Answer:
Yes it does
Explanation:
Gravity is pushing down on the pencil but the coffee mug is also pushing the pencil up with the same amount of force so they both don't move
Answer:i dont remember
Explanation: ?>>> confused
Answer:
Velocity, v = 0.239 m/s
Explanation:
Given that,
The distance between two consecutive nodes of a standing wave is 20.9 cm = 0.209 m
The hand generating the pulses moves up and down through a complete cycle 2.57 times every 4.47 s.
For a standing wave, the distance between two consecutive nodes is equal to half of the wavelength.

Frequency is number of cycles per unit time.

Now we can find the velocity of the wave.
Velocity = frequency × wavelength
v = 0.574 × 0.418
v = 0.239 m/s
So, the velocity of the wave is 0.239 m/s.
Answer:
The correct answer is 
Explanation:
The formula for the electron drift speed is given as follows,

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.
Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is
. Converting this number to m³ using very elementary unit conversion we get
. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

if we convert the area from mm³ to m³ we get
.So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

which is our final answer.
Answer:
241.7 s
Explanation:
We are given that
Charge of particle=
Kinetic energy of particle=
Initial time=
Final potential difference=
We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.
We know that

Using the formula


Initial voltage=

Using the formula





Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.