Answer: Option A : Technician A
Explanation:
The statement/observation, "that the starter motor used to crank diesel engines can draw up to 400 amps of current" made by Technician A is correct.
A diesel engine uses up to 400+ Amperes of electricity to start up a diesel engine in the ignition chamber of motor engine.
Answer:
6010.457N
Explanation:
Centripetal acceleration = a= V²/R
At a radius of 3.6m and velocity of 16.12m/s,
Acceleration is
a = 16.12²/ 3.6 = 72.182 m/s²
Force = Mass (m) * Acceleration (a)
36 = m * 72.182
m = 36/72.182
At breaking point
Radius = 0.468 m and Velocity = 75.1 m/s
a = V²/R = 75.1²/0.468
a = 12051.3 m/s
F = Mass(m) * Acceleration (a)
F = m * 12051.3
m = F/ 12051.3
Settings the ratio of mass equal
m = m
=> 36/72.182 = F/12051.3
F = 12051.3 * 36/72.182
F = 6010.457N
Answer:
a) 19.4 m/s
b) 19 m/s
Explanation:
a) In the given question,
the potential energy at the initial point = Ui = 0
the potential energy at the final point = Uf = mgh
the kinetic energy at the initial point = Ki = 1/2 mv₀².
the kinetic energy at the final point = Kf = 0
work done by air= Ea= fh = 0.262 N
Now, using the law of conservation of energy
initial energy= final energy
Ki +Ui = Kf + Uf +Ea
1/2 mv₀² + 0 = 0 + mgh + fh
1/2 mv₀² = mgh + fh
h = v₀²/ 2g (1 +f/w)
calculate m
m= w/g = 5.29 /9.8
= 0.54 kg
h = 20 ²/ (2 x9.80) x (1 0.265/5.29)
h = 19.4 m.
b) 1/2 mv² + 2fh = 1/2 mv₀²
Vg = 19 m/s
Answer: 7022.2kg/m³, yes, I was cheated
Explanation:
Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;
Density = Mass/Volume
Note that the unit of both mass and volume must be standard unit.
Given mass = 0.0158kg
Dimension of the metal = 5mm×15mm×30mm
Note that 1mm = 0.001m
The volume of the metal will be
0.005×0.015×0.03
= 0.00000225m³
Density = 0.0158/0.00000225
Average density of the metal = 7022.2kg/m³
Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.
Answer:opposite
Explanation:for a capacitor to discharge (after charging) the polarities of the current and voltage have to be reversed
