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dem82 [27]
3 years ago
11

The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s2. Neptune has mass 1.02×1026 kg and radiu

s 2.46×104 km and rotates once around its axis in a time of about 16 h.
a)What is the gravitational force on a 4.60-kg object at the north pole of Neptune? Express your answer with the appropriate units.
b)What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.) Express your answer with the appropriate units.

Physics
1 answer:
larisa [96]3 years ago
5 0

Answer: (a) The gravitational force on the object at the North Pole of Neptune is 51.7N

(b) The apparent weight of the object at Neptune's equator is 50.4N

Explanation: Please see the attachments below

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An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the
alexandr402 [8]

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

5 0
3 years ago
A car moves along a curved road of diameter 2 km. If the maximum velocity for safe driving on this path is 30 m/s, at what angle
Leto [7]
The maximum velocity in a banked road, ignoring friction, is given by;

v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.

Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°

Therefore, the road has been banked at 5.24°.
4 0
3 years ago
If you're driving towards the sun late in the afternoon, you can reduce the glare from the road by wearing sunglasses that permi
AveGali [126]

Correct answer choice is:

C. Polarized in a vertical plane.

Explanation:

Polarized sunglasses give excellent glare shield, particularly on the water. Polarized lenses include a specific filter that prevents this type of strong reflected light, diminishing glare.

This is because when you angle one polarized lens to different perpendicularly, they prevent glare both horizontally and vertically. The polarized lenses are enduringly tinted sunglasses that exceedingly decrease glare.

7 0
3 years ago
What is the volume of an object that has a density of 65g/cm3 and a mass of 130g.
lora16 [44]

Density <em>ρ</em> is mass <em>m</em> per unit volume <em>v</em>, or

<em>ρ</em> = <em>m</em> / <em>v</em>

Solving for <em>v</em> gives

<em>v</em> = <em>m</em> / <em>ρ</em>

So the given object has a volume of

<em>v</em> = (130 g) / (65 g/cm³) = 2 cm³

5 0
3 years ago
Planet a and planet b are in circular orbits around a distant star. planet a is 7.8 times farther from the star than is planet
gogolik [260]
To find the ratio of planetary speeds Va/Vb we need the orbital velocity formula: 

V=√({G*M}/R), where G is the gravitational constant, M is the mass of the distant star and R is the distance of the planet from the star it is orbiting. 

So Va/Vb=[√( {G*M}/Ra) ] / [√( {G*M}/Rb) ], in our case Ra = 7.8*Rb 

Va/Vb=[ √( {G*M}/{7.8*Rb} ) ]  / [√( {G*M}/Rb )], we put everything under one square root by the rule: (√a) / (√b) = √(a/b) 

Va/Vb=√ [ { (G*M)/(7.8*Rb) } / { (G*M)/(Rb) } ], when we cancel out G, M and Rb we get:

Va/Vb=√(1/7.8)/(1/1)=√(1/7.8)=0.358 so the ratio of Va/Vb = 0.358.  
6 0
3 years ago
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