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3 years ago

Which of the following conversion factors would you use to change 18 kilometers to meters?

Physics

2 answers:

Tcecarenko [31]3 years ago

6 0

Use choice B. 1 km / 1000 m

from what result that 18 km = 18 000 m

hope helped

Shtirlitz [24]3 years ago

6 0

Answer:

The answer is A) 1,000 meters = 1 Kilometer.

Explanation:

To convert 18 Kilometers into meters, one would have to multiply the figure given in Kilometer by 1000.

Therefore we have 18 x 1000

= 18,000.

So 18 Kilometers is also 18,000 meters when converted.

Cheers!

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3 years ago

(b) If you decrease the length of the pendulum by 25%, how does the new period TN compare to the old period T?

malfutka [58]

Answer:

The new period will be reduced by 50%

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The period of pendulum is given by;

$T= 2\pi\sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g} }\\\\(\frac{T}{2\pi} )^2 = {\frac{L}{g}}\\\\\frac{T^2}{4\pi ^2} = {\frac{L}{g}}\\\\T^2(\frac{g}{4\pi ^2}) = L\\\\ \frac{g}{4\pi ^2}= \frac{L}{T^2}\\\\\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}$

When the length is decreased by 25%, the new length L₂ is given by;

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$\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{T_1^2L_2}{L_1} \\\\T_N^2 = \frac{T^2(0.25L_1)}{L_1}\\\\ T_N^2 =0.25T^2\\\\T_N = \sqrt{0.25T^2}}\\\\T_N = 0.5 T$

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3 years ago

PLEASE HELP

Oksana_A [137]

Answer:

The answer is A

Explanation: hope this helps :)

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3 years ago

Read 2 more answers

4.72 A full-wave bridge-rectifier circuit with a 1-k load operates from a 120-V (rms) 60-Hz household supply through a 12-to-1 s

melisa1 [442]

Answer:

a) 12.74 V

b) Two pairs of diode will work only half of the cycle

c) 8.11 V

d) 8.11 mA

Explanation:

The voltage after the transformer is relationated with the transformer relationshinp:

$V_o=Vrms*\frac{1}{12}\\V_o=10Vrms$

the peak voltage before the bridge rectifier is given by:

$V_{op}=Vo*\sqrt{2}\\V_{op}=14.14V$

The diodes drop 0.7v, when we use a bridge rectifier only two diodes are working when the signal is positive and the other two when it's negative, so the peak voltage of the load is:

$V_l=V_{op}-2(0.7)\\V_l=12.74V$

As we said before only two diodes will work at a time, because the signal is half positive and half negative,so two of them will work only half of the cycle.

The averague voltage on a full wave rectifier is given by:

$V_{avg}=2*\frac{V_l}{\pi}\\V_{avg}=8.11V$

Using Ohm's law:

$I_{avg}=\frac{V_{avg}}{R}\\\\I_{avg}=8.11mA$

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3 years ago