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balu736 [363]
3 years ago
10

Assume that a magnetic field exists and its direction is known. Then assume that a charged particle moves in a specific directio

n through that field with velocity (v). Which rule do you use to determine the direction of force on that particle?
A. first right-hand rule
B. second right-hand rule
C. third right-hand rule
D. fourth right-hand rule
Physics
1 answer:
katen-ka-za [31]3 years ago
4 0
 The first right-hand rule determines the directions of magnetic force, conventional current and the magnetic field.  Given any two of theses, the third can be found. 
The second Right-Hand Rule determines the direction of the magnetic field around a current-carrying wire and vice-versa<span> </span>
So, assuming that a magnetic field <span>exists and its direction is known and assuming that a charged particle moves in a specific direction through that field with velocity (v(, to determine the direction of force on the particle we should use the second right-hand rule.</span>
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DochEvi [55]

Answer:

b. 3 times

Explanation:

Lets take

Coefficient for ordinary glass = α₁

Coefficient for pyrex glass = α₂

 Given that α₁ = 3 α₂

Initial length of both glasses are equal = L

Change in the temperature is also same .= ΔT

We know that change in the length  given as

ΔL =  L α ΔT

Therefore

\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{L\alpha_1\Delta T}{L\alpha_2\Delta T}

\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{3\alpha_2}{\alpha_2}

ΔL₁ = 3ΔL₂

Therefore change in the length of original glass is three time of pyrex glass.

b. 3 times

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A circular pipe of 25-mm outside diameter is placed in an airstream at 25C and 1-atm pressure. The air moves in cross flow over
kifflom [539]

Answer:

f_D = =3.24 N/m

Explanation:

data given

properties of air\nu\ of air =19.31*10^{-6} m2/s

\rho = 1.048 kg/m3

k = 0.0288 W/m.K

WE KNOW THAT

Reynold's number is given as

Re =\frac{VD}{\nu}

      = \frac{ 15*0.025}{19.31*10^{-6}}

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drag coffecient is given as

C_D = \frac{f_D}{A_f\frac{\rho v^2}{2}}

solving for f_D

f_D = C_D A_f*\frac{\rho v^2}{2}

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f_D = 1.1*0.025 *\frac{1.048*15^2}{2}

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4 0
2 years ago
A 4,000 kg boat floats with one-third of its volume submerged. if two more people get into the boat, each of whom weighs 690 n,
AveGali [126]

 

The weightiness of the added water displaced is equivalent to the joined weight of the two extra people who come to be into the boat:


<span>m water g                   = 2 x 690 N</span>

<span>                                   = 1,380 N</span>

<span>
</span>

The mass of the water displace is then


<span>m water g                   = 1,380 N</span>

<span>                                   = 1,380 N / 9.8 m/s^2</span>

<span>                                   = 141 kg</span>

<span>
</span>

Compute the calculation for density for the volume of water displace and practice this outcome for the mass of the water displace to get the answer:


<span>p water                      = mass of water / volume of water</span>

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<span>volume of water        = mass of water / p water</span>

<span>                                  = 141 kg / 1000 kg /m^3 eliminate kilogram</span>

<span>                                  = 0.14 m^3 the additional volume of water that is displaced</span>

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Answer:

A................

Explanation:

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