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Anastaziya [24]
3 years ago
13

A vector AP is rotated about the Z-axis by 60 degrees and is subsequently rotated about X-axis by 30 degrees. Give the rotation

matrix that accomplishes these rotations in the given order.
Engineering
1 answer:
Musya8 [376]3 years ago
4 0

Answer:

R = \left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]

Explanation:

The mappings always involve a translation and a rotation of the matrix. Therefore, the rotation matrix will be given by:

Let \theta and \alpha be the the angles 60⁰ and 30⁰ respectively

that is \theta = 60⁰ and

\alpha = 30⁰

The matrix is given by the following expression:

\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]

The angles can be evaluated and left in the surd form.

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A binary star system consists of two stars of masses m1m1m_1 and m2m2m_2. The stars, which gravitationally attract each other, r
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Answer:

          a_c_2=\dfrac{a_c_1\times m_1}{m_2}

Explanation:

The question is: <em>Find the magnitude of the centripetal acceleration of the star with mass m₂</em>

The <em>centripetal acceleration</em> is the quotient of the centripetal force and the mass.

                a_c=\dfrac{F_c}{m}

Thus, you can write the equations for each star:

     

       a_c_1=\dfrac{F_c_1}{m_1}

       a_c_2=\dfrac{F_c_2}{m_2}

As per Newton's third law, the centripetal forces are equal in magnitude. Then:

       a_c_1\times m_1=a_c_2\times m_2

Now you can clear a_c_2:

          a_c_2=\dfrac{a_c_1\times m_1}{m_2}

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4 years ago
Is a gas turbine a heat engine?
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3 years ago
A 25 kVA transformer has an iron loss of 200 W and a full-load copper loss of 350 W. Calculate the transformer efficiency for th
Oksanka [162]

Answer:

a. 97.32 percent

b. 97.92 percent

c. 95.91 percent

Explanation:

given parameters

apparent power, S = 25 kVA

Iron Loss, Piron = 200 W

copper Loss at full load, P cufl = 350 W

let the transformer efficiency at full load be E

let the real power be P out

a.

at full load and 0.8 power factor

P out = Scos∅

        = 25 x 10³ x 0.8

        =20000 W or 20 kW

efficiency at full load is

E =  (P out)/(P out + P iron + P cufl) x 100%

   =  (20000)/(20000 + 200 + 350) x 100%

   = 97.32%

b.

at 70% of full load at unity power factor of 1

Pout = 70% x Scos∅

        = 0.7 x 25 x 10³ x 1

        =  17,500W or 17.5kW

copper loss at 70% full load

P cu0.7fl = (70%)² x 350

                 = (0.7)² x 350

                 = 171.5 W

Iron loss remain the same, P iron = 200 W

Efficiency of transformer at 70% of full load

E =  (P out0.7)/(P out0.7 + P iron + P cufl0.7) x 100%

  =  (17500)/(17500 + 200 + 171.5) x 100%

  = 97.92%

c.

at 40% of full load at a power factor of 0.6

P out = 40% x Scos∅

= 0.4 x 25 x 10³ x 0.6

        =  6000 W or 6 kW

copper loss at 40% full load

P cu0.7fl = (40%)² x 350

                 = (0.4)² x 350

                 = 56 W

Iron loss remain the same, P iron = 200 W

Efficiency of transformer at 40% of full load

E =  (P out0.4)/(P out0.4 + P iron + P cu0.4) x 100%

  =  (6000)/(6000 + 200 + 350) x 100%

  = 95.91%

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