Ask question

Ask question

All categories

2 years ago

13

Why does the ring on saturn spin

2 answers:

spayn [35]2 years ago

8 0

Answer: THERE IS NO GRAVITY IN SPACE SO ROCKS SPIN

Explanation:

Mnenie [13.5K]2 years ago

3 0

Answer:

The particles that make up Saturn's rings revolve around the planet. ... As Saturn spins around, so do its rings, rotating with the planet. More exactly, it is the millions of particles that make up these rings which rotate in the same direction as Saturn

Explanation:

You might be interested in

Milk has a density of as much as 64.6 lb/ft3. What is the gage pressure at the bottom of the straw 6.1 inches deep in the milk?

gregori [183]

Answer:

Explanation:

1 inch is 0.0833333feet

6.1 inches is 0.5083 feet

Density = mass/volume

64.6 = mass/0.50833

mass = 64.6 x 0.5083 =32.83618lb

3 0

3 years ago

A three-point bending test is performed on a glass specimen having a rectangular cross section of height d = 5.4 mm (0.21 in.) a

Fudgin [204]

Answer:

5.21e-2mm

Explanation:

Please see attachment

8 0

3 years ago

Because of ____________ people must make choices, and when they choose, they incur a(n)______________.

lbvjy [14]

There must be a photo for me to answer!

3 0

3 years ago

An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an in

Anika [276]

Answer:

Equilibrium Temperature is 382.71 K

Total entropy is 0.228 kJ/K

Solution:

As per the question:

Mass of the Aluminium block, M = 28 kg

Initial temperature of aluminium, $T_{a} = 140^{\circ}C$ = 273 + 140 = 413 K

Mass of Iron block, m = 36 kg

Temperature for iron block, $T_{i} = 60^{\circ}C$ = 273 + 60 = 333 K

At 400 k

Specific heat of Aluminium, $C_{p} = 0.949\ kJ/kgK$

At room temperature

Specific heat of iron, $C_{p} = 0.45\ kJ/kgK$

Now,

To calculate the final equilibrium temperature:

Amount of heat loss by Aluminium = Amount of heat gain by Iron

$MC_{p}\Delta T = mC_{p}\Delta T$

$28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)$

Thus

$T_{e} = 109.71^{\circ}C$ = 273 + 109.71 = 382.71 K

where

$T_{e}$ = Equilibrium temperature

Now,

To calculate the changer in entropy:

$\Delta s = \Delta s_{a} + \Delta s_{i}$

Now,

For Aluminium:

$\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}$

$\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K$

For Iron:

$\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}$

$\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K$

Thus

$\Delta s =-2.025 + 2.253 = 0.228\ kJ/K$

6 0

3 years ago

Which is one of the aspects in PR game marketing?

mr_godi [17]

C, I took the test already.

7 0

3 years ago

Read 2 more answers