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Ede4ka [16]
3 years ago
15

In tokyo ghoul book 2 who is the main character and who taught ken kaneki the kagune

Physics
2 answers:
Viktor [21]3 years ago
4 0
......................
Inessa05 [86]3 years ago
3 0

Answer:by book 2 do you mean the original or re, if you you mean who taught kaneki to control his kagune it was yoshimura (the manager) but if your talking about how to fight with control of his kagune its was Yomo

Explanation: if I'm wrong correct me and i'll do more research

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A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi
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To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

W = F * d

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V_f^2-V_i^2 = 2a\Delta X

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W = 50*9.8*(0.3*30)

W = 4.41kJ

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V_f^2-V_i^2 = 2a\Delta X

Here,

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0-0.9^2=2a(30*0.3)

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a = -\frac{0.9^2}{2*30*0.3}

a = -45*10^{-3}m/s^2

At this point we can calculate the time, which is,

t = \frac{\Delta V}{a}

t = \frac{0-0.9}{-45*10^{-3}}

t = 20s

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

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3 years ago
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3 years ago
The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on
alekssr [168]

Answer:

The change in temperature is \Delta T  = 1795 K

Explanation:

From the question we  are told that

   The temperature coefficient is  \alpha  =  4 * 10^{-3 }\  k^{-1 }

The resistance of the filament is mathematically represented as

           R  =  R_o [1 + \alpha  \Delta T]

Where R_o is the initial resistance

Making the change in temperature the subject of the formula

     \Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]

Now from ohm law

           I = \frac{V}{R}

This implies that current varies inversely with current so

           \frac{R}{R_o} = \frac{I_o}{I}

Substituting this we have

       \Delta T  = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]

From the question we are told that

    I  = \frac{I_o}{8}

Substituting this we have

   \Delta T  = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]

=>     \Delta T  = \frac{1}{3.9 * 10^{-3}} (8 -1 )

        \Delta T  = 1795 K

6 0
4 years ago
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