The alternating current frequency used in the United States is

To measure the voltage drop across a resistor, a _______________ must be placed in _______________ with the resistor. Ammeter; S

gulaghasi [49]

Answer:

The given blanks can be filled as given below

Voltmeter must be connected in parallel

Explanation:

A voltmeter is connected in parallel to measure the voltage drop across a resistor this is because in parallel connection, current is divided in each parallel branch and voltage remains same in parallel connections.

Therefore, in order to measure the same voltage across the voltmeter as that of the voltage drop across resistor, voltmeter must be connected in parallel.

4 0

3 years ago

Refrigerant 22 flows in a theoretical single-stage compression refrigeration cycle with a mass flow rate of 0.05 kg/s. The conde

Debora [2.8K]

Answer:

a. The work done by the compressor is 447.81 Kj/kg

b. The heat rejected from the condenser in kJ/kg is 187.3 kJ/kg

c. The heat absorbed by the evaporator in kJ/kg is 397.81 Kj/Kg

d. The coefficient of performance is 2.746

e. The refrigerating efficiency is 71.14%

Explanation:

According to the given data we would need first the conversion of temperaturte from C to K as follows:

Temperature at evaporator inlet= Te=-16+273=257 K

Temperatue at condenser exit=Te=48+273=321 K

Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg

Enthalpy at evaporator exit of Te 48=i1=260.51 Kj/Kg

b. To calculate the the heat rejected from the condenser in kJ/kg we would need to calculate the Enthalpy at the compressor exit by using the compressor equation as follows:

w=i4-i3

W/M=i4-i3

i4=W/M + i3

i4=2.5/0.05 + 397.81

i4=447.81 Kj/kg

a. Enthalpy at the compressor exit=447.81 Kj/kg

Therefore, the heat rejected from the condenser in kJ/kg=i4-i1

the heat rejected from the condenser in kJ/kg=447.81-260.51

the heat rejected from the condenser in kJ/kg=187.3 kJ/kg

c. Temperature at evaporator inlet= Te=-16+273=257 K

The heat absorbed by the evaporator in kJ/kg is Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg

d. To calculate the coefficient of performance we use the following formula:

coefficient of performance=Refrigerating effect/Energy input

coefficient of performance=137.3/50

coefficient of performance=2.746

the coefficient of performance is 2.746

e. The refrigerating efficiency = COP/COPc

COPc=Te/(Tc-Te)

COPc=255/(321-255)

COPc=3.86

refrigerating efficiency=2.746/3.86

refrigerating efficiency=0.7114=71.14%

8 0

4 years ago

To make an even better electrical junction, what should you do?

docker41 [41]

Answer:

A:Solder it.

Explanation:

Hopefully this helps!

4 0

3 years ago

An experiment compares the initial speed of bullets fired from two handguns: a 9 mm and a 0.44 caliber. The guns are fired into

olasank [31]

Answer:

1.176

Explanation:

When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.

v2 * (M + mb) = v1 * mb

Where

v1: muzzle velocity of the bullet

M: mass of the bob

mb: mass of the bullet

v2: mass of the bob with the bullet after being hit

v2 = v1 * mb / (M + mb)

Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.

Ek = 1/2 (M + mb) * v2^2

Ep = (M + mb) * g * h

Ek = Ep

1/2 (M + mb) * v2^2 = (M + mb) * g * h

1/2 * (v1 * mb / (M + mb))^2 = g * h

1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h

v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)

$v2 = \sqrt{\frac{g *h * (M+ mb)^2}{\frac{1}{2} * mb^2}}$