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Alchen [17]
3 years ago
12

2. A particular planet has a moment of inertia of 9.74 × 1037 kg•m2 and a mass of 5.98 × 1024 kg. Based on these values, what is

the planet’s radius?
Physics
1 answer:
mash [69]3 years ago
3 0

Answer:

6.38\cdot 10^6 m

Explanation:

The planet can be thought as a solid sphere rotating around its axis. The moment of inertia of a solid sphere rotating arount the axis is

I=\frac{2}{5}MR^2

where

M is the mass

R is the radius

For the planet in the problem, we have

M=5.98\cdot 10^{24} kg

I=9.74\cdot 10^{37} kg\cdot m^2

Solving the equation for R, we find the radius of the planet:

R=\sqrt{\frac{5I}{2M}}=\sqrt{\frac{5(9.74\cdot 10^{37}}{2(5.98\cdot 10^{24}}}=6.38\cdot 10^6 m

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A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this
Novay_Z [31]

Answer:

SEE EXPLANATION

Explanation:

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7 0
3 years ago
Given that a pipe having a diameter of 1.5 feet and a height of 10 feet, what is the psi at 5 feet ?
grin007 [14]

Answer:

The pressure is 2.167 psi.

Explanation:

Given that,

Diameter = 1.5 feet

Height = 10 feet

We need to calculate the psi at 5 feet

Using formula of pressure at a depth in a fluid

Suppose the fluid is water.

Then, the pressure is

P=\rho g h

Where, P = pressure

\rho = density

h = height

Put the value into the formula

P=1000\times9.8\times1.524

P=14935.2\ N/m^2

Pressure in psi is

P=2.166167621\ psi

P=2.167\ psi

Hence, The pressure is 2.167 psi.  

4 0
3 years ago
A baseball m=.34kg is spun vertically on a massless string of length l=.52m. the string can only support a tension of tmax=9.9n
larisa86 [58]
<span>4.5 m/s This is an exercise in centripetal force. The formula is F = mv^2/r where m = mass v = velocity r = radius Now to add a little extra twist to the fun, we're swinging in a vertical plane so gravity comes into effect. At the bottom of the swing, the force experienced is the F above plus the acceleration due to gravity, and at the top of the swing, the force experienced is the F above minus the acceleration due to gravity. I will assume you're capable of changing the velocity of the ball quickly so you don't break the string at the bottom of the loop. Let's determine the force we get from gravity. 0.34 kg * 9.8 m/s^2 = 3.332 kg m/s^2 = 3.332 N Since we're getting some help from gravity, the force that will break the string is 9.9 N + 3.332 N = 13.232 N Plug known values into formula. F = mv^2/r 13.232 kg m/s^2 = 0.34 kg V^2 / 0.52 m 6.88064 kg m^2/s^2 = 0.34 kg V^2 20.23717647 m^2/s^2 = V^2 4.498574938 m/s = V Rounding to 2 significant figures gives 4.5 m/s The actual obtainable velocity is likely to be much lower. You may handle 13.232 N at the top of the swing where gravity is helping to keep you from breaking the string, but at the bottom of the swing, you can only handle 6.568 N where gravity is working against you, making the string easier to break.</span>
7 0
3 years ago
Read 2 more answers
You make a capacitor by cutting the 16.0-cm-diameter bottoms out of two aluminum pie plates, separating them by 3.25 mm, and con
dalvyx [7]

Answer:

The capacitance of your capacitor is 5.476 x 10⁻⁵ μF

Explanation:

Given;

diameter of the aluminum pie plates = 16 cm = 0.16 m

separation distance, d = 3.25 mm = 0.00325 m

voltage across the parallel plates = 6 V

C = \frac{\epsilon A}{d}

where;

C is the capacitance of your capacitor

ε is the permittivity of free space = 8.85 x 10⁻¹²  (F/m)

d is separation distance

A is the area of the plate = ¹/₄ (πd²) = 0.25 (π x 0.16²) = 0.02011 m²

C = \frac{8.85*10^{-12} *0.02011}{0.00325} = 5.476 * 10^{-11} \ F = \ 5.476 * 10^{-5} \mu F

Therefore, the capacitance of your capacitor is 5.476 x 10⁻⁵ μF

3 0
3 years ago
A particle of mass m moves under a conservative force with potential energy V ( x )= cx/(x2+a2),where c and a are positive const
Anvisha [2.4K]

Answer:

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V(x) = \frac{c x}{a^2+x^2}

We first look for a position of stable equilibrium. This posiiton must satisfy two considtions, that the first derivative of the potential must vanish at this point and the second derivative must be positive.

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And:

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Therefore:

a)

The position of stable equilibrium is -a

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(c/(ma^3))^1/2

b)

Let's find the maximum amplitude if the particle starts at this point with velocity v

If this is the case, the total energy will be:

(mv^2)/2

And the maximum amplitude will be

x = a^3/c mv^2 = (m v^2 a^3)/ c

7 0
3 years ago
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