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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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Answer:

his acceleration rate is -0.00186 m/s²

Explanation:

Given;

initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)

time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s

final position of the car, x₁ = 150 miles = 241,350 m

time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s

The initial velocity is calculated as;

u = 160, 900 m / 3,600 s

u = 44.694 m/s

The final velocity is calculated as;

v = 241,350 m / 6,000 s

v = 40.225 m/s

The acceleration is calculated as;

a = \frac{\Delta V}{\Delta t} = \frac{v- u}{t_1 - t_ 0} = \frac{40.225  - 44.694}{6000-3600} = -0.00186 \ m/s^2\\\\

Therefore, his acceleration rate is -0.00186 m/s²

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Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example
iragen [17]

Given Information:

Magnetic field = B = 1×10⁻³ T

Frequency = f = 72.5 Hz

Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m

Required Information:

Maximum Emf = ?

Answer:

Maximum Emf = 20.66×10⁻¹² volts

Explanation:

The maximum emf generated around the perimeter of a cell in a field is given by

Emf = BAωcos(ωt)

Where A is the area, B is the magnetic field and ω is frequency in rad/sec

For maximum emf cos(ωt) = 1

Emf = BAω

Area is given by

A = πr²

A = π(d/2)²

A = π(7.60×10⁻⁶/2)²

A = 45.36×10⁻¹² m²

We know that,

ω = 2πf

ω = 2π(72.5)

ω = 455.53 rad/sec

Finally, the emf is,

Emf = BAω

Emf = 1×10⁻³*45.36×10⁻¹²*455.53

Emf = 20.66×10⁻¹² volts

Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts

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