Answer: 0.642mm
Explanation: F= force = 5.2×10^-16 N,
v = velocity of electron = 1.2×10^7 m/s,
m = mass of electron = 9.11×10^-31 kg.
We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.
Recall that f = ma.
Where a = acceleration
This acceleration of vertical because it occurred when the object deflected.
5.2×10^-16 = 9.11×10^-31 (ay)
ay = 5.2×10^-16 / 9.11×10^-31
ay = 5.71×10^14 m/s²
For the horizontal motion, x = vt
Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,
By substituting the parameters, we have that
0.019 = 1.27×10^7 × t
t = 0.019 / 1.27 × 10^7
t = 1.5×10^-9 s
The vertical distance (y) is gotten by using the formulae below
y = ut + at²/2
but u = 0
y = at²/2
y = 5.71×10^14 × (1.5×10^-9)²/2
y = 0.00128475/2
y = 0.000642m = 0.642mm
Answer:
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Answer:
Psm = 30.66 [Psig]
Explanation:
To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.
P * v = R * T
where:
P = pressure [Pa]
v = specific volume [m^3/kg]
R = gas constant for air = 0.287 [kJ/kg*K]
T = temperature [K]
<u>For the initial state</u>
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P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)
T1 = -2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)
Therefore we can calculate the specific volume:
v1 = R*T1 / P1
v1 = (0.287 * 270.4) / 266.8
v1 = 0.29 [m^3/kg]
As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.
V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.
T2 = 43 + 273 = 316 [K]
P2 = R*T2 / V2
P2 = (0.287 * 316) / 0.29
P2 = 312.73 [kPa]
Now calculating the manometric pressure
Psm = 312.73 -101.325 = 211.4 [kPa]
And converting this value to Psig
Psm = 30.66 [Psig]
The starting and ending points of the motion are the same.. . . . .
