Answer:
Each portable fire extinguisher should only be used for its specific type of fire. Class A extinguishers could cause a Class B fire to spread or electrocution in a Class C fire. A Class B extinguisher could fail to completely extinguish a Class A fire, causing the flame to re-ignite later.
Given Information:
Angular velocity = ω = 4 rad/s
Angular acceleration = α = 5 rad/s²
Center deceleration = a₀ = 2 m/s
Required Information:
Acceleration of point A at this instant = ?
Answer:
Acceleration of point A at this instant = 5.94 m/s²
Explanation:
Refer to the attached diagram of the question,
The acceleration of point A is given by
a = a₀ + rα - rω²
Where r is the radial distance between the center and point A, a₀ is the deceleration of center, α is the angular acceleration and ω is the angular velocity.
a = -2i + 0.3j*5k - 0.3j*4²
a = -2i + 1.5(j*k) - 0.3j*16
a = -2i + 1.5(-i) - 4.8j
a = -2i - 1.5i - 4.8j
a = -3.5i - 4.8j
The magnitude of acceleration vector is
a = √(-3.5)² + (-4.8)²
a = √35.29
a = 5.94 m/s²
Therefore, the acceleration of point A is 5.94 m/s²
The angle is given by
θ = tan⁻¹(y/x)
θ = tan⁻¹(-4.8/-3.5)
θ = 53.9°
Answer:
Explanation:
A smaller clearance volume means a higher compression. A higher compression means better thermal efficiency. However a compression ratio too high might be troublesome, as it can cause accidental ignition of the fuel-air mix. This is the reason why Otto cycle engines have lower compressions that Diesel engines. In a Diesel engine the mix ignites by compression instead of a spark.
Answer:
248.756 mV
49.7265 µA
Explanation:
The Thevenin equivalent source at one terminal of the bridge is ...
voltage: (100 V)(1000/(1000 +1000) = 50 V
impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω
The Thevenin equivalent source at the other terminal of the bridge is ...
voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V
impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω
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The open-circuit voltage is the difference between these terminal voltages:
(50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage
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The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:
(50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA
