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12345 [234]
3 years ago
11

block is attached to an oscillating spring. The function below shows its position (cm) vs. time (s). What is the angular frequen

cy ( ω ω ) of oscillation? x ( t ) = 1.5 cos ( 20 t ) x(t)=1.5cos⁡(20t)
Physics
1 answer:
faltersainse [42]3 years ago
7 0

Answer:

Angular frequency is 20 rad/s.      

Explanation:

Given that,

A block is attached to an oscillating spring. The function below shows its position (cm) vs. time (s) is given by :

x(t)=1.5\cos(20\ t).....(1)

The general equation of oscillating particle is given by :

x(t)=A\cos(\omega t).......(2)

Compare equation (1) and (2) we get :

\omega=20\ rad/s

So, the angular frequency of the oscillation is 20 rad/s.

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What is the magnitude of speed of a person goes 75 meters east followed by 20 meters west in 4 seconds
marshall27 [118]

Answer:23.75 m/sec

Explanation:

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23.75 m/s

4 0
3 years ago
For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

4 0
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Answer:

Number of protons equals the number of neutrons

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Solution:

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<span>v = 0.8c = 2.4E8 m/s</span>


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Mrrafil [7]

water is correct

have a great day


5 0
3 years ago
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