Answer:
Explanation:
force constant of spring k = force / extension
= 35.6 / 0.5
k = 71.2 N / m
angular frequency ω of oscillation by spring mass system
where m is mass of the body attached with spring
Putting the values
ω = 3.77 radian / s
The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2
so the equation for displacement from equilibrium position that is middle point can be given as follows
x = .5 sin ( ω t + π / 2 )
= 0.5 cos ω t
= 0.5 cos 3.77 t .
x = 0.5 cos 3.77 t .
Continental deflections, the Coriolis effect and global winds all affect surface ocean currents.
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:
Current = 10 Amperes.
Explanation:
Given the following dat;
Quantity of charge, Q = 36 kilocoulombs (KC) = 36 * 1000 = 36000C
Time = 1 hour to seconds = 60*60 = 3600 seconds
To find the current;
Quantity of charge = current * time
Substituting in the equation
36000 = current * 3600
Current = 36000/3600
Current = 10 Amperes.
Hi There,
This is False.
Hope this helped!
