Answer:
The
electrons are moving through the superconductor per second.
Explanation:
Given :
Current
A
Charge of electron
C
Time
sec
From the formula of current,
Current is the number of charges flowing per unit time.
![I = \frac{ne}{t}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7Bne%7D%7Bt%7D)
Where
number of charges means in our case number of electrons
![n = \frac{It}{e}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7BIt%7D%7Be%7D)
![n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B1%20%5Ctimes%2010%5E%7B5%7D%20%7D%7B1.6%20%5Ctimes%2010%5E%7B-19%7D%20%7D)
![n = 6.25 \times 10^{23}](https://tex.z-dn.net/?f=n%20%3D%206.25%20%5Ctimes%2010%5E%7B23%7D)
Therefore,
electrons are moving through the superconductor per second.
Answer: the waves travel in an horizontal direction while the strings vibrate in a vertical direction.
The object does not move.
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = ![ma_a (0.35)](https://tex.z-dn.net/?f=ma_a%20%280.35%29)
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :
![{1.1 * 1200 \ a} - 14700 = 420 a](https://tex.z-dn.net/?f=%7B1.1%20%2A%201200%20%5C%20a%7D%20-%2014700%20%3D%20420%20a)
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
I am pretty sure the answer would be a