Answer:
C. Technician B
Explanation:
Excessive Galvanic activity:
To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.
Electrolysis problem:
When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.
In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.
Answer:
vector with direction equal to the axis X.
Explanation:
We use the Gauss Law and the superposition law in order to solve this problem.
<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:
<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:
Then:
This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.
Answer:
true
Explanation:
The law of conservation of charge states that whenever electrons are transferred between objects, the total charge remains the same.
Answer:
There are four main ways of doing that :-
- Velocity
- Acceleration
- Momentum
- Kinetic energy
Hope it helps!
Kinetic of automobile
Mass m = 1,250 Kg; V = 11 m/s
Formula: K.E = 1/2 mV²
K.E = 1/2(1,250 Kg)(11 m/s)²
K.E = 75,625 J
Speed required for insect to have the same kinetic energy as automobile
Mass of insect = 0.72 g convert to Kg m = 7.2 x 10⁻⁴ Kg
K.E = 1/2 mV² Derive V =?
V = 2 K.E/m
V = √2(75,625 J)/7.2 x 10⁻4 Kg
V = √2.1 x 10⁸ m²/s²
V = 14,491.34 m/s (velocity of insect)
