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Gre4nikov [31]
4 years ago
5

Which feature of the sun appears in cycles of about 11 years?

Physics
1 answer:
goldfiish [28.3K]4 years ago
8 0
The Sun's magnetic field goes through a cycle, called the solar cycle. Every 11 years or so, the Sun's magnetic field completely flips. This means that the Sun's north and south poles switch places. Then it takes about another 11 years for the Sun's north and south poles to flip back again.
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The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any wo
wariber [46]

Explanation:

Expression for energy balance is as follows.

        \Delta E_{system} = E_{in} - E_{out}

or,          E_{in} = E_{out}

Therefore,  

         m(h_{1} \frac{v^{2}_{1}}{2}) = m (h_{2} \frac{V^{2}_{2}}{2})

          h_{1} + \frac{V^{2}_{1}}{2} = h_{2} + \frac{V^{2}_{2}}{2}

Hence, expression for exit velocity will be as follows.

           V_{2} = [V^{2}_{1} + 2(h_{1} - h_{2})^{0.5}

                      = V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}

As C_{p} for the given conditions is 1.007 kJ/kg K. Now, putting the given values into the above formula as follows.

       V_{2} = V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}                    

                  = [(350 m/s)^{2} + 2(1.007 kJ/kg K) (30 - 90) K \frac{1000 m^{2}/s^{2}}{1 kJ/kg}]^{0.5}

                 = 40.7 m/s

Thus, we can conclude that velocity at the exit of a diffuser under given conditions is 40.7 m/s.

5 0
3 years ago
Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m
Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5  *10⁻⁹C

r₁=0.840m

r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}

sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246

cos\beta =\frac{1}{\sqrt{1.64} } =0.7808

Calculation of the electric field at point P due to q1

Ep₁x=0

Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2}  } =\frac{8.99*10^{9}*2.9*10^{-9}  }{0.84^{2} } =36.95\frac{N}{C}

Calculation of the electric field at point P due to q2

Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2}  } =-21.4\frac{N}{C}

Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2}  } =-17.11\frac{N}{C}

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0
3 years ago
A visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simp
masha68 [24]

To solve this problem we must rely on the equations of the simple harmonic movement that define the period as a function of length and gravity as

T = 2\pi \sqrt{\frac{l}{g}}

Where

l = Length

g = Gravity

Re-arrange to find L,

L = g (\frac{T}{2\pi})^2

Our values are given as

g = 9.81m/s

T = 10.1s

Replacing,

L = g (\frac{T}{2\pi})^2

L = (9.81) (\frac{10.1}{2\pi})^2

L = 25.348m

Therefore the height would be 25.348m

5 0
3 years ago
Explain what kind of heat transfer occurs when you burn yourself on a hot car seat in the summer.
liubo4ka [24]
Conduction is a mode of transfer of heat there
6 0
4 years ago
Complete the sentences to describe the difference between speed and velocity.
xeze [42]

Answer:

velocity =displacement/time

and speed =distance/time

6 0
3 years ago
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