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Lelu [443]
3 years ago
13

Particles at the very outer edge of Saturn’s A Ring are in a 7:6 orbital resonance with the moon Janus. If the orbital period of

Janus is 16 hours 41 minutes, what is the orbital period of the outer edge of Ring A?
Physics
1 answer:
vivado [14]3 years ago
3 0

Answer:

14 hours 18 minutes.

Explanation:

ratio of number of orbits, so it completes 7 orbits in the time Janus does 6.

(16*60+41)*6/7=858 minutes or 14 hours 18 minutes

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Can someone pls help me?!!!! i need it asap!! i’m very stressed
Mariulka [41]

Answer:

Acceleration

Explanation:

Its speed or velocity change

6 0
3 years ago
What is the force acting on a 10kg object that accelerates from 5 m/s to 20 m/s in 5s?
vitfil [10]

Answer:

Option C

Explanation:

v= u + at

20 = 5 + a(5)

15= a(5)

a= 3 m/s²

Force = mass × acceleration

= 10 × 3

= 30 N

3 0
3 years ago
How much work done when .0080 C is moved through a potential difference of 1.5 V? Use W = qV. A.
grin007 [14]

Answer:

0.012 J

Explanation:

We are given:

q = 0.0080C

Potential difference =  1.5V

W=qV

Substituting the values into the equation:

W=0.0080*1.5= 0.012J

8 0
3 years ago
A circular loop of flexible iron wire has an initial circumference of 167 cm, but its circumference is decreasing at a constant
Kisachek [45]

Answer:

Part a)

EMF = 5.6 \times 10^{-3} V

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

Explanation:

As we know that magnetic flux linked with the coil is given as

\phi = \pi r^2 B

now the rate of change in flux is given as

\frac{d\phi}{dt} = 2\pi r \frac{dr}{dt} B

now we know that circumference is decreasing at rate of 15 cm/s

so here we know the length of circumference as

C = 2\pi r

So rate of change in circumference is

\frac{dC}{dt} = 2\pi \frac{dr}{dt}

\frac{1}{2\pi}(15 cm) = \frac{dr}{dt}

final length of circumference at t = 8 s

C = 167 - (15)(8) = 47

Part a)

Now the induced EMF is given as

EMF = (2\pi r)(\frac{1}{2\pi})(0.15)(0.5)

EMF = (0.47)(\frac{1}{2\pi})(0.15)(0.5)

EMF = 5.6 \times 10^{-3} V

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

5 0
3 years ago
The planet Neptune is the farthest planet from the Sun. A satellite is orbiting around Neptune at a distance of 180 km. As the s
stich3 [128]
2,992² + r² = ( r + 180 )²
8,952,064 + r² = r² + 360 r + 32,400
360 r = 8,952,064 - 32,400
360 r = 8,919,664
r = 8,919,664 : 360
r = 24,776.844 km
d = 24,776.844  ·  2 = 49,553688 ≈ 49,554 km
Answer:
The approximate diameter of the planet Neptune is 49,554 km.
4 0
3 years ago
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