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fgiga [73]
2 years ago
11

The toy car is about 3 inches long is an example of a ?

Physics
1 answer:
balandron [24]2 years ago
8 0

Answer:

a quantitative observation because it includes numerical data

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The per-unit impedance of a single-phase electric load is 0.3. The base power is 500 kVA, and the base voltage is 13.8 kV. a. Fi
Leto [7]

Answer:

114.26

Explanation:

a)Formula for per unit impedance for change of base is

Zpu2= Zpu1×(kV1/kV2)²×(kVA2/kVA1)

Zpu2: New per unit impedance

Zpu1: given per unit impedance

kV1: give base voltage

kV2: New bas votlage

kVA1: given bas power

kVA2: new base power

In the question

Zpu2=??

Zpu1= 0.3

kV2=24kV

kV1= 13.8 kV

kVA2= 1MVA ×1000= 1000 kVA

kVA1=500kVA

Zpu2= 0.3(13.8/24)²×(1000/500)

Zpu2= 0.198

b) to find ohmic impedance we will first calculate base value of impedance(Zbase). So,

Zbase= kV²/MVA

  Zbase= 13.8²/(500/1000)

  Zbase=380.88

Now that we have base value of impedance, Zbase, we can calculate actual ohmic value of impedance(Zactual) by using the following formula:

Zpu=Zactual/Zbase

0.3= Zactual/380.88

Zactual= 114.26 ohms

8 0
3 years ago
VERY EASY QUESTION!!!!!! 20 POINTS!!!!!!!!
slamgirl [31]
C. at a science fair
4 0
2 years ago
Read 2 more answers
Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in
postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is \rho = \frac{m}{V}, and the volume of a pyramid is (confusingly written on the proposal) V=\frac{1}{3} Ah, so we can write:

m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h

Where s is the side of the base, being s^2 the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that \frac{1m}{100cm}=1, and we can use this factor to convert anything written in cm to anything written in m. Example:

500cm=500cm\frac{1m}{100cm}=5m

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3

Where we have used the fact that 1^3=1 (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg

And now we will convert to short tons using two conversion factors at the same time:

m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0
2 years ago
A.
Brut [27]

Answer:

\boxed{\sf Work \ done = 4 \ J}

Given:

Force = 8 N

Distance covered by the body = 50 cm = 0.5 m

Explanation:

Work Done = Force × Distance covered by the body

= 8 × 0.5

= 4 J

6 0
2 years ago
6. A satellite is orbiting Earth just above its surface. The centripetal force making the satellite follow a circular trajectory
Svetllana [295]

Answer:

Engular velocity: w=1,24*10^{-3}/s

Linear velocity: V=7905 m/s

The time it takes:

t=5060s=84min

Explanation:

The magnitude of the centripetal acceleration can be related to the angular velocity and radius as:

(1)a=r*w^{2}

Solving for w:

(2)w=\sqrt{\frac{a}{r} }

Replacing a=9,8m/s2 and r=6,375,000m:

(3)w=\sqrt{\frac{9,8m/s^{2}}{6375000m} }=1,24*10^{-3}/s

And the angular velocity relates to the linear velocity:

V=w*r=1,24*10^{-3}/s*6375000m=7905 m/s

The perimeter of the orbit is:

P=\pi *2*r=\pi *2*6375000m=40.05*10^{6}m

The time it takes:

t=\frac{P}{V} =\frac{40.05*10^{6}m}{7905 m/s}=5060s=84min

5 0
2 years ago
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