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kakasveta [241]

3 years ago

10

Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M and radius R about an axis perpendicular to the ho

op's plane at an edge.

1 answer:

mote1985 [20]3 years ago

7 0

Answer:

I = 2 MR²

Explanation:

Given that

Radius of the hollow ring ( hoop ) = R

The mass of the hoop = M

We know that mass moment of inertia of a hoop about its center is given as

Io= M R²

By using theorem ,mass moment of inertia at distance d from center is given as

I= Io + m d²

Here ,M= m ,d =R

Now by putting the values in the above equation we get

I = M R² + M R²

I = 2 MR²

Therefore the mass moment of inertia will be 2 M R².

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Express 9/15 as a percentage

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The answer is 60%. If you divide the 9 and 5, you will get 0.6. So you just move the decimal point to the right twice. (:

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When studying atoms, scientists can ignore the ______ force between charged particles that make up the atoms because it is many

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Answer:

Gravitational

Explanation:

When studying atoms, scientists can ignore the gravitational force between charged particles that make up the atoms because it is many millions of times smaller than other forces in the atom.

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You are designing a new solenoid and experimenting with material for each turn. The particular turn you are working with is a ci

andriy [413]

Answer:

The magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.

Explanation:

Given;

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Magnetic field at the center of circular loop is given as;

B = μ₀I / 2R

Where;

μ₀ is constant = 4π x 10⁻⁷ T.m/A

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I is the current in the loop

Substitute the given values in the above equation and calculate the magnitude of the magnetic field;

B = (4π x 10⁻⁷ x 12)/ 0.03

B = 5.0272 x 10⁻⁴ T

Therefore, the magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.

8 0

3 years ago

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3 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

a.

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

b.

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

7 0

3 years ago