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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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Answer:

2m head start or else you done for

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2 years ago
A 4.5-kg, three legged stool supports a 89-kg person. If each leg of the stool has a cross-sectional diameter of 2.8 cm and the
kkurt [141]

Answer:

4.96 × 10⁵ Pa

Explanation:

F = mg

F = (m_{person}+m_{stool})g\\\\F =  (4.5 + 89)*9.8\\\\F = 916.3 N

This force is evenly distributed on the three leg

radius, r = d/2

= 2.8 / 2

= 1.4 cm = 0.014 m

total cross sectional area of the three legs, A = 3*pi*r^2

= 3\times\pi \times0.014^2\\\\= 1.847\times10^-^3m^2

Pressure due to weight,

P = Weight/A

P = / 1.847 × 10⁻³\\P = \frac{ 916.3N}{1.847\times10^-^3} \\\\P= 496032.9Pa

P = 4.96 × 10⁵ Pa

8 0
2 years ago
What are the two ost important processes in the oxygen in and out of the atmospher
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Decomposing - When plants and animals die, they decompose. This process uses up oxygen and releases carbon dioxide. Rusting - This is also called oxidation. When things rust they use up oxygen.
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3 0
3 years ago
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Why was newton's invention of calculus significant?
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8 0
3 years ago
A parallel-plate capacitor with circular plates of radius R is being charged by a battery, which provides a constant current. At
NikAS [45]

To solve this problem it is necessary to apply the concepts related to the magnetic field.

According to the information, the magnetic field INSIDE the plates is,

B=\frac{1}{2} \mu \epsilon_0 r

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\mu =Permeability constant

\epsilon_0 =Electromotive force

r = Radius

From this deduction we can verify that the distance is proportional to the field

B \propto r

Then the distance relationship would be given by

\frac{r}{R} = \frac{B}{B_{max}}

r =\frac{B}{B_{max}} R

r = \frac{0.5B_{max}}{B_{max}}R

r = 0.5R

On the outside, however, it is defined by

B = \frac{\mu_0 i_d}{2\pi r}

Here the magnetic field is inversely proportional to the distance, that is

B \not\propto r

Then,

\frac{r}{R} = \frac{B_{max}{B}}

r = \frac{B_{max}{B}}R

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7 0
3 years ago
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