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3 years ago

A jet makes a landing traveling due east with a speed of 120 m/s .

Physics

1 answer:

vesna_86 [32]3 years ago

7 0

Average acceleration over a time interval lasting $\Delta t$ is

$a_{\rm ave}=\dfrac{\Delta v}{\Delta t}$

where $\Delta v$ is the difference in the jet's final and initial velocities. It's coming to a rest, so

$a_{\rm ave}=\dfrac{0-120\frac{\rm m}{\rm s}}{13.5\,\rm s}=-8.9\dfrac{\rm m}{\mathrm s^2}$

so the average acceleration has magnitude 8.9 m/s^2 and is pointing West (the direction opposite the jet's movement, which should make sense because the jet is slowing down).

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What is the difference between a cannabinoid and a trichome?

Sedbober [7]

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3 years ago

He power output of a cyclist moving at a constant speed of 5.3 m/s on a level road is 120 w. (a) what is the force exerted on th

EleoNora [17]

P=F*V
F=P/V
F=120/5.3
F=22.6N

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3 years ago

Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. Th

zlopas [31]

When they meet the 40-kg boy would have moved a distance of 6 m.

<h3>Distance moved by the 40 kg boy</h3>

Apply the principle of center mass;

Take the 40 kg mass as the reference point;

X(40 kg) = (40kg x 0 + 60kg x 10 m)/(40 kg + 60 kg)

X(40 kg) = (600)/(100)

X(40 kg) = 6 m

Thus, when they meet the 40-kg boy would have moved a distance of 6 m.

Learn more about distance here: brainly.com/question/2854969

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2 years ago

A day on a distant planet observed orbiting a nearby star is 21.5 hr. Also, a year on the planet lasts 69.3 Earth days. In other

serg [7]

Answer:

Part A

The angular speed of rotation of the plane is 8.11781 × 10⁻⁵ rad/s

Part B

The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s

Explanation:

The parameters of the planet are;

The duration of a day on the distant planet = 21.5 hr.

The duration of a year on the distant planet = 69.3 Earth days

Part A

The duration of a day = The time to make one complete revolution of 2·π radians

∴ The average angular speed about its axis, $\omega_{rotation}$ = Angle turned/Time

∴ $\omega_{rotation}$ = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s

The average angular speed of the planet about its own axis, $\omega_{rotation}$ = 8.11781 × 10⁻⁵ rad/s

The angular speed of rotation of the plane $\omega_{rotation}$ = 8.11781 × 10⁻⁵ rad/s

Part B

The time it takes the planet to revolve round the neighboring star once = 69.3 Earth days

Therefore, the average angular speed of the planet around its neighboring star, $\omega _{Star}$ , is given as follows;

$\omega _{Orbit}$ = 2·π/((69.3 × 24 × 60 × 60) s) = 1.04938 × 10⁻⁶ rad/s

The average angular speed of orbit, $\omega _{Orbit}$ = 1.04938 × 10⁻⁶ rad/s

The angular speed of orbit of the planet, $\omega _{Orbit}$ = 1.04938 × 10⁻⁶ rad/s.

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3 years ago

The energy required to dislodge electrons from cesium metal via the photoelectric effect is 207 kJ/mol . What wavelength (in nm)

ki77a [65]

Answer:

So wavelength in nm will be $\lambda =0.956\times 10^{-19}nm$

Explanation:

We have given that the energy = 207 KJ/mol $=207\times 10^3j/mol$

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

Plank's constant $h=6.6\times 10^{-34}J-s$

According to plank's rule energy of the photon is given by

$E=\frac{hc}{\lambda }$

$207\times 10^3=\frac{6.6\times 10^{-34}\times 3\times 10^8}{\lambda }$

$\lambda =0.956\times 10^{-28}m$

So wavelength in nm will be $\lambda =0.956\times 10^{-19}nm$

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3 years ago