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3 years ago

Why does a compass work differently on a moon then it would on earth

1 answer:

8 0

A compass would work differently on the moon than the Earth because Earth's magnetic field is between 25,000 and 65,000 NanoTeslas. Whereas, the Moons magnetic field is about 2,500-6,500 and some parts of the moon have nonexistent magnetic fields.

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A pivoted beam supports a load P at one end and a load Q at the other end.

Alecsey [184]

Answer:

P Q

c) 30 10

Explanation:

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3 years ago

Please helo me to 1st question

k0ka [10]

Explanation:

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6×60

<h2>360 seconds </h2><h2 /><h2 />

6 min in hours:

<h2>0.1 hours</h2>

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3 years ago

Roberta is going over the form with a patient; the form includes such information as the name of the patient, the

velikii [3]

Answer:

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4 years ago

A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be

lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × $10^{-3}$ m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ) ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v / 2 ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 ) / 2

0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 = 108

ω² = 108 / 7.160 × $10^{-5}$

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

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4 years ago

A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-g

dsp73

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

$f = \frac{4.04}{60} =0.067 rev/s$

here we know that angular frequency is given as

$\omega = 2\pi f$

$\omega = 2\pi(0.067)$

$\omega = 0.42 rad/s$

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

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3 years ago