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Trava [24]
3 years ago
15

A machine part has the shape of a solid uniform sphere of mass 205 g and diameter 4.10 cm. It is spinning about a frictionless a

xle through its center, but at one point on its equator, it is scraping against metal, resulting in a friction force of 0.0200 N at that point.
(a) Find its angular acceleration. Let the direction the sphere is spinning be the positive sense of rotation.
(b) How long will it take to decrease its rotational speed by 22.0 rad/s?

Physics
1 answer:
kondor19780726 [428]3 years ago
3 0

Answer:

Answer is in the following attachment.

                                                                                                                                                                                                 

Explanation:

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The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 77.0 nC. The plates are in va
WINSTONCH [101]

Answer:

Part A: 7500 V

Part B: 2.899×10⁻³ m²

Part C: 10.27 pF or 10.27×10⁻¹² F

Explanation:

Part A:

Applying,

E = V/d................ Equation 1

Where E = electric field intensity between the plates, V = potential difference between the plates, d = distance of separation between the plates

make V the subject of the equation above,

V = Ed............. Equation 2

Given: E = 3.0×10⁶ V/m, d = 2.5 mm = 2.5×10⁻³ m

Substitute into equation 2

V =  3.0×10⁶ (2.5×10⁻³ )

V = 7.5×10³ V

V = 7500 V

Part B:

Using,

E = Q/(e₀A).................... Equation 3

Where Q = Charge on each plate of the capacitor, A = Area of each plate, e₀ = constant = dielectric = permitivity of free space

make A the subject of the equation,

A = Q/(e₀E).............. Equation 4

Given: Q = 77 nC = 77×10⁻⁹ C, E = 3.0×10⁶ V/m

Constant: e₀ = 8.854×10⁻¹² F/m

Substitute into equation 4

A = 77×10⁻⁹/(8.854×10⁻¹²× 3.0×10⁶)

A = 77×10⁻⁹/(26.562×10⁻⁶)

A = 2.899×10⁻³ m²

A = 2.899×10⁻³ m².

Part C:

Using,

Q = CV.................. Equation 5

Where C = Capacitance of the capacitor

make C the subject of the equation

C = Q/V.............. Equation 6

Given: Q = 77 nC = 77×10⁻⁹ C, V = 7500 V

Substitute into equation 6

C = 77×10⁻⁹/7500

C = 10.27×10⁻¹² F

C = 10.27 pF

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3 years ago
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Answer:

La rapidez con que el fármaco es absorbido en el torrente sanguíneo depende, en parte, del suministro de sangre al músculo: cuanto menor sea el aporte de sangre, más tiempo necesitará el fármaco para ser absorbido.

Para la administración por vía intravenosa se inserta una aguja directamente en una vena

Explanation:

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The correct answer is C
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