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Dmitry_Shevchenko [17]
3 years ago
14

The acceleration due to gravity at the north pole of Neptune is approximately 10.7m/s2. Neptune has mass 1026kg and radius 25000

km and rotates once around its axis in a time of about 16h.
Part A

What is the gravitational force on an object of mass 5.5kg at the north pole of Neptune?

Part B

What is the apparent weight of this same object at the Neptune's equator? (Note that Neptune
Physics
1 answer:
Scilla [17]3 years ago
7 0

Answer:

(A) Force on the object will be 58.85 N

(B) Apparent weight will be equal to 57.21 N

Explanation:

We have given mass of Neptune m = 1026 kg

Radius R = 25000 Km

Time period = 16 hour

We know that 1 hour = 3600 sec

So 16 hour = 16×86400 = 57600 sec

Acceleration due to gravity at north pole of Neptune g=10.7m/sec^2

(A) Mass of the object m = 5.5 kg

So gravitational force on the object F_g=mg=5.5\times 10.7=58.85N

(B) Velocity will be equal to v=\frac{2\pi R}{T}=\frac{2\times 3.14\times25000\times 1000}{57600}=2725.694m/sec

Apparent weight of the object will be equal to w=f_g-\frac{mv^2}{r}

W=58.85-\frac{5.5\times (2725.694)^2}{25000\times 1000}=57.21N

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Which describes the average velocity of an ant traveling at a constant speed
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Answer: A. The total displacement divided by the time and  C. The slope of the ant's displacement vs. time graph.

Explanation:

Hi! The question seems incomplete, but I found the options on the internt:

A. The total displacement divided by the time.

B. The slope of the ant's acceleration vs. time graph.

C. The slope of the ant's displacement vs. time graph.

D. The average acceleration divided by the time.

Now, since we know the ant is travelling at a constant speed, its average velocity V will be expressed by the following equation:

V=\frac{d}{t}

Where:

d is the ant's total displacement

t is the time it took to the ant to travel to the kitchen

Hence one of the correct options is: A. The total displacement divided by the time

On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is: C. The slope of the ant's displacement vs. time graph.

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3 years ago
Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros
kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

v^2-u^2=2gh

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v = 0 is the velocity at the maximum height

u is the initial velocity

g=-9.8 m/s^2 is the acceleration of gravity

Solving for u,

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The time needed to reach the maximum height can now be found by using the equation

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Solving for t,

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Using again the equation

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we find

v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s

So the time to go from h' to h is

\Delta t = t-t'=0.52-0.32=0.20 s

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

\Delta t=2\cdot 0.20 = 0.40 s

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This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

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v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

\Delta t = 2\cdot 0.04 =0.08 s

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3 years ago
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An astronaut holds a rock 100m above the surface of Planet X. The rock is then thrown upward with a speed of 15m/s, as shown in
Harlamova29_29 [7]

The acceleration due to gravity of the planet X is 1 m/s².

The given parameters;

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  • initial velocity of the rock, u = 15 m/s
  • time of motion of the rock, t = 10 s

The acceleration due to gravity is calculated as follows;

h = ut - \frac{1}{2} gt^2\\\\100 = 15(10) - (0.5\times 10^2)g\\\\100 = 150 - 50g\\\\50g = 150-100\\\\50g = 50\\\\g = 1 \ m/s^2

Thus, the acceleration due to gravity of the planet X is 1 m/s²

Learn more here: brainly.com/question/24564606

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2 years ago
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