Answer: Mid-ocean ridges are geologically important because they occur along the kind of plate boundary where new ocean floor is created as the plates spread apart. Thus the mid-ocean ridge is also known as a "spreading center" or a "divergent plate boundary." The plates spread apart at rates of 1 cm to 20 cm per year.
Answer:
<u>Question 2</u>
<u>Part (a)</u>
Chlorine: type of compound = chloride
Oxygen: type of compound = oxide
<u>Part (b)</u>
The iron reacts with water and oxygen to form rust.
A water molecule is made up of two hydrogen atoms joined to one oxygen atom: Di-hydrogen oxide.
<u>Question 3</u>
This circuit is in parallel.
The current in a parallel circuit splits into different branches then combines again before it goes back into the supply.
We are told that A₁ = 0.8 A
As the lamps have <u>equal resistance</u>, the current splits equally:
A₂ = 0.4 A
A₃ = 0.4 A
Then combines again:
A₄ = 0.8 A
Answer:
a = 3.125 [m/s^2]
Explanation:
In order to solve this problem, we must use the following equation of kinematics. But first, we have to convert the speed of 90 [km/h] to meters per second.
where:
Vf = final velocity = 25 [m/s]
Vi = initial velocity = 0
a = acceleration [m/s^2]
t = time = 8 [s]
The initial speed is zero as the bus starts to koverse from rest. The positive sign of the equation means that the bus increases its speed.
25 = 0 + a*8
a = 3.125 [m/s^2]
Answer: The person will still have a mass of 90kg on Mars
Explanation: The Truth is, the mass of a body remains constant from place to place. It is the weight which is equal to {mass of body * acceleration due to gravity{g}} that varies from place to place since it is dependent on {g}.
In this case the person will have a Weight of 90*9.8 = 882N on Earth.
{ "g" on Earth is 9.8m/s²}
And a Weight of 90*3.3 = 297N on Mars.
{ From the question "g" on Mars is {9.8m/s²}/3 which is 3.3m/s²}
From this analysis you notice that the WEIGHT of the person Varies but the MASS remained Constant at 90kg.
Answer:
E = 0 r <R₁
Explanation:
If we use Gauss's law
Ф = ∫ E. dA = 
/ ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
E = 0 r <R₁
