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
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Answer:
The shortest distance is
Explanation:
The free body diagram of this question is shown on the first uploaded image
From the question we are told that
The speed of the bicycle is 
The distance between the axial is 
The mass center of the cyclist and the bicycle is
behind the front axle
The mass center of the cyclist and the bicycle is
above the ground
For the bicycle not to be thrown over the
Momentum about the back wheel must be zero so

=> 
=> 
Here 
So 
Apply the equation of motion to this motion we have

Where 
and
since the bicycle is coming to a stop

=>
Answer: Hazmat products are allowed in your FC are:
- A GPS unit (lithium batteries)
- A subwoofer (magnetized materials)
Explanation:
Hazmat products consist of flammable, corrosive and harmful substances which are actually very hazardous to human health and environment.
Hazardous material allowed in FC are as follows.
- Magnetized material products like as speakers.
- Non-spillable battery products like toy cars.
- Lithium-ion battery containing products like laptops, mobile phones etc.
- Non-flammable aerosol.
So, hazmat allowed products are GPS unit (lithium batteries) and subwoofer (magnetized materials).
Thus, we can conclude that hazmat products are allowed in your FC are:
- A GPS unit (lithium batteries)
- A subwoofer (magnetized materials)
Answer:
11.25 amps
Explanation:
For transformers, the magnetic flux

Therefore;

Ф = Фmax (cosωt) = 0.21·(cos(5·t))
From Faraday's law of induction, we have;
ε = -N × dΦ/dt
Which gives;
dΦ/dt = -1.05(sin (5t)
)
ε = -N × dΦ/dt = -50× -1.05(sin (5t)
)
ε = 52.5(sin (5t)
)
I = ε/R = 52.5(sin (5t)
)/3.3 = 15.9091(sin (5t)
) amps
The peak current is therefore = 15.9091 amps
The rms current = Peak current /√2 = 15.9091/(√2) = 11.25 amps.
Answer:
the answer if im not mistaken is A
Hope this helps:)