Answer:
The molecular weight will be "28.12 g/mol".
Explanation:
The given values are:
Pressure,
P = 10 atm
= 
= 
Temperature,
T = 298 K
Mass,
m = 11.5 Kg
Volume,
V = 1000 r
= 
R = 8.3145 J/mol K
Now,
By using the ideal gas law, we get
⇒ 
o,
⇒ 
By substituting the values, we get
As we know,
⇒ 
or,
⇒ 
Answer:
The speed of shaft is 1891.62 RPM.
Explanation:
given that
Amplitude A= 0.15 mm
Acceleration = 0.6 g
So
we can say that acceleration= 0.6 x 9.81
We know that
So now by putting the values
We know that
ω= 2πN/60
198.0=2πN/60
N=1891.62 RPM
So the speed of shaft is 1891.62 RPM.
Answer:
The amount of energy transferred to the water is 4.214 J
Explanation:
The given parameters are;
The mass of the object that drops = 5 kg
The height from which it drops = 86 mm (0.086 m)
The potential energy P.E. is given by the following formula
P.E = m·g·h
Where;
m = The mass of the object = 5 kg
g = The acceleration de to gravity = 9.8 m/s²
h = The height from which the object is dropped = 0.086 m
Therefore;
P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J
Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;
The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.
Answer:
I = 0.625 A
Explanation:
Given that,
Power of the light bulb, P = 75 W
Voltage of the circuit, V = 120 V
We need to find the current flowing through it. We know that, Power is given by :
I is the electric current
So, the current is 0.625 A.
Answer:
Technician B only is correct
Explanation:
Voltage drop testing is a method used to find the amount of electrical resistance available in an high amperage circuit that involves connecting the leads of the meter in parallel to the circuits being tested such that disassembly is not required
In voltage drop test, the red voltmeter lead and black voltmeter lead are placed at two points on the same side of the circuit connection such that the leads are in between two positive connection or two negative connection (rather than connecting the red to the positive and the black to the negative sides of the circuit) and digital voltmeter is used for the voltage drop measurement across the lead while the connection is under load.
Therefore, Technician B only is correct
