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3 years ago

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A body oscillates with simple harmonic motion along the x-axis. Its displacement varies with time according to the equation x =

0.5 sin (pt+p/3). The acceleration (in m/s2) of the body at t = 1.0 s is approximately
a. 3.5
b. 49
c. 14
d. 43
e. 4.3

1 answer:

abruzzese [7]3 years ago

6 0

I'll tell you how I look at this, although I may be missing something important.

Position = x(t) = 0.5 sin(pt + p/3)

Speed = position' = x'(t) = 0.5 p cos(pt + p/3)

Acceleration = speed' = position ' ' = x ' '(t) = -0.5 p² sin(pt + p/3)

At (t = 1.0),

x ' '(t) = -0.5 p² sin( 4/3 p )

In order to evaluate this, don't I still have to know what 'p' is ? ?

I don't think it can be evaluated with the information given in the question.

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Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

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(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

$U=-\frac{GMm}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

$\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

and so, substituting:

$R=6370 km\\h_A = 5970 km\\h_B = 21200 km$

We find

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(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

$K=\frac{1}{2}\frac{GMm}{r}$

So, the ratio between the two kinetic energies is

$\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

$E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}$

For satellite A, we have

$E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J$

For satellite B, we have

$E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J$

So, satellite B has the greater total energy (since the energy is negative).

(d) $-2.57\cdot 10^8 J$

The difference between the energy of the two satellites is:

$E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J$

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Answer:

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where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

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$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

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soldi70 [24.7K]

<h2>Resultant is 235.54 pounds at an angle 44.16° to X axis.</h2>

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That is force 1 is 100 pound with x axis at 20°

F₁ = 100 cos 20 i + 100 sin 20 j

F₁ = 93.97 i + 34.20 j

That is force 2 is 150 pound with x axis at 60°

F₂ = 150 cos 60 i + 150 sin 60 j

F₂ = 75 i + 129.90 j

F₁ + F₂ = 93.97 i + 34.20 j + 75 i + 129.90 j

F₁ + F₂ = 168.97 i + 164.10 j

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3 years ago