Spacecraft used is "Friendship 7". Hope it helps.
Answer:
Technician A says that this is the normal operation of the ETC self -test is the correct answer.
Explanation:
An engine control unit (ECU), also widely referred to as an engine control module (ECM), is a type of electronic control device that controls an internal combustion engine with a series of actuators to ensure maximum engine performance.
It achieves so by reading values from a multitude of sensors within the engine bay, translating data using multidimensional feedback maps (the so-called lookup tables) and modifying the actuators.
Mechanically fixed and dynamically regulated by mechanical and pneumatic means, air-fuel combination, ignition time, and idle speed were before ECUs.
As soon as the system gets battery voltage, after ignition is turned, the efi computer makes a self-test of all the actuators and sensors, included the ETC.
They go in the boxes in this order:
density
2.meter
3.matter
4.hypothesis
5.control
6.kilogram
Answer
given,
vertical speed of stone,v = 12 m/s
height of the cliff = 70 m
a) time taken by the stone to reach at the bottom of the cliff
We know that,
S = u t + 1/2 a t²
- 70 = 12 t - 0.5 x 9.8 t²
4.9 t² - 12 t - 70 = 0
solving the equation
t = 5.2 s (neglecting the negative value)
b) again using equation of motion
v = u + a t
v = 12 - 9.8 x 5.2
v = -38.96 m/s
ignoring the negative sign
magnitude of velocity is equal to 38.96 m/s
c) total distance travel by the stone
vertical distance covered by the stone
v² = u² + 2 g h
0 = 12² - 2 x 9.8 x h
h = 7.34 m
to reach the stone to the same level distance travel be doubled.
Total distance travel by the stone
H = h + h + 70
H = 7.34 x 2 + 70
H = 84.7 m.
Answer:
-0.1875 V.
Explanation:
Using
E₂ = MdI₁/dt........................ Equation 1
Where E₂ = Voltage induced in the second coil, M = mutual inductance of both coil, dI₁ = change in current in the first coil, dt = change in time.
Given: M = 3.00 mH = 0.003 H, dI₁ = (0-2.50) = -2.5 A, dt = 40 ms = 0.04 s.
Substitute into equation 1
E₂ = 0.003(-2.5)/0.04
E₂ = -0.1875 V.
Hence the induced emf = -0.1875 V.
