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dezoksy [38]
3 years ago
8

Which one of the following planets does not have rings surrounding it? A. Saturn B. Uranus C. Mars D. Jupiter

Physics
2 answers:
natali 33 [55]3 years ago
7 0

Answer:

mars aint got rings cuh

Explanation:

MrMuchimi3 years ago
5 0
Mars -No 
b. Uranus - Uranus has 9 brighter rings as well as several fainter rings. 
c. Neptune - Neptune has several faint rings around it. 
d. Saturn - Saturn has bright ice rings, 
e. Jupiter - Jupiter has faint, narrow rings.
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Light wave travel faster than sound waves
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Calculate the gravitational attraction between two objects which masses are: mass A: 2.5kg, mass B: 5kg. The distance between th
jonny [76]

From the calculation, the gravitational force of attraction is  1.33 * 10^-14 N.

<h3>What is the gravitational force?</h3>

The gravitational force is an attractive force that acts between any two masses.

It is given by;

F = Gm1m2/r^2

F = 6.67 * × 10−11 * 2.5 * 5/(250)^2

F  = 83.4  × 10−11 /62500

F= 1.33 * 10^-14 N

Learn more about gravitational force:brainly.com/question/12528243

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4 0
2 years ago
An ice skater has a moment of inertia of 5.0 kgm2 when her arms are outstretched. at this time she is spinning at 3.0 revolution
givi [52]
With arms outstretched,
Moment of inertia is I = 5.0 kg-m².
Rotational speed is ω = (3 rev/s)*(2π rad/rev) = 6π rad/s
The torque required is
T = Iω = (5.0 kg-m²)*(6π rad/s) = 30π 

Assume that the same torque drives the rotational motion at a moment of inertia of 2.0 kg-m².
If u = new rotational speed (rad/s), then
T = 2u = 30π
u = 15π rad/s
   = (15π rad/s)*(1 rev/2π rad)
   = 7.5 rev/s

Answer: 7.5  revolutions per second.

7 0
3 years ago
The specific heat of fat is roughly 1,700 J/(kgoC) whereas that for water is around 4200 J/(kgoC). A 0.63 kg solution of lipids
BartSMP [9]

Answer:

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

Explanation:

Total heat content of the fat = heat content of water +heat content of the lipids

Let it be Q

the Q= (mcΔT)_lipids + (mcΔT)_water

total mass of fat  M= 0.63 Kg

Q= heat supplied = 100 W in 5 minutes

ΔT= 20°C

c_lipid= 1700J/(kgoC)

c_water= 4200J/(kgoC)

then,

100\times5\times60= m(1700)20+(0.63-m)(4200)20

solving the above equation we get

m= 0.46 kg

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

3 0
3 years ago
Question 4 of 10
kiruha [24]
B hey what do u know i took that test to
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