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 average force ≅ 519.44 N.
Explanation:
Impulse = change in momentum of a body
i.e Ft = m(v - u)
where F is the force, t is the time, m is the mass of the body, v is the final velocity and u is the initial velocity.
m = 55.0 g (0.055 Kg), t = 0.00360 s, v = 34.0 m/s, since the ball was initially at rest; u = 0 m/s
So that,
F x 0.00360 = 0.055(34 - 0)
F x 0.00360 = 0.055 x 34
= 1.87
F = 
= 519.4444
The average force exerted on the ball by the club is approximately 519.44 N.
Resultant is the correct answer!
Refer to the diagram shown below.
Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.
The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.
The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s
The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°
Answer: 10.2 m/s at 21.5° downstream.
The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

We can apply the first Newton's law in x and y-direction.
If we do a free body diagram of the system we will have:
x-direction
All the forces acting in this direction are:
(1)
Where:
- T(1) is the tension due to the rope 1
- T(2) is the tension due to the rope 2
Here we just conclude that T(1) = T(2)
y-direction
The forces in this direction are:
(2)
Here W is the weight of the steel beam.
We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.
Knowing that T(1) = T(2) and W = mg, we have:



T(1) must be equal to 5479 N, so we have:


Therefore, the maximum angle allowed is θ = 37.01°.
You can learn more about tension here:
brainly.com/question/12797227
I hope it helps you!