I’m not from USA and in our physics literature, weight and gravitational force are considered as two different forces. Gravitational force acting any body on Earth (or anywhere else) is basically described as the force with which Earth pulls objects towards itself, and it’s equal to the mass of...
Fine but I wanted to take a more theoretical approach. It would only be reasonable to me that weight has to be equal to the gravity in that case in order to provide normal reaction of surface that is equal to and opposite of gravity and balances it? But it it were true, it would look like we are...
I’ll reformulate the question to make it more precise.
Between any two bodies with mass (according to Newton), there arises attractive force (gravity).
The bodies in Earth’s atmosphere “fall” due to gravity.
The bodies that are standing on surface experience the force of gravity (Earth...
A relatively simple question has been bothering me for some time: Why is the weight of an object (the force the object exerts on a surface/rope) equal to the force of gravity that the body experiences in Earth’s gravitational field?
Earth’s gravitational field accelerates all bodies in its...
Would it be accurate if I interpreted derivatives this way:
The rate of change of linear functions tells us how many times y changes if we increase x by 1.
So: ##y = mx + b##
Where m is the initial value and we will set it to zero:
##y = mx##
So solving for m we get:
##m = \dfrac{y}{x}##...
Thanks. I would like to ask you one more thing because I’m struggling with understanding what derivative is.
I’ve seen everywhere that definition “derivative is instantaneous rate of change”. I am so confused by this definition I can’t even explain the part I don’t understand, it just sounds...
If the formula for instantaneous velocity is:
##v = \lim_{\Delta t \to 0} \dfrac{\Delta s}{\Delta t}##
Why the result of equation isn’t infinity? It’s said that if we divide something by number very close to zero, it results in infinitely large number. But how does this equation work then? It...
How does this oppose the fact that pressure of water in pool at certain depth will be equal to the sum of static (atmospheric) and hydrostatic pressure?
If that’s the case, why do we use this formula for pressure of water inside some container:
##p = p_0 + \rho g h##
Where the second term is hydrostatic pressure. And the first is obviously static (atmospheric) pressure.
When dealing with problem of water flowing through small opening on a...