Lenz's Law: The polarity of the induced emf is such that it produces a current whose magnetic field opposes the change in magnetic flux through the loop.
Answer:
Approximately
(assuming that
.)
Explanation:
Let
denote the force that this spring exerts on the object. Let
denote the displacement of this spring from the equilibrium position.
By Hooke's Law, the spring constant
of this spring would ensure that
.
Note that the mass of the object attached to this spring is
. Thus, the weight of this object would be
.
Assuming that this object is not moving, the spring would need to exert an upward force of the same magnitude on the object. Thus,
.
The spring in this question was stretched downward from its equilibrium by:
.
(Note that
is negative since this displacement points downwards.)
Rearrange Hooke's Law to find
in terms of
and
:
.
Answer:
B. Electron
Static charge is produced by electron transfer.
Explanation:
Answer A: The neutron does not possess a charge and is said to be neutral.
Answer D: Protons and neutrons never move from object to object.
Only negative charges can move freely from one object to another. The energy that comes from these charges particles is called electrical energy.
....
Hope this answer can help you.
Answer:
J = 0.693 N.s
Explanation:
The impulse of one single drop is given by:
J1 = m*(Vf - Vo) where Vf = 0
![J1 = -9.25*10^{-4}N.s](https://tex.z-dn.net/?f=J1%20%3D%20-9.25%2A10%5E%7B-4%7DN.s)
The magnitude of the total impulse will be:
Jt = J1 * 150 * 5
Jt = 0.693 N.s
Answer:
The frequency is ![F = 325 Hz](https://tex.z-dn.net/?f=F%20%20%3D%20325%20Hz)
Explanation:
From the question we are told that
The frequency for the first note is ![F_1 = 330 Hz](https://tex.z-dn.net/?f=F_1%20%20%20%3D%20%20%20330%20Hz)
The beat frequency of the first note is ![f_b = 5 \ Hz](https://tex.z-dn.net/?f=f_b%20%20%3D%20%205%20%5C%20Hz)
The frequency for the second note is ![F_2 = 350 \ H_z](https://tex.z-dn.net/?f=F_2%20%20%3D%20%20350%20%5C%20H_z)
The beat frequency of the first note is ![f_a = 25 \ Hz](https://tex.z-dn.net/?f=f_a%20%20%3D%2025%20%5C%20Hz)
Generally beat frequency is mathematically represented as
![F_{beat} = | F_a - F_b |](https://tex.z-dn.net/?f=F_%7Bbeat%7D%20%3D%20%20%7C%20F_a%20-%20F_b%20%20%7C)
Where
are frequencies of two sound source
Now in the case of this question
For the first note
![f_b = F_1 - F \ \ \ \ \ ...(1)](https://tex.z-dn.net/?f=f_b%20%20%3D%20F_1%20-%20%20F%20%5C%20%5C%20%5C%20%5C%20%5C%20...%281%29)
Where F is the frequency of the string note
For the second note
![f_a = F_2 - F \ \ \ \ \ ...(2)](https://tex.z-dn.net/?f=f_a%20%20%3D%20F_2%20-%20%20F%20%5C%20%5C%20%5C%20%5C%20%5C%20...%282%29)
Adding equation 1 from 2
![f_b + f_a = F_1 + F_2 + ( - F) + (-F) )](https://tex.z-dn.net/?f=f_b%20%20%2B%20f_a%20%20%3D%20%20F_1%20%2B%20%20F_2%20%20%2B%20%28%20-%20F%29%20%20%2B%20%28-F%29%20%29)
![f_b + f_a = F_1 + F_2 -2F](https://tex.z-dn.net/?f=f_b%20%20%2B%20f_a%20%20%3D%20%20F_1%20%2B%20%20F_2%20%20-2F)
substituting values
![5 +25 = 330 + 350 -2F](https://tex.z-dn.net/?f=5%20%2B25%20%20%3D%20%20330%20%2B%20350%20%20-2F)
=> ![F = 325 Hz](https://tex.z-dn.net/?f=F%20%20%3D%20325%20Hz)