Anode
Explanation:
The anode in the gas discharge tube used by Thomson in his experiment was the positively charged electrode.
Using the gas discharge tube, Thomson made the remarkable discovery of cathode rays.
The rays moves from the negatively charged cathode to the positively charged anode. This indicated that the rays carry positive charges.
Some parts of the tube are:
- Cathode - negatively charged electrode
- Power source
- Gas at low pressure
- Outlet to vacuum pump
Learn more:
cathode brainly.com/question/12747250
#learnwithBrainly
A geologist is studying rock layers in an old river bed, and he finds a fossil of a fish and a horsetail rush in the same rock layer. According to the law of faunal and floral succession, the geologist can assume that the rock containing the fossils may date back as far as the <span>Devonian period</span>.
Answer:
No, the farmer is not able to move the mule.
Explanation:
Mass =100 kg
Force=F=800 N
The coefficient between the mule and the ground=

Static friction force,f=
Normal force=N=mg
Static friction force,f=
Using 
F<f
Static friction force is greater than applied force.
Therefore , the farmer is not able to move the mule.
Answer:
a) ω = 9.86 rad/s
b) ac = 194. 4 m/s²
c) minimum coefficient of static friction, µs = 19.8
Explanation:
a) angular speed, ω = 2πf, where f is frequency of revolution
1 rps = 6.283 rad/s, π = 3.142
ω = 2 * 3.14 * 0.25 * 6.28
ω = 9.86 rad/s
b) centripetal acceleration, a = rω²
where r is radius in meters; r = 200 cm or 2 m
a = 2 * 9.86²
a = 194. 4 m/s²
c) µs = frictional force/ normal force
frictional force = centripetal force = ma; where a is centripetal acceleration
normal force = mg; where g = 9.8 m/s²
µs = ma/mg = a/g
µs = 194.4 ms⁻²/9.8 ms⁻²
c) minimum coefficient of static friction, µs = 19.8
In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

The wavelength of the 2nd harmonic is:

The wavelength of the 4th harmonic is:

It is not possible to find any integer n such that
, therefore the correct options are A, B and D.