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larisa86 [58]
3 years ago
13

Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I0e-t/τ, where I0 is the i

nitial current (at t = 0), and τ is a constant having dimensions of time. Consider a fixed observation point within the conductor.
Required:
a. How much charge passes this point between t = 0 and t = τ?
b. How much charge passes this point between t = 0 and t = 10 τ?
c. What If ? How much charge passes this point between t = 0 and t = [infinity]?

Physics
1 answer:
ELEN [110]3 years ago
5 0

Answer:

Pls see attached file

Explanation:

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How many earths could fit inside jupiter
Snezhnost [94]

1,300 or more.

hope this helped :)

8 0
4 years ago
La masa de un camión es de
Nutka1998 [239]

Answer:

45000 kg and 45 tons  

Explanation:

The expression in kilograms and tons is shown below;

As we know that

1 gr is 0.001 kg

So, 45000000 = 45,000 kg

And,

1 kg = 0.001 tons  

So, 45000 kg = 45 tons    

Therefore the same would be considered

7 0
3 years ago
A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and
alexira [117]

Answer:

A) \omega_f=17.503\ rad.s^{-1}

B) t=55.6822\ s

C) \theta=1312\ rad

Explanation:

Given:

  • mass of flywheel, m=40\ kg
  • diameter of flywheel, d=0.72\ m
  • rotational speed of flywheel, N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}
  • duration for which the power is off, t_0=35\ s
  • no. of revolutions made during the power is off, \theta=180\times 2\pi=360\pi\ rad

<u>Using equation of motion:</u>

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2

\alpha=-0.8463\ rad.s^{-2}

Negative sign denotes deceleration.

A)

Now using the equation:

\omega_f=\omega_i+\alpha.t

\omega_f=15\pi-0.8463\times 35

\omega_f=17.503\ rad.s^{-1} is the angular velocity of the flywheel when the power comes back.

B)

Here:

\omega_f=0\ rad.s^{-1}

Now using the equation:

\omega_f=\omega_i+\alpha.t

0=15\pi-0.8463\times t

t=55.6822\ s is the time after which the flywheel stops.

C)

Using the equation of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2

\theta=1312\ rad revolutions are made before stopping.

3 0
3 years ago
When atoms _____, they break down into atoms of different elements and release energy.
SCORPION-xisa [38]

Answer:

It may be combine?

Do you have multiple choice i can see?

Explanation:

5 0
3 years ago
A box of unknown mass is sliding with an initial speed vi = 4.70 m/s across a horizontal frictionless warehouse floor when it en
malfutka [58]

Here Change in Kinetic Energy = Work Done by Friction

Therefore, substituting the given values to the equation, we get

0.5 * m * (vFinal^2 - vInitial^2) = µ m g * d

Therefore

0.5*( 5.90^2 - Vfinal^2 ) = 0.100*9.8*2.10

Therefore

vfinal = 5.54 m/sec

<span> </span>

8 0
3 years ago
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