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Aleksandr [31]
3 years ago
12

A small asteroid with a mass of 1500 kg moves near the earth. At a particular instant the asteroid’s velocity is ⟨3.5 × 104, −1.

8 × 104, 0⟩ m/s, and its position with respect to the center of the earth is ⟨8 × 106, 9 × 106, 0⟩ m. (The center of the earth is at the origin of the coordinate system.) What is the (approximate) new momentum of the asteroid 1.5 × 103 seconds later?
Physics
1 answer:
zalisa [80]3 years ago
4 0

Answer:

P_{f} =(5.7 x 10^{7 i - 2.24 x 10^{7 j) kgm/s

Explanation:

Due to earths gravity, force on asteroid is given by:

F= \frac{Gm_{1}m_{2} }{r^{2} } r^

Plugging in the values, we have

F= [(6.67x10^{-11})(1500)(5.97 x 10^{24})(8x10^{6}i + 9x10^{6 j)] / ((8x10^{6})² + (9x10^{6 )²)^{1.5}

F= 2736 i^ + 3078 j^

In order find the final momentum of the Asteroid, apply impulse momentum theorem

P_{f} = P_{i + FΔt

P_{f} = 1500(3.5 x 10^{4 i - 1.8x10^{4 j) + (2736i + 3078j)(1.5x10^{3)

P_{f} =(5.7 x 10^{7  i- 2.24 x 10^{7 j)kgm/s

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Answer:

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Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

y = \frac{\lambda L}{d}\\\\d  =\frac{\lambda L}{y}\\\\

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\

<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>

7 0
3 years ago
Dimension equation of work
kkurt [141]

Answer:

Explanation:

Work

Other units Foot-pound, Erg

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Idk if this is what u are looking for but i hope this help.:)

3 0
3 years ago
There. That is better.
mihalych1998 [28]

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This will really help you learn a lot.

6 0
3 years ago
A projectile is launched with an initial velocity of (40 m/s), at an angle of (30°) above
zlopas [31]

Answer:

330.5  m

Explanation:

In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .

The maximum height will be calculated as;

h=\frac{v^2_isin^2\alpha }{2g}

where ∝ is the angle of launch = 30°

vi= initial launch velocity = 40 m/s

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h={1600*0.4132 }/ 20

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8 0
3 years ago
does increasing the frequency of a wave also increase its wavelength if not how are these quantities related​
Aleks [24]

Answer: Increasing the frequency does not increase the wavelength. They are inversely related.

Explanation:

As wavelength increases, frequency decreases. If you look at a transverse wave and it has a long wavelength, there only a few waves produce. Which means there is less frequency produced. So as wavelength increases, frequency decreases. The other way around can work to. As frequency increases, wavelength decreases. They are inversely related.

8 0
3 years ago
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