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Vinil7 [7]
3 years ago
11

21. (a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500-μC charge and flies due west at a

speed of 660. m/s over Earth’s south magnetic pole, where the 8.00 × 10−5 − T magnetic field points straight down into the ground. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect.

Physics
1 answer:
beks73 [17]3 years ago
6 0

Answer:

Explanation:

The picture attached shows the full explanation and i hope it helps. Thank you

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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.

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3 years ago
Calculate the potential difference across a 10 ohm register carrying a current of 2.5 ampere​
Bas_tet [7]

Answer:

using ohm's law

V=IR

V= 10 X 0.2

V = 2 Volt.

Explanation:

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