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patriot [66]
3 years ago
13

Which planets have rings

Physics
2 answers:
Arlecino [84]3 years ago
8 0
Jupiter, Saturn, Uranus, and Neptune all have rings. 
skelet666 [1.2K]3 years ago
5 0
Saturn, Neptune, Uranus 
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The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom ris
Andrew [12]

Explanation:

Let us assume that forces acting at point B are as follows.

        \sum F_{x} = 0

        T + F_{AB} Sin 60 = 0 ...... (1)

       \sum F_{y} = 0

       F_{AB} Cos 60 + W = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

               \sigma_{allow} = \frac{T}{A}

       T = (20 \times 1000) \frac{\pi}{4} \times (0.5)^{2}

          = 3925 kip

From equation (1),   F_{AB}Sin (60^{o})  = -3925

               F_{AB} \times -0.304 8 = -3925

             F_{AB} = 12877.29 kip

From equation (2),    -12877.29 (Cos 60) + W = 0

         -12877.29 kip \times \frac{1}{2} + W = 0

                           W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

5 0
3 years ago
A 30 g bullet moving a horizontal velocity of 500 m/s comes to a stop 12 cm within a solid wall. (a) what is the change in the b
Sidana [21]
M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet

By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
      = 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.

The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J

If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN

Answer:
(a) 3750 J
(b) 62.5 kN

7 0
4 years ago
Please help me important will give brainliest
irinina [24]
Pretty sure it’s C. all the others are speeding up. acceleration means gradually (over time) getting faster. So it’s C.
7 0
3 years ago
Setting goals that match your lifestyle helps improve chances that they will be reached.
Tasya [4]

Answer:

TRUE!!!!!!

Explanation:

7 0
3 years ago
Read 2 more answers
a. How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter t
Agata [3.3K]

Answer:

Explanation:

The electric field outside the sphere is given as,

E = k Q /r²

here Q = n x 1.6 x 10⁻¹⁹ C

where n is the number of electons

if the dimeter of sphere d= 25 cm= 0.25 m

then the radius r = 0.125 m

we get

n= E r²/ k x 1.6 x 10⁻¹⁹ C

n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)

n = 14664731646

5 0
3 years ago
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