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3 years ago

How long did Skylab orbit the earth?

Physics

1 answer:

maria [59]3 years ago

4 0

Answer:

for abt 36 seconds

Explanation: cuz thats how long it was

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Anyone know how to do this?

Gala2k [10]

Answer:

I think, (remember think) it might be 2.0 m/s

Explanation:

If it's wrong I'm truly sorry.

6 0

3 years ago

The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which

emmainna [20.7K]

Answer:

Explanation:

A ) angular velocity ω = 2π / T

= 2 x 3.14 / 60

= .10467 rad / s

linear velocity v = ω R

= .10467 x 50

= 5.23 m / s

centripetal force = m v² / R

= mg v² / gR

= 834 x 5.23² / 9.8 x 50

= 46.55 N

B )

apparent weight

= mg - centripetal force

= 834 - 46.55

= 787.45 N

C ) apparent weight

= mg + centripetal force

= 834 + 46.55

= 880.55 N.

D )

For apparent weight to be zero

centripetal force = mg

mg = mv² / R

v² = gR

= 9.8 x 50

= 490

v = 22.13 m /s

time period of revolution

= 2π R /v

2 x 3.14 x 50 / 22.13

= 14.19 s

8 0

3 years ago

The 4.0 -kg head of an ax is moving at 3 m/s when it strikes a log and penetrates 0.01m into the log. What is the average force

Elina [12.6K]

Answer:

the average force the blade exerts on the log is 1791.05 N.

Explanation:

Given;

mass of the ax head, m = 4 kg

speed of the ax, v = 3 m/s

depth traveled into the log, d = 0.01 m

The time to traveled through the depth;

$s = (\frac{u+v}{2} )t\\\\0.01 = (\frac{0+3}{2} )t\\\\0.01 = 1.5t\\\\t = \frac{0.01}{1.5} \\\\t = 0.0067 \ s$

The average force the blade exerts on the log;

$F = ma=\frac{mv}{t} = \frac{4 \times 3}{0.0067} = 1791.05 \ N \\\\$

Therefore, the average force the blade exerts on the log is 1791.05 N.

5 0

3 years ago

Way back in the _____________, there lived a man named Isaac Newton who really loved physics.

Elden [556K]

Answer:

i can't click the answer bottom but the answer is "17th to 18th century" i hope this helps

7 0

3 years ago

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each othe

deff fn [24]

Answer:

$\theta_2 - \theta_1 = 156.93 degree$

Explanation:

As we know that the displacement of the particle from the mean position is 1/5 times of its amplitude

so we have

$y = A sin\omega t$

$y = \frac{A}{5}$

so now we have

$\frac{A}{5} = A sin\omega t$

now we have

$\theta_1 = 11.53 degree$

so the phase other particle in opposite direction is given as

$\theta_2 = 180 - 11.53 = 168.46 degree$

so we have phase difference given as

$\theta_2 - \theta_1 = 168.46 - 11.53$

$\theta_2 - \theta_1 = 156.93 degree$

7 0

3 years ago