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Sophie [7]
2 years ago
11

Two spherical shells have their mass uniformly distrubuted over the spherical surface. One of the shells has a diameter of 2 met

ers and a mass of 1 kilogram. The other shell has a diameter of 1 meter. What must the mass m of the 1-meter shell be for both shells to have the same moment of inertia about their centers of mass? m = nothing kg
Physics
1 answer:
Gekata [30.6K]2 years ago
3 0

Answer:

the mass m of the 1-meter shell = 4kg

Explanation:

The moment of inertia of a spherical shell is given as;

I = mr²

From the question, we want the two sphere's to have the same moment of inetia: Thus,

I1 = I2

Where I1 is the moment of inertia of the first spherical shell while I2 is the moment of Inertia of the second spherical shell.

Thus;

m1•r1² = m2•r2²

Where;

m1 is mass of the first spherical shell which is 1kg

r1 is radius of first shell which is = 2/2 = 1

m2 is mass of the second spherical shell which is unknown.

r2 is radius of second shell which is = 1/2 = 0.5

Now,we are are asked to find the mass of the 1m diameter shell and in this case it's m2.

Thus, let's make m2 the subject of the formula;

m2 = m1 (r1/r2)²

Plugging in the relevant values ;

m2 = 1(1/0.5)² = 4kg

For both to have same moment of inertia, mass of 1m shell must be 4kg

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K=E[(m+M)/M]

Kmin=4.4
8 0
2 years ago
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

5 0
2 years ago
A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva
AleksandrR [38]

Answer:

Explanation:

For the problem, we should have same reynolds number

ρvd/mu = constant

1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600

d = 25.66 cm

5 0
2 years ago
A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i
victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

F= \frac{kqQ}{r^2}

Where,

k = 8.99 * 10^9 N m^2 / C^2.

q = charges of the objects

r= distance/radius

Our values are previously given, so

q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59

Replacing,

F=\frac{kqQ}{r^2}

F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}

F= 4.8423N

The force acting on the block are given by,

F-mgsin\theta = ma

a = \frac{F-mgsin\theta}{m}

a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}a = 10.31m/s^2

Therefore the box is accelerated upward.

3 0
2 years ago
Daniel is 50.0 meters away from a building. He observes that his line-of-sight to the tip of the building makes an angle of 63.0
frez [133]
Since his line of sight 63 degrees makes with the tip of the building

Tan63° =  height of building / Horizontal distance

tan63° =  Height / 50

50tan63° = Height

Height = 50tan63°

Height ≈ 50*1.9626

Height ≈ 98.13 m

Height of the building is ≈ 98.13 m. Mind you in solving for this height we have neglected the height of Daniel.

The height of building actually should be 98.13 m plus the height of Daniel. Since the 63° was measured from his eye level.
4 0
2 years ago
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