%5Calpha_%7B2%7D)
=
Hence, ![\tau_{1} = \tau_{2}](https://tex.z-dn.net/?f=%5Ctau_%7B1%7D%20%3D%20%5Ctau_%7B2%7D)
![4 \alpha_{1} = \alpha_{2}](https://tex.z-dn.net/?f=4%20%5Calpha_%7B1%7D%20%3D%20%5Calpha_%7B2%7D)
Thus, we can conclude that the new rotation is
times that of the first rotation rate.