Perform the following mathematical operation and report the answer to the correct number of significant figures 143.6/21.2

13) The condenser pressure is 417.4 psig for r-410A and the condenser outlet temperature is 108f. how much subcooling is there i

Tanzania [10]

Answer:

12°F

Explanation:

Calculation for how much subcooling is there in the condenser

Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

Amount of Condenser subcooling=12°F

Therefore the amount of subcooling that is there in the condenser will be 12°F

8 0

2 years ago

A 2.00 kg block hangs from a spring. A 300 g body hung below the block stretches the spring 2.00 cm farther. (a) What is the spr

Leokris [45]

The spring constant is 147 N/m

Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm

We need to find the spring constant

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.

The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring

We know that F = kx

300(9.8)= k (0.02)

k = 147.15 N/m

Rounding off to the nearest is 147N/m

The spring constant is 147N/m

Learn more about Hooke's law here

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7 0

2 years ago

The amplitude of a simple harmonic oscillator will be doubled by:a) doubling only the initial speedb) doubling the initial displ

OLEGan [10]

Answer:

When both initial speed and initial displacement is doubled then amplitude will be doubled.

Explanation:

Given that :- Amplitude of simple harmonic Oscillator is doubled.

So,

Formula of Simple harmonic oscillator is $X=A\sin\ (2\pi ft +\phi)$ ...........(1)

Where X = Position in (m,cm,km.....)

A = Amplitude in (m,cm,km.....)

F = Frequency in (Hz)

T = Time in (sec.)

Ф = Phase in (rad)

For initial displacement taking t=0 we get,

Initial displacement = $A\sin(\phi)$ .................(2)

Now taking equation (1) and differentiating it w.r.t to (t) we get

$\frac{dx}{dt} = 2\pi fA\cos\ (2\pi ft+\phi)$

$V= 2\pi fA\cos\ (2\pi ft+\phi)$

taking t=0 for initial speed then we get,

Initial speed = $2\pi fA\cos\phi$ ...............(3)

observing equation (2) & (3) that the initial displacement and initial speed depends on the Amplitude of the Oscillator.

Hence,

when both initial speed and displacement is doubled then amplitude will be doubled.

4 0

3 years ago

John pushes a box with a constant force as shown in the graph below.

andrew11 [14]

From the graph, it can be seen that the constant force that John exerted in order to move the object is 14N. Work is calculated by multiplying the force with the distance to which the object moves in parallel with the direction of the force.
Work = Force x displacement
Work = (14 N) x (8 m)
Work = 112 J
The closest value is 110J. Thus, the answer to this item is the second choice.

4 0

3 years ago

7) Three resistors having resistances of 4.0 Ω, 6.0 Ω, and 10.0 Ω are connected in parallel. If the combination is connected in

viva [34]

Answer:

A, 0.59A

Explanation:

The total resistance in the circuit is the resistances in parallel plus that in series.

Total resistance for those in parallel is;

1/(1/4 +1/6 +1/10) = 1/ (15+10+6 /60)

1/(31/60)= 60/31 ohms

Hence total resistance of the circuit is;

60/31 + 2 = (60+62)/31 = 122/31=3.94 ohms

To calculate the current flowing through the 10ohm resistance we need to know the voltage drop by subtracting the voltage drop in the 2ohm resistance from the total voltage drop.

Voltage drop on the 2 ohm resistance is;

Current on the 2 ohm resistor × 2 ohms

V = I ×R ; I - current

R - resistance

Current drop on the 2ohm resistance is;

Total voltage in the circuit/ total resistance in the circuit

12/3.94= 3.05A

Voltage drop on the 2 ohm resistance;

3.05 × 2 = 6.10volts

Hence voltage drop on the parallel resistance would be ;

12-6.10= 5.90V

Now voltage drop in a parallel circuit is the same hence 5.90v is dropped in each of the parallel resistance.

That said, the current drop on the 10 ohm resistor would be;

5.90/10 = 0.59A

Remember V= I× R so that I = V/R

6 0

3 years ago