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3 years ago

Need help on this please

Physics

1 answer:

Ratling [72]3 years ago

8 0

question 2 answer is ALL OF THE ABOVE

question 3 answer is WARM AND MOIST.

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A circuit consists of a resistance R, and an inductance L in parallel connected, which is in series with a second resistance R2.

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Answer:

The impedance of the inductor is ω L:

This is in parallel wth R giving a resistance of the parallel combination:

1 / Rp = 1 / ω L + 1 / R

Rp = (ω L * R) / (ω L + R) for the impedance of the parallel combination

Total resistance (impedance) Rt = Rp + R2 = Z

One can use the term impedance to show that the resulting current is not in phase with the resulting voltage

phase angle φ - cos φ = ω L / Rt

V = I Z shows the relation of voltage and urrent

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Two people are pushing a couch from the same side. One person pushes with a force of 2 N. The other person pushes with a force o

SashulF [63]

Answer:

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Explanation:

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3 years ago

During the water cycle, solar energy causes water to change from a _____ to a ______.

tankabanditka [31]

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3 years ago

Read 2 more answers

A catapult with a radial arm 3.81 m long accelerates a ball of mass 18.2 kg through a quarter circle. The ball leaves the appara

saw5 [17]

Answer:

(a) $\alpha = 53.73 m/s^2$

(b) I =428 $kgm^2$

(c) $\tau = 428 \times 53.73 = 22996 .44Nm$

Explanation:

GIVEN

mass = 18.2 kg

radial arm length = 3.81 m

velocity = 49.8 m/s

mass of arm = 22.6 kg

we know using relation between linear velocity and angular velocity

$\omega = \frac{v}{l}$

$\omega = \frac{49.8}{3.81} \\\omega = 12.99 rad/s$

for angular acceleration, use the following equation.

$\omega _{f}^2 = \omega_{i}^2+2\alpha\theta$

since $\omega _{i} = 0$

here for one circle is 2 π radians. therefore for one quarter of a circle is π/2 radians

so for one quarter $\theta = \frac{\pi }{2}$

$(12.99)^2 = 2\alpha(\frac{\pi }{2})$

on solving

$\alpha = \frac{168.74}{\pi }\\\alpha = 53.73 m/s^2$

(b)

For the catapult,

moment of inertia

$I = \frac{1}{2}MR^2$

$I = \frac{1}{2} \times 22.6\times 3.81 \times 3.81\I = 164kg m^2$

For the ball,

$I = MR^2$

$I = 18.2 \times 14.51$

$I = 264 kgm^2$

so total moment of inertia = 428 $kgm^2$

(c)

$\tau = I\alpha$

$\tau = 428 \times 53.73 = 22996 .44Nm$

3 0

4 years ago