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Stella [2.4K]
3 years ago
13

How do you do this question?

Physics
1 answer:
PIT_PIT [208]3 years ago
8 0

Explanation:

Assuming a uniform mass, let's say ρ is the mass per area density.

ρ = M / (πR²)

Let's look at this as the difference of two disks, a large one and a small one.

The moment of inertia of the large disk is:

I = 1/2 MR²

The mass of the small disk is:

m = ρ πr²

m = M / (πR²) πr²

m = M (r/R)²

Using parallel axis theorem, the moment of inertia of the small disk is:

I = 1/2 mr² + md²

I = 1/2 M (r/R)² r² + M (r/R)² d²

I = 1/2 M (r²/R)² + M (rd/R)²

The total moment of inertia is:

I = 1/2 MR² − 1/2 M (r²/R)² − M (rd/R)²

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3 years ago
The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC
enyata [817]

Answer:

The y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

Explanation:

<u>Given:</u>

  • Electric field in the region, \vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.
  • Charge placed into the region, q = 10.4\ nC = 10.4\times 10^{-9}\ C.

where, \hat i,\ \hat j are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.

Thus, the y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

3 0
3 years ago
Problem 1 Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed
ANTONII [103]

Answer:

markers are 29.76 m far apart in the laboratory

Explanation:

Given the data in the question;

speed of particle = 0.624c

lifetime = 159 ns = 1.59 × 10⁻⁷ s

we know that; c is speed of light which is equal to 3 × 10⁸ m/s

we know that

distance = vt

or s = ut

so we substitute

distance = 0.624c × 1.59 × 10⁻⁷ s

distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s

distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s

distance =  29.76 m

Therefore, markers are 29.76 m far apart in the laboratory

3 0
3 years ago
What is the momentum of an 18-kg object moving at 0.1 m/s ?
Anit [1.1K]

Answer:

1.8 m/s

Explanation:

here's the solution : -

momentum = mass × velocity

=》18 × 0.1

=》1.8 m/s

6 0
3 years ago
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