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3 years ago

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A small spaceship with a mass of only 2.6 ✕ 103 kg (including an astronaut) is drifting in outer space with negligible gravitati

onal forces acting on it. If the astronaut turns on a 12 kW laser beam, what speed will the ship attain in 1.0 day because of the momentum carried away by the beam?

1 answer:

vladimir1956 [14]3 years ago

6 0

Answer:

v = 2.66 10⁻³ m/s

Explanation:

To work this problem we must use the relationship between the amount of movement and energy for an electromagnetic wave

p = U / c

They give us the power of the laser, which is the relationship between work and time. Work is the variation of energy, in a laser beam that is an electromagnetic wave. We have its power

P = W / t

W = P t

Work is the variation of energy W = U

p = P t / c

½ m v = P t / c

v = 2 P t / m c

Let's reduce time to SI units

t = 1.0 day (24 h / 1 day) (3600 s / 1 h) = 86400 s

Let's calculate

v = 2 12 10³ 86400 / (2.6 10³ 3 10⁸)

v = 2.66 10⁻³ m / s

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Answer:

The angle of incident ray is 40°.

Explanation:

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3 years ago

Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

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Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

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$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

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$I=\dfrac {7}{12}\lambda_ 0 L^3$

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$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

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3 years ago

Say I have a series circuit with 20v and four 65 ohm resistors, what is the current in each resistor?

Komok [63]

Data:

$E = 20 V$
$R_{1} = 65\Omega$
$R_{2} = 65\Omega$
$R_{3} = 65\Omega$
$R_{4} = 65\Omega$

<span>Now that we have all the values we need properly identified, simply calculate the equivalent total resistance of the circuit and the intensity of the total electric current using the Ohm's Law:

</span> $R_{T} = R_{1} + R_{2} + R_{3} + R_{4}$
$R_{T} = 65 + 65 + 65+ 65$
$R_{T} = 260\Omega$

<span>Like this:
</span>
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$I_{T} = \frac{20}{ 260 }$
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$\boxed{\boxed{I_{T} \approx 0,07A}}$
Answer:
<span>The intensity of the total electric current
</span> $\boxed{\boxed{I_{T} \approx 0,07A}}$

P.S:. Since the association is in series, the current of 0.07A is the same for all resistors.

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