Answer:
a) F = 3.2 10⁻¹⁰ N
, b) v = 9.9 10⁷ m / s
Explanation:
a) The electric force is
F = q E
The electric field is related to the potential reference
V = E d
E = V / d
Let's replace
F = e V / d
Let's calculate
F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²
F = 3.2 10⁻¹⁰ N
b) For this part we can use kinematics
v² = v₀ + 2 a d
v = √ 2 ad
Acceleration can be found with Newton's second law
e V / d = m a
a = e / m V / d
a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²
a = 3,516 10⁻¹⁷ m / s²
Let's calculate the speed
v = √ (2 3,516 10¹⁷ 1.4 10⁻²)
v = √ (98,448 10¹⁴)
v = 9.9 10⁷ m / s
Answer:
Explanation:
Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J
Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J
But ,
4 * 1875000 = 7500000
so the KE has increased by 4 times.
Answer:
a) For y = 102 mA, R = 98.039 ohms
For y = 97 mA, R = 103.09 ohms
b) Check explanatios for b
Explanation:
Applied voltage, V = 10 V
For the first measurement, current 
According to ohm's law, V = IR
R = V/I
Here, 
For the second measurement, current 
b) ![y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}](https://tex.z-dn.net/?f=y%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dy_%7B1%7D%20%26y_%7B2%7D%20%5Cend%7Barray%7D%5Cright%5D%20%5E%7BT%7D)
![y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]](https://tex.z-dn.net/?f=y%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dy_%7B1%7D%20%5C%5Cy_%7B2%7D%20%5Cend%7Barray%7D%5Cright%5D)
![y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]](https://tex.z-dn.net/?f=y%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D102%2A10%5E%7B-3%7D%20%5C%5C97%2A10%5E%7B-3%7D%20%20%5Cend%7Barray%7D%5Cright%5D)
A linear equation is of the form y = Gx
The nominal value of the resistance = 100 ohms
![x = \left[\begin{array}{ccc}100\end{array}\right]](https://tex.z-dn.net/?f=x%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D100%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D102%2A10%5E%7B-3%7D%20%5C%5C97%2A10%5E%7B-3%7D%20%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DG_%7B1%7D%20%5C%5CG_%7B2%7D%20%20%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D100%5Cend%7Barray%7D%5Cright%5D%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DG_%7B1%7D%20%5C%5CG_%7B2%7D%20%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D102%2A10%5E%7B-5%7D%20%5C%5C97%2A10%5E%7B-5%7D%20%20%5Cend%7Barray%7D%5Cright%5D)
As per the question, the velocity of the airplane [v] = 660 miles per hour.
The total time taken by airplane [t] = 3.5 hours.
We are asked to determine the total distance travelled by the airplane during that period.
The distance covered [ S] by a body is the product of velocity with the time.
Mathematically distance covered = velocity × total time
S = v × t
= 660 miles/hour ×3.5 hours
= 2310 miles.
Hence, the total distance travelled by the airplane in 3.5 hour is 2310 miles.
