Vf = Vi + at
Vf = 0 + 5.4•28
= 151.2m/s..
not sure if its right
The correct answer is - C. Crustal plates, crustal plates.
The movement of the crustal plates of the Earth is the reason why the geologic processes are occurring on our planet. With their movement, the crustal plates manage to create lot of pressure, open up gaps for the magma from the mantle, as well as make adjustments deeper into the ground.
Because of the adjustments deep into the ground, or rather the crust, lot of force is released, manifested as strong and very quick vibrations, better known as earthquakes.
The gaps that are opening between the plates let the magma reach the surface, thus enabling the volcanic activities on the planet.
The pressure from the plates' collision makes the land lift up slowly, and over time a mountain range starts to form because of the continuous lifting up of the area.
The electrostatic force is directly proportional to the product of the charges, by Coulomb's law.
F α Qq
If the charges are now half the initial charges:
<span>F α (1/2)Q *(1/2)q
</span>
F α (1/4)Q<span>q
The new force when the charges are each halved is (1/4) the first initial force experienced at full charge.</span>
Answer:
The electric charge, q (in coulomb units) = 5004 C
Given:
The charge stored as printed on NiMH battery, q = 1390 mAh
Solution:
To express the amount of electric charge printed on the battery in milli-ampere-hour (mAh) in coulomb, we will do simple conversion of milli amperes in ampere and hours in seconds:
1 mA = 
1 hour = 
Also, we know that the rate of flow of charge is electric current, I:
I = 
⇒ q = [tex]I\times t[tex] (1)
where
q = electric charge
I = current
t = time taken for flow of current
Using eqn(1), we get:
q = [tex]1390\times 10^{-3}\times 60\times 60[tex]
q = 5004 A-s = 5004 C
The area of the plates must have is(A)= 9.91×10⁷ m²
<h3 /><h3>How can we calculate the value of a area of a capacitor?</h3>
To calculate the the value of a area of the plates of a capacitor, we are using the formula,
C=
Or, A= 
Here we are given,
C= The desired capacitance of a capacitor.
= 0.23F
d=distance of separation between the plates.
=3.8mm= 0.0038m.
= permittivity of the vacuum.
=8.854×10⁻¹²F/m
We have to calculate the area of the plates must have = A m².
Now we put the known values in the above equation, we can get
A= 
Or, A=
Or, A= 9.91×10⁷ m²
From the above calculation, we can conclude that the area of the plates must have is(A)= 9.91×10⁷m²
Learn more about Capacitor:
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