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galina1969 [7]
3 years ago
14

Question 5 (1 point)

Physics
1 answer:
Tom [10]3 years ago
4 0

Answer: 18000 coulombs

Explanation:

Given that:

Current, I = 5.0A

Electric charge Q = ?

Time, T = 1 hour

(The SI unit of time is seconds. So, covert 1.0 hour to seconds)

If 1 hour = 60 minutes and 60 seconds = 1 minute

Then, 1.0 hour = (60 x 60)

= 3600 seconds

Since electric charge, Q = current x time

i.e Q = I x T

Q = 5.0 A x 3600 seconds

Q = 18000 coulombs

Thus, 18000 coulombs of charge flows through the lamp in this time.

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Answer:

\mathbf{V_x = 3.25 \ cm/s}

\mathbf{V_y = 1.29\ cm/s}

Explanation:

Given that:

The radius of the table r = 16 cm  = 0.16 m

The angular velocity = 45 rpm

= 45 \times \dfrac{1}{60}(2 \pi)

= 4.71 rad/s

However, the relative velocity of the bug with turntable is:

v = 3.5 cm/s = 0.035 m/s

Thus, the time taken to reach the bug to the end is:

t = \dfrac{r}{v}

t = \dfrac{0.16}{0.035}

t = 4.571s

So the angle made by the radius r  with the horizontal during the time the bug gets to the end is:

\theta = \omega t

\theta = 4.712 \times 4.571

\theta = 21.54^0

Now, the velocity components of the bug with respect to the table is:

V_x = Vcos \theta

V_x = 0.035 \times cos (21.54^0)

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\text {V_x = 3.25 \ cm/s}\mathbf{V_x = 3.25 \ cm/s}

Also, for the vertical component of the velocity V_y

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Is it true that mproving overall fitness may not be possible for people who have physical limitations.
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A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If
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Answer:

vcyl / vsph = 1.05

Explanation:

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       K_{trans} = \frac{1}{2}* M* v_{cm} ^{2}  (1)

  • The rotational part can be expressed as follows:

       K_{rot} = \frac{1}{2}* I* \omega ^{2}  (2)

  • where I = moment of Inertia regarding the axis of rotation.
  • ω = angular speed of the rotating object.
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       v = \omega * R (3)

  • For a solid cylinder, I = M*R²/2 (4)
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       K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2}  (5)

  • Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

       K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2}  +\frac{1}{4}* M* v_{cmc}^{2}  =  \frac{3}{4}* M* v_{cmc} ^{2} (6)

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        I_{sph} = \frac{2}{3} * M* R^{2} (7)  

       K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2}  (8)

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       \frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)

  • This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.
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