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3 years ago

Question 4 (1 point)

Physics

1 answer:

irga5000 [103]3 years ago

6 0

Answer:

ω₁ = 8 rad/s

Explanation:

Assuming

1) the clay sticks to the disc we can apply conservation of angular momentum

2) That the disc moment of inertia is 1.5 kg•m²

Initial angular momentum = L = I₀ω₀ = 1.5(14) = 21 kg•m²/s

After the clay joins the party, the moment of inertia is

I₁ = 1.5 + 0.5(1.5²) = 2.625 kg•m²

21 = 2.625ω₁

ω₁ = 8 rad/s

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Explanation:

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3 years ago

During which segments is kinetic energy decreasing?<br> A)1,2,3<br> B) 4,5<br> C)1,3,5<br> D)2,4

Nutka1998 [239]

Answer:

Explanation:

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3 years ago

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A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago

A 165-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,015 A. If the conductor is copper

elena55 [62]

Answer:

22.1 years

Explanation:

Since the current in the wire is I = nevA where n = electron density = 8.50 × 10²⁸ electrons/cm³ × 10⁶ cm³/m³= 8.50 × 10³⁴ electrons/m³, e = electron charge = 1.602 × 10⁻¹⁹ C, v = drift velocity of electrons and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.00 cm = 2 × 10⁻² m

Making v subject of the formula, we have

v = I/neA

So, v = I/neπd²/4

v = 4I/neπd²

Since I = 1,015 A, substituting the values of the other variables into the equation, we have

v = 4I/neπd²

v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π ×(2 × 10⁻² m)²]

v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π × 4 × 10⁻⁴ m²]

v = (1,015 A)/[42.779 × 10¹¹ electronsC/m]

v = 23.73 × 10⁻¹¹ m/s

v = 2.373 × 10⁻¹⁰ m/s

Since distance d = speed, v × time, t

d = vt

So, the time it takes one electron to travel the full length of the cable is t = d/v

Since d = distance moved by free charge = length of transmission line = 165 km = 165 × 10³ m and v = drift velocity of charge = 2.373 × 10⁻¹⁰ m/s

t = 165 × 10³ m/2.373 × 10⁻¹⁰ m/s

t = 69.54 × 10⁷ s

t = 6.954 × 10⁸ s

Since we have 365 × 24 hr/day × 60 min/hr × 60 s/min = 31536000 s in a year = 3.1536 × 10⁷ s

So, 6.954 × 10⁸ s = 6.954 × 10⁸ s × 1yr/3.1536 × 10⁷ s = 2.21 × 10 yrs = 22.1 years

It will take one electron 22.1 years to travel the full length of the cable

3 0

3 years ago