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3 years ago

a jet fighter accelerates at 17.7 m/s^2 , increasing its velocity from 119 m/s to 233 m/s. how much time does that take?

Physics

1 answer:

Aleksandr-060686 [28]3 years ago

7 0

Answer:

6.44 s

Explanation:

Given:

v₀ = 119 m/s

v = 233 m/s

a = 17.7 m/s²

Find: t

v = at + v₀

(233 m/s) = (17.7 m/s²) t + (119 m/s)

t = 6.44 s

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Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

Circular velocity for the particle in the Encke Division:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

Circular velocity for the particle in D Ring:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago

A 65-kg person stands on a scale in a moving elevator while holding a 5.0 kg mass suspended from a massless spring with spring c

disa [49]

Answer:

The question is incomplete. However, I believe, it is asking for the acceleration of the elevator. This is 3.16 m/s².

Explanation:

By Hooke's law, $F = ke$

F is the force on a spring, k is the spring constant and e is the extension or compression.

From the question,

$F = (1.08\text{ kN/m}) \times (6.0 \times 10^{-2}\text{ m}) = 64.8 \text{ N}$

This is the force on the mass suspended on the spring. Its acceleration, a, is given by

$F = ma$

$a = \dfrac{F}{m}$

$a = \dfrac{64.8 \text{ N}}{5\text{ kg}} = 12.96\text{ m/s}^2$

This acceleration is more than the acceleration due to gravity, g = 9.8 m/s². Hence the elevator must be moving up with an acceleration of

12.96 - 9.8 m/s² = 3.16 m/s²

7 0

3 years ago

X-rays with frequency 3 ⨉ 10 18 Hz shine on a crystal, producing an interference pattern. If the first bright spot is observed a

FrozenT [24]

Answer:

$5.1645\times 10^{-11}\ m$

Explanation:

n = Order = 1

c = Speed of light = $3\times 10^8\ m/s$

f = Frequency = $3\times 10^{18}\ Hz$

$\theta$ = Angle = $75.5^{\circ}$

Lattice spacing is given by

$d=\dfrac{n\lambda}{2\sin\theta}\\\Rightarrow d=\dfrac{n\times c}{f2\sin\theta}\\\Rightarrow d=\dfrac{1\times 3\times 10^{8}}{3\times 10^{18}\times 2\times \sin75.5^{\circ}}\\\Rightarrow d=5.1645\times 10^{-11}\ m$