Acceleration in m/s^2 = 2/10 = 0.2 m/s^2
Answer:
a) Therefore 2.6km is greater than 2.57km.
Statement A is greater than statement B.
b) Therefore 5.7km is equal to 5.7km
Statement A is equal to statement B
Explanation:
a) Statement A : 2.567km to two significant figures.
2.567km 2. S.F = 2.6km
Statement B : 2.567km to three significant figures.
2.567km 3 S.F = 2.57km
Therefore 2.6km is greater than 2.57km.
Statement A is greater than statement B.
b) statement A: (2.567 km + 3.146km) to 2 S.F
(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km
Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).
2.567km to 2 S.F = 2.6km
3.146km to 2 S.F = 3.1km
2.6km + 3.1km = 5.7km
Therefore 5.7km is equal to 5.7km
Statement A is equal to statement B
Attenuation is the correct answer.
Answer:
I will say that the the potential energy will be at its maximum.
Explanation:
potential energy deals with gravity and gravity deals with height, so when a object is in its maximum height it will have the maximum potential energy.
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance 
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q =
= 
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is .
