Answer: The spinning of Earth's outer core causes the lithosphere to break into individual plates and collide with each other.
Earth's surface forms because of this reason:
The spinning of Earth's outer core causes the lithosphere to break into individual plates and collide with each other.
Answer:
acceleration 8 km/h/s south
Explanation:
First of all, let's remind that a vector quantity is a quantity which has both a magnitude and a direction.
Based on this definition, we can already rule out the following two choices:
distance: 40 km
speed: 40 km/h
Since they only have magnitude, they are not vectors.
Then, the following option:
velocity: 5 km/h north
is wrong, because the car is moving south, not north.
So, the correct choice is
acceleration 8 km/h/s south
In fact, the acceleration can be calculated as

where
v = 40 km/h is the final velocity
u = 0 is the initial velocity
t = 5 s is the time
Substituting,

And since the sign is positive, the direction is the same as the velocity (south).
Answer:
Given:
Thermal Kinetic Energy of an electron, 
= Boltzmann's constant
Temperature, T = 1800 K
Solution:
Now, to calculate the de-Broglie wavelength of the electron,
:

(1)
where
h = Planck's constant = 
= momentum of an electron
= velocity of an electron
= mass of electon
Now,
Kinetic energy of an electron = thermal kinetic energy



(2)
Using eqn (2) in (1):

Now, to calculate the de-Broglie wavelength of proton,
:

(3)
where
= mass of proton
= velocity of an proton
Now,
Kinetic energy of a proton = thermal kinetic energy



(4)
Using eqn (4) in (3):

Refer to the diagram shown below.
In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s
The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m
Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.
The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s
The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.
Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.