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3 years ago

When lining up the song on the tempo grid it is important to allow

Engineering

1 answer:

Reika [66]3 years ago

6 0

Tempo decides the speed at which the music is played.

<u>Explanation:</u>

The Tempo of a bit of music decides the speed at which it is played, and is estimated in beats per minute (BPM). The 'beat' is dictated when mark of the piece, so 100 BPM in 4/4 compares to 100 quarter notes in a single moment.

A quick tempo, prestissimo, has somewhere in the range of 200 and 208 beats for each moment, presto has 168 to 200 beats for every moment, allegro has somewhere in the range of 120 and 168 beats for every moment, moderato has 108 to 120 beats for every moment, moderately slow and even has 76 to 108, adagio has 66 to 76, larghetto has 60 to 66, and largo, the slowest rhythm, has 40 to 60.

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Saturated liquid water at 150 F is put under pressure to decrease the volume by 1% while keeping the temperature constant. To wh

Romashka-Z-Leto [24]

Answer:

Between 5 & 10 MPa in Table B.1.4

Explanation:

150F = 65°C

State 1:

T = 65°C , x = 0.0; Table B.1.1: v = 0,001020 m^3 /kg

Process: T = constant = 65°C

State 2:

T, v = 0.99 x v f (65°C) = 0.99 x 0,001020 = 0.0010098 m^3 /kg

Between 5 & 10 MPa in Table B.1.4

5 0

3 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$

$X_{C} =3*V^2/Q_{C}$

$X_{C} =3*(480)^2/14*10^3$ Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is

$I_{cap} =V/X_{C} =480/49.37=9.72 A$

Line current flowing from the source is

$I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A$

8 0

3 years ago

A capillary tube is immersed vertically in a water container. Knowing that water starts to evaporate when the pressure drops bel

Anuta_ua [19.1K]

Answer:

$d=0.414\times 10^{-4}\ m$

Explanation:

Given that

P = 4 KPa

Contact angle = 6°

Surface tension = 1 N/m

Lets assume that atmospheric pressure = 100 KPa

Lets take that density of water = $1000\ kg/m^3$

So the capillarity rise h

$h=\dfrac{\Delta P}{\rho g}$

$h=\dfrac{100\times 1000-4\times 1000}{1000\times 10}$

h= 9.61 m

We know that for capillarity rise h

$h=\dfrac{2\sigma cos\theta }{r\rho g}$

$r=\dfrac{2\sigma cos\theta }{h\rho g}$

$r=\dfrac{2\times 1 cos4^{\circ} }{9.61\times 1000\times 10}$

$r=0.207\times 10^{-4}\ m$

$d=0.414\times 10^{-4}\ m$

3 0

3 years ago

In this lab, your task is to configure the external vEthernet network adapter with the following IPv6 address: Prefix: 2620:14F0

OverLord2011 [107]

Here is the complete question

You work as the IT administrator for a small corporate network. You need to create a separate subnet to use for testing. The test subnet needs access to the rest of the network through a router, but it should not have any local access to production machines.

You have installed Windows Server 2016 on CorpRTR, which you plan to use to isolate the test segment from the rest of the network. You'll use traditional routing or NAT.

In this lab, your task is to add the necessary role and role services to meet the stated requirements. Do not add unnecessary role services.

Answer & Explanation:

Complete this lab as follows:

1.In the notification area, right-click the Network icon and select Open Network and Sharing Center.

2.On the left, select Change adapter settings.

3.Right-click the vEthernet (External)adapter and select Properties.

4.Select Internet Protocol Version 6 (TCP/IPv6).

5.Select Properties.

6.Select Use the following IPv6 address.

7.In the IPv6 address field, enter 2620:14F0:45EA:0001:192:168:0:10as the IPv6 address.

8.In the Subnet prefix length field, enter 64.

9.Click OK.

10.Click Close.

11.In the Search the web and Windows field, enter cmd.

12.Under Best match, right-click Command Prompt and select Run as Administrator.

13.At the command prompt, type ipconfig /all and press Enter to view the IPv6 address.

4 0

4 years ago

9. Calculate the total resistance and current in a parallel cir-

Taya2010 [7]

Answer:

d. 2.3 ohms (5.3 amperes)

Explanation:

The calculator's 1/x key makes it convenient to calculate parallel resistance.

Req = 1/(1/4 +1/8 +1/16) = 1/(7/16) = 16/7 ≈ 2.3 ohms

This corresponds to answer choice D.

<em>Additional comment</em>

This problem statement does not tell the applied voltage. The answer choices suggest that it is 12 V. If so, the current is 12/(16/7) = 21/4 = 5.25 amperes.

5 0

3 years ago