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sweet [91]
3 years ago
12

Partes de un transformador

Engineering
1 answer:
BartSMP [9]3 years ago
6 0
Está constituido por dos bobinas de material conductor, devanadas sobre un núcleo cerrado de material ferromagnético, pero aisladas entre sí eléctricamente. ... Las bobinas o devanados se denominan primario y secundario según correspondan a la entrada o salida del sistema en cuestión, respectivamente.
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Astronauts who landed on the moon during the Apollo 15, 16, and 17 missions brought back a large collection of rocks to the eart
stiks02 [169]

Answer: a) W(earth) = 935.62 lbs

b) Mass of rocks in slugs = 29.06 slugs

Explanation:

a) From Newton's law, W = mg. Whether on the moon or on earth. Although, the mass of the rocks everywhere is the same, that is, mass of rocks on the moon = mass of rocks on earth.

W(moon) = mg(moon)

W(moon) = 154 lbs

g(moon) = 5.30 ft/s2

m = W(moon)/g(moon) = 154/5.3 = 29.06 lb.s2/ft

W(earth) = m g(earth)

g(earth) = 32.2 ft/s2

W(earth) = 29.06 × 32.2 = 935.62 lbs.

b) A slug = 1 lb.s2/ft, therefore the mass of the rocks in slugs is 29.06 slugs.

QED!

8 0
3 years ago
Describe the Range of a set of numbers and how it is calculated.
allochka39001 [22]

Answer:

The range of a set of data is the difference between the highest and lowest values in the set. To find the range, first order the data from least to greatest. Then subtract the smallest value from the largest value in the set.

Explanation:

7 0
3 years ago
Gold and silver rings can receive an arc and turn molten. True or False
liubo4ka [24]
The answer is False!
The answer is false
8 0
3 years ago
Read 2 more answers
If the average speed of a car is 45 km/jr, how far can it travel in 40 minutes?
Sholpan [36]

Answer:

30km

Explanation:

speed = distance ÷ time

3 0
2 years ago
The acceleration of a particle as it moves along a straight line is given
BARSIC [14]

Answer:

V_t=6 = 32 m/s

Explanation:

The origin is at point 0 with the positive motion to the right  

The instantaneous acceleration is change of velocity measured at infinitesimal interval of time, so the expression for instantaneous acceleration is:  

a=dv/dt

From here we can express dv as:

dv = a dt

Replace a by 2t — 1

dv = (2t — 1) dt

Integrate both sides of equation  

\int\limits^v_a  {2t-1} \, dv

v=t

a=t_0

putting these value in integral

<em>v-v_0=(t^2-t)-(t_0^2-t_0)</em>

We know that v_0 = 2 at t_0 = 0, so we'll replace t_0 and v_0 by their values

v — 2 = (t^2 — t) — (0^2 — 0)

From here we can write the expression for v as:  

v_t=6=6^2-6+2                             (1)  

So the velocity at t = 6 s is:

v_t=6 = 32 m/s

V_t=6 = 32 m/s

In order to determine the total distance travelled, we must check how maw times the particle has changed its direction, i.e. how many times its speed was equal to zero  

To do that, we'll just replace v by 0 in expression (1)

0 = t^2 — t + 2

The roots of the quadratic equation are:

t_1/2=1±  √(1^2-4*2*1)/2

Since 1^2-4*2*1 < 0, the quadratic equation have no real roots, so we can say that the velocity is always positive, i.e. to the right  

Now that we have all the details, we can correctly draw the path of the particle  

We can see from the sketch that the total distance traveled is:  

s^T=Δs_0-1

s^T=| s_1 - s_0 |

Replace s_0 by its value  

s^T=| s_1 - 1 |                                        (2)  

In order to determine the position of particle at t = 6 s, we'll need to determine the expression for s as function of time  

Since we have already wrote expression for v as function of time (step 2), we'll use expression  to get the expression for s

v= ds/dt  

Multiply both sides of equation by dt

v dt = ds

Replace v by expression (1)

(t^2 — t + 2) dt = ds

Integrate both sides of equation  

\int\limits^t_b {x} \, dx

t=s

b=(s=0)

x=(t^2 — t + 2)

dx=ds

putting these value in integral

(t^3/3-t^2/2+t)-(t_0^3/3-t_0^2/2+t_0)= s-s_0

Since s = 1 m at t = 0, and we want to determine the position s at t = 6, we'll replace so by 1, t_0 by 0 and t by 6  

(6^3/3-6^2/2+6)-(0^3/3-0^2/2+0)=s_t=6-1

4 0
3 years ago
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