The object is maintaining a constant velocity
Answer:
The resulting magnetic force on the wire is -1.2kN
Explanation:
The magnetic force on a current carrying wire of length 'L' with current 'I' in a magnetic field B is
F = I (L*B)
Finding (L * B) , where L = (2, 0, 0)m , B = (30, -40, 0)
L x B = ![\left[\begin{array}{ccc}i&j&k\\2&0&0\\30&-40&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C2%260%260%5C%5C30%26-40%260%5Cend%7Barray%7D%5Cright%5D)
= (0, 0, -80)
we can now solve
F = I (L x B) = I (-80)
F = -1200 kmN
F = -1200 kN * 10⁻³
F = -1.2kN
Answer:F=1.7802
Explanation:
Since we've been given the mass to be .18kg,we are asked to find the Force of which the formulae is
F=ma where f-force,m-mass and a-acceleration due to gravity
So we can just substitute
F-?.m=.18 and a9.89
F=.18×9.89
F=1.7802N
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:
75 rotations
Explanation:
f0 = 0, f = 3000 rpm = 50 rps, t = 3 s
(a) use first equation of motion for rotational motion
w = w0 + α t
2 x 3.14 x 50 = 0 + α x 3
α = 104.67 rad/s^2
(b) Let θ be the angular displacement
use second equation of motion for rotational motion
θ = w0 t + 1/2 α t^2
θ = 0 + 0.5 x 104.67 x 3 x 3
θ = 471.015 rad
The angle turn in one rotation is 2 π radian.
Number of rotation = 471.015 / (2 x 3.14) = 75 rotations
