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Nostrana [21]
2 years ago
15

The function x = (1.2 m) cos[(3πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 9.7 s, what are the (a) di

splacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?
Physics
2 answers:
nlexa [21]2 years ago
8 0

Answer and Explanation:

Let:

x(t)=Acos(\omega t+ \phi)

The equation representing a simple harmonic motion, where:

x=Displacement\hspace{3}from\hspace{3}the\hspace{3}equilibrium\hspace{3}point\\A=Amplitude \hspace{3}of\hspace{3} motion\\\omega= Angular \hspace{3}frequency\\\phi=Initial\hspace{3} phase\\t=time

As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

v(t)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)

Also, you may know these fundamental formulas:

f=\frac{\omega}{2 \pi} \\\\T=\frac{2 \pi}{\omega}

Now, using the previous information and the data provided by the problem, let's solve the questions:

(a)

x(9.7)=1.2 cos((3 \pi *(9.7))+\frac{\pi}{5} ) \approx -0.70534m

(b)

v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s

(c)

a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+\frac{\pi}{5} )\approx -62.653m/s^2

(d)

We can extract the phase of the motion, the angular frequency and the amplitude from the equation provided by the problem:

\phi = \frac{\pi}{5}

(e)

f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz

(f)

T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s

riadik2000 [5.3K]2 years ago
7 0

Answer:

a) -0.705 m ; b) -9,15 m/s ; c) a = -62,65 m/s² ; d) 92,05 rad ; e) 1,5 Hz ; f) 0,666 s

Explanation:

By definition of sum of angles: sin(α-β) = sin(α)*cos(β) - sin(β)*cos(α)

In the particular case of: β = π/2, then: sin(α-π/2) =

In the expression:

<em>x = 1.2*cos((3πrad/s)*t+π/5)</em>

where: ∅ = (3πrad/s)*t+π/5 , es called PHASE.

By definition:

(velocity) v = dx/dt ; (aceleration) a = dv/dt = d²x/dt²

If, we express x like:

x = A*cos(w*t+π/5) (with A = 1.2 m , w = 3π rad/s)

Then:

v = dx/dt = -A*w*sin(w*t+π/5)

a = d²x/dt² = -A*w²*cos(w*t+π/5)

reemplacing in the expresions t = 9.7 seconds:

x = -0.705 m

v = -9,15 m/s

a = -62,65 m/s²

∅ = 92,05 rad

<u>Frequency (f):</u>

f = w/(2*π)

Then reeplacing values:

f = 3*π/2*π = 1,5 Hz

<u>Period:</u>

T = 1/f = 0,666 seconds

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Answer:

Lifting force, F = 21240 N

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It is given that,

Mass of the helicopter, m = 1800 kg

It rises with an upward acceleration of 2 m/s². We need to find the lifting force  supplied by its rotating blades. It is given by :

F = mg + ma

Where

mg is its weight

and "ma" is an additional acceleration when it is moving upwards.

So, F=1800\ kg(9.8\ m/s^2+2\ m/s^2)

F = 21240 N

So, the lifting force supplied by its rotating blades is 21240 N. Hence, this is the required solution.

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Two competing models attempt to explain the motions and changing brightness of the planets: Ptolemy's geocentric model and Coper
AlekseyPX

Answer:

A. Geocentric: This model is Earth Centered . Retrograde motion is explained by epicycles .

B. Heliocentric: This model is Sun centered.  Retrograde motion is explained by the orbital speeds of planets

C. Both geocentric and heliocentric: Epicycles and deferents help explain planetary motion . Planets move in circular orbits and with uniform motion . The brightness of a planet increases when the planet is closest to Earth.

Explanation:

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A 76-W incandescent light bulb operates at 120 V. How many electrons and coulombs flow through the bulb in one day?
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Answer:

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Explanation:

7 0
3 years ago
A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3
Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,  

\text { Electric power }(p)=\frac{V^{2}}{R}

\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}

\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}

For calculating number of turns

\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}

Given that,

80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }

\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}

\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .

We need to find the number of turns in the secondary winding \left(N_{S}\right) to run the bulb at 120W \left(P_{2}\right)

Firstly find the secondary voltage in the transformer use, \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}

\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}

V_{2}^{2}=\frac{120^{2} \times 120}{80}

V_{2}^{2}=\frac{1728000}{80}

V_{2}^{2}=21600

V_{2}=\sqrt{21600}

V_{2}=146.9 \mathrm{V}=V_{S}

Now, finding the number of turns in secondary coil. Use, \frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}

\frac{30}{N_{S}}=\frac{65}{146.9}

N_{S}=\frac{30 \times 146.9}{65}

N_{S}=\frac{4407}{65}N_{S}=67.8

The number of turns in the secondary winding are 67.8 turns.

6 0
3 years ago
* CRIMINOLOGY*
Setler [38]

I think the answer is B.

Hope this helps.

5 0
3 years ago
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