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Andrew [12]
2 years ago
9

A marching band consists of rows of musicians walking in straight, even lines. When a marching band performs in an event, such a

s a parade, and must round a curve in the road, the musician on the outside of the curve must walk around the curve in the same amount of time as the musician on the inside of the curve. This motion can be approximated by a disk rotating at a constant rate about an axis perpendicular to its plane. In this case, the axis of rotation is at the inside of the curve.
Knowing that If Beth travels a distance s during time Δt, how far does Alf travel during the same amount of time= (1/4)s. If Alf moves with speed v, what is Beth's speed? Speed in this case means the magnitude of the linear velocity, not the magnitude of the angular velocity.

a. 4v
b. v
c. v/4
Physics
1 answer:
Svetllana [295]2 years ago
7 0

Answer:

a. 4v

Explanation:

Alf moves with speed v

Alf travel during the same amount of time that is Δt = (1/4)s

v = (1/4)s /  Δt = s / 4 Δt

s / Δt  = 4 v

Beth travels a distance s during time Δt,

speed of Beth = s / Δt = 4 v .

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If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat
Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, m_p = 1.67 x 10⁻²⁷ kg

mass of electron, m_e = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

F = \frac{k(q_p)(q_e)}{r^2}

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N

Acceleration of proton is given by;

F = ma

F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2

Acceleration of the electron is given by;

F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2

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The average Intensity of the electromagnetic field due to sunlight at the surface of the earth is 1400 W/m^2. Calculate the maxi
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Answer:

1027 N/C

3.42 x 10⁻⁶ T

Explanation:

I = Intensity of electromagnetic field = 1400 W/m²

E₀ = Maximum value of electric field

Intensity of electromagnetic field is given as

I = (0.5) ε₀ E₀² c

1400 =  (0.5) (8.85 x 10⁻¹²) (3 x 10⁸) E₀²

E₀ = 1027 N/C

B₀ = maximum value of magnetic field

using the equation

E₀ = B₀ c

1027 = B₀ (3 x 10⁸)

B₀ = 3.42 x 10⁻⁶ T

4 0
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