Answer: its an Ignition coil
Answer: True
Explanation: Injector orifice is the factor which describes the size of the opening of the injector .There are different pattern and size of the opening for the injector which affects the mixture of the chemical substance that is used for the production of the energy that is known as propellant.
The pattern and size of the orifice will define the variation in the amount of energy that could be produced.Thus the statement given is true.
Answer:
total amount of water after 2 min will be 84.4 kg/s
Explanation:
Given data:
one tank inflow = 0.1 kg/s
2nd tank inflow = 0.3 kg/s
3rd tank outflow = 0.03 kg/s
Total net inflow in tank is = 0.3 +0.1 =0.4 kg/s
From third point, outflow is 0.03 kg/s
Therefore, resultant in- flow = 0.4 - 0.03
Resultant inflow is = 0.37 kg/s
Tank has initially 40 kg water
In 2 min ( 2*60 sec), total inflow in tank is 0.37*60*2 = 44.4 kg
So, total amount of water after 2 min will be = 40+44.4 = 84.4 kg
Answer:
a) v = +/- 0.323 m/s
b) x = -0.080134 m
c) v = +/- 1.004 m/s
Explanation:
Given:
a = - (0.1 + sin(x/b))
b = 0.8
v = 1 m/s @ x = 0
Find:
(a) the velocity of the particle when x = -1 m
(b) the position where the velocity is maximum
(c) the maximum velocity.
Solution:
- We will compute the velocity by integrating a by dt.
a = v*dv / dx = - (0.1 + sin(x/0.8))
- Separate variables:
v*dv = - (0.1 + sin(x/0.8)) . dx
-Integrate from v = 1 m/s @ x = 0:
0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5
0.5v^2 = 0.8cos(x/0.8) - 0.1x - 0.3
- Evaluate @ x = -1
0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3
v = sqrt (0.104516)
v = +/- 0.323 m/s
- v = v_max when a = 0:
-0.1 = sin(x/0.8)
x = -0.8*0.1002
x = -0.080134 m
- Hence,
v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134
v = sqrt (0.504)
v = +/- 1.004 m/s
