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S_A_V [24]
3 years ago
8

A 75 kg pilot flies a plane in a loop. At the top of the loop, where the plane is completely

Physics
1 answer:
erica [24]3 years ago
6 0

Answer:

1.47 km

Explanation:

When the pilot hangs freely in the seat and does not push against the  seat belt, his centripetal acceleration would counter balance gravitational acceleration g = 9.81m/s2 = a

We can calculate the radius of the loop using velocity v = 120m/s and a

a = \frac{v^2}{r}

r = \frac{v^2}{a} = \frac{120^2}{9.81} = 1467.89 m or 1.46 km

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The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and 589.6 nm. What are the freque
stiv31 [10]

Answer:

Explanation:

Given

wavelength of emissions are

\lambda _1=589 nm

\lambda _2=589.6 nm

Energy is given by

E=\frac{hc}{\lambda }

where h=Planck's constant

x=velocity of Light

\lambda=wavelength of emission

E_1=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589\times 10^{-9}}

E_1=3.374\times 10^{-19} J

E_1 in kJ/mol

E_1=203.2 kJ/mol

frequency corresponding to this emission

\nu =\frac{c}{\lambda }

\nu _1=\frac{3\times 10^8}{589\times 10^{-9}}

\nu _1=5.09\times 10^{14} Hz

Energy corresponding to \lambda _2

E_2=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589.6\times 10^{-9}}

E_2=3.371\times 10^{-19} J

E_2=203.02 kJ/mol

frequency corresponding to this emission

\nu =\frac{c}{\lambda }

\nu _1=\frac{3\times 10^8}{589.6\times 10^{-9}}

\nu _1=5.088\times 10^{14} Hz

6 0
3 years ago
The area under acceleration time garph represents?​
Keith_Richards [23]

Answer:

Change in Velocity because

at = v

Explanation:

Remeber area is length times Width. In this case, the area under a accleraton vs time graph is Accleration Times Time. Which is velocity

8 0
2 years ago
Read 2 more answers
A rollerblader is blading along the sidewalk. Which forms of measurement would be the best to use to determine the rollerblader'
Greeley [361]
You would use distance an time formula to mathmaticly solve

3 0
3 years ago
SHOW WORK
Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\

For Point b:

E=\frac{1}{2} m a^2 w^2\\\\

   =\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J

For Point C:

V_{max}= a w

        = (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}

For point D:

X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm

5 0
3 years ago
You are riding a bicycle up a gentle hill. It is fairly easy to increase your potential
Dahasolnce [82]
True or false: while riding a bicycle up a gentle hill, it fairly easy to increase your potential energy, but to increase your kinetic energy would ...
3 0
2 years ago
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