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alexandr1967 [171]
3 years ago
7

6x+y=6 8x-2y=-1 con solución de sistemas de ecuaciones por favor

Physics
1 answer:
sveticcg [70]3 years ago
5 0

6x+y=6

y=6-6x

8x-2y=-1

8x-2(6-6x)=-1

8x-12+12x=-1

20x=11

x=11/20

x=0.55

y=6-6x

y=6-6(0.55)

y=6-3.3

y=2.7

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Answer:

0° FROM or 180° TO

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A course deviation indicator or the CDI may be defined as an avionics tool or instrument that is used in the aircraft navigation which is used to determine an aircraft's lateral position with relation to a course.

While using the VOT service, we should tune the VOT frequency of the VOR receiver. And with the CDI leveled centered, the omnibearing selector or the OBS should read as 0° with the TO/FROM indicator showing 'FROM'. Else the OBS should read as 180° with the TO/FROM indication showing 'TO.'

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A 500kg car is driven forward with a thrust force of 1500N. Air resistance and friction acts against the motion of the car with
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2m/s^2, this is because F=ma, meaning a is also equal to F/m. The car applies 1500N in one direction and outside sources apply a total of -500N, meaning the 500kg car is moving forward with a total of 1000N of force. Taking the total 1000N and dividing it by 500kg gives you and acceleration of 2m/s^2. Hope this helps!
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At what altitude above the earth's surface would the acceleration due to gravity be 4.9 m/s^2? Assume the mean radius of the ear
Mandarinka [93]
We apply the gravity calculation expressed in the formula: g=GM/r2 
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A small sandbag is dropped from rest from a hovering hot-air balloon. (assume the positive direction is upward.) (a) after 1.5 s
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solution:

We know v0 = 0, a = 9.8, t = 4.0. We need to solve for v

so,

we use the equation:

v = v0 + at

v = 0 + 9.8*4.0

v = 39.2 m/s

Now we just need to solve for d, so we use the equation:

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3 0
2 years ago
A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =
SashulF [63]

Answer:

The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

The magnetic field along z axis is

B_{z}=3.484\times10^{-10}\ T

Explanation:

Given that,

Length of the current element dl=(0.5\times10^{-3})j

Current in y direction = 5.40 A

Point P located at \vec{r}=(-0.730)i+(0.390)k

The distance is

|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}

|\vec{r}|=0.827\ m

We need to calculate the magnetic field

Using Biot-savart law

B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}

Put the value into the formula

B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}

We need to calculate the value of \vec{dl}\times\vec{r}

\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k

\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})

\vec{dl}\times\vec{r}=0.000175i+0.000365k

Put the value into the formula of magnetic field

B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}

B=1.670\times10^{-10}i+3.484\times10^{-10}k

Hence, The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

The magnetic field along z axis is

B_{z}=3.484\times10^{-10}\ T

7 0
3 years ago
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