Answer:
Frequency, 
Explanation:
Given that,
The wavelength of the x-rays, 
We need to find the frequency of an x-ray. All electromagnetic wave travel with a speed of light. It is given by the formula as :

f is the frequency

So, the frequency of an x-ray is
. Hence, this is the required solution.
Potential energy due to gravity
To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

Where
Angular acceleration
Angular velocity
t = Time
Our values are




Replacing at the previous equation we have that the angular velocity is



Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be



Therefore the angular acceleration of a point on the outer edge of the tires is 
Is the divine injunction found in Genesis 1:28, in which God, after having created the world and all in it, ascribes to humankind the tasks of filling, subduing, and ruling over the earth.
Answer:
Turn the heater on
Explanation:
There are two main forces involved in a balloon flight
The downward force is the total weight of the balloon: the air it contains, the gas bag, the basket, the passengers, etc.
The upward force is the weight of the of the air the balloon displaces.
During level flight
,
buoyant force = weight of displaced air - total weight of balloon
If you increase the temperature of the air in the bag, the air molecules spread out and leave through the bottom of the bag.
The balloon still has the same volume, so the weight of displaced outside air stays the same.
However, the balloon has lost some hot inside air, so its total weight decreases.
The upward force is greater than the downward force, so the balloon rises.