The prefix for oxygen in As2O5 is PENTA.
Penta is used to show that the number of oxygen atoms present in the compound is five. Penta means five, while mono mean one, di means two, tri means three and tetra mans four and so on. The chemical name for the given compound is Arsenic pentoxide.
Units of impulse: N • s, kg • meters per second
Explanation:
Impulse is defined in two ways:
1)
Impulse is defined as the product between the force exerted in a collision and the duration of the collision:
![I=F\Delta t](https://tex.z-dn.net/?f=I%3DF%5CDelta%20t)
where
F is the force
is the time interval
Since the force is measured in Newtons (N) and the time is measured in seconds (s), the units for the impulse are
![[I] = [N][s]](https://tex.z-dn.net/?f=%5BI%5D%20%3D%20%5BN%5D%5Bs%5D)
So,
N • s
2)
Impulse is also defined as the change in momentum experienced by an object:
![I=\Delta p](https://tex.z-dn.net/?f=I%3D%5CDelta%20p)
where the change in momentum is given by
![\Delta p = m\Delta v](https://tex.z-dn.net/?f=%5CDelta%20p%20%3D%20m%5CDelta%20v)
where m is the mass and
is the change in velocity.
The mass is measured in kilograms (kg) while the change in velocity is measured in metres per second (m/s), therefore the units for impulse are
![[I]=[kg][m/s]](https://tex.z-dn.net/?f=%5BI%5D%3D%5Bkg%5D%5Bm%2Fs%5D)
so,
kg • meters per second
Learn more about impulse:
brainly.com/question/9484203
#LearnwithBrainly
Acceleration of cheetah (a) = 4m/s²
time = 10s
initial velocity(u) = 0
final velocity = v
distance travelled = s
v = u +at = 0 + 10×4 = 40m/s
s = (v²-u²)/2a = 40²/(2×4) = 1600/8 = 200m
Answer:
C. ![\frac{3F}{8}](https://tex.z-dn.net/?f=%5Cfrac%7B3F%7D%7B8%7D)
Explanation:
Let initial charges on both spheres be,![q](https://tex.z-dn.net/?f=q)
![F=\frac{Kq^2}{d^2} \ \ \ \ \ \ \ \ \ \ \_i](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BKq%5E2%7D%7Bd%5E2%7D%20%20%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%20%5C%20%5C_i)
When the sphere C is touched by A, the final charges on both will be,![\frac{q}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7B2%7D)
#Now, when C is touched by B, the final charges on both of them will be:
![q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\](https://tex.z-dn.net/?f=q_c%3Dq_d%3D%5Cfrac%7Bq%2F2%2Bq%7D%7B2%7D%5C%5C%5C%5C%3D%5Cfrac%7B3q%7D%7B4%7D%5C%5C)
Now the force between A and B is calculated as:
![F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}](https://tex.z-dn.net/?f=F%5Cprime%3D%5Cfrac%7Bk%5Ctimes%5Cfrac%7Bq%7D%7B2%7D%5Ctimes%20%5Cfrac%7B3q%7D%7B4%7D%7D%7Bd%5E2%7D%5C%5CF%5Cprime%3D%5Cfrac%7B3F%7D%7B8%7D)
Hence the electrostatic force becomes 3F/8