The typical unit for a period used with Kepler's third law is

1 answer:

6 0

Well, if you're using the law to work with periods of Earth satellites,
then the most convenient unit is going to be 'hours' for the largest
orbits, or 'minutes' for the LEOs.

But if you're using it to work with periods of planets, asteroids, or
comets, then you'd be working in days or years.

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A resistor uses 100 w of power when connected to 120 v emf. What is the current through the same resistor when connected to a 22

cricket20 [7]

Answer:

Explanation:

Let the resistance of resistor be R .

Power of resistor V² / R , where V is potential applied .

V² / R = 100

120² / R = 100

R = 120² / 100

= 144 ohm .

Now potential diff applied = 220 V

current = potential diff / resistance

= 220 / 144

= 1.53 A approx .

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The position of a 55 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. determine the velocity at t=0.4

NNADVOKAT [17]

The position of the mass is given by (in cm):
$x(t)=2 \cos (10 t)$
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Substituting t=0.40 s, we can find the velocity at this time:
$v(0.40 s)= -20 \sin (10 \cdot 0.4)=15 cm/s=15 \cdot 10^{-2}m/s$

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2 years ago

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An object of mass 0.400.40 kg, hanging from a spring with a spring constant of 8.08.0 N/m, is set into an up-and-down simple har

sineoko [7]

Answer:

Acceleration will be equal to $2m/sec^2$