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Pie
3 years ago
12

How can the pilot determine, for an ILS runway equipped with MALSR, that there may be a penetration of the obstacle identificati

on surfaces (OIS), and care should be taken in the visual segment to avoid any obstacles?
Physics
1 answer:
Stells [14]3 years ago
6 0

Explanation:

Dissolve 2 -naphthol in ethanol in a round bottom flask.

Add sodium hydroxide pellets and boiling chips. Attach a condenser.

Heat under reflux for 20 minutes, until all the base dissolves.

Add 1 -bromobutane and reflux for an additional hour.

Pour the contents fo the round bottom flask into a beaker of ice

Collect the precipitate by vacuum filtration on a Bachner funnel

Rinse the product with cold water and dry it on the filter.

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The image produced by a convex lens depends upon ____.
Karolina [17]
I recently did this topic in science class - the answer is A ;)
7 0
3 years ago
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Simplified. The absorption spectra of ions have been used to identify the presence of the elements in the atmospheres of the sun
Nookie1986 [14]

Answer:

The answer is "3.83 \times 10^9 \ m"

Explanation:

Z=2, so the equation is E= \frac{-4B}{n^2}

Calculate the value for E when:  

n=2 and n=9

The energy is the difference in transformation, name the energy delta E Deduct these two energies  

In this transition, the wavelength of the photon emitted is:

\Delta E=2.18 \times  10^{-18} ( \frac{1}{4}- \frac{1}{81})

\lambda = \frac{h c}{\Delta E}

h ( Planck's\  constant) = 6.62 \times  10^{(-34)} \ Js \\\\ speed \ of \ light = 3 \times 10^{8} \ \frac{m}{s}\\\\= \frac{6.62 \times 10^{(-34)} \times 3 \times 10^ {8}}{2.18 \times  10^{-18}} (\frac{1}{4}- \frac{1}{81}) \\\\=3.83 \times 10^9 \ m\\\\

6 0
3 years ago
During normal conditions, we only see the Sun’s __________, but during a solar eclipse, we get to see the _______________.
liberstina [14]
I would say A. because during the solar eclipse you cannot see the suns Photosphere at all, and the corona is the light that emits around the moon during the eclipse.
8 0
3 years ago
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A car moves at a constant velocity of 30 m/s and has 3.6 × 105 J of kinetic energy. The driver applies the brakes and the car st
LekaFEV [45]

The force needed to the stop the car is -3.79 N.

Explanation:

The force required to stop the car should have equal magnitude as the force required to move the car but in opposite direction. This is in accordance with the Newton's third law of motion. Since, in the present problem, we know the kinetic energy and velocity of the moving car, we can determine the mass of the car from these two parameters.

So, here v = 30 m/s and k.E. = 3.6 × 10⁵ J, then mass will be

K.E = \frac{1}{2} * m*v^{2}  \\\\m = \frac{2*KE}{v^{2} } = \frac{2*3.6*10^{5} }{30*30}=800 kg

Now, we know that the work done by the brake to stop the car will be equal to the product of force to stop the car with the distance travelled by the car on applying the brake.Here it is said that the car travels 95 m after the brake has been applied. So with the help of work energy theorem,

Work done = Final kinetic energy - Initial kinetic energy

Work done = Force × Displacement

So, Force × Displacement = Final kinetic energy - Initial Kinetic energy.

Force * 95 = 0-3.6*10^{5}\\ \\Force =\frac{-3.6*10^{5} }{95}=-3.79 N

Thus, the force needed to the stop the car is -3.79 N.

5 0
3 years ago
A research-level Van de Graaff generator has a 2.15 m diameter metal sphere with a charge of 5.05 mC on it. What is the potentia
Llana [10]

Answer:

42.3 MV

Explanation:

d = diameter of the metal sphere = 2.15 m

r = radius of the metal sphere

diameter of the metal sphere is given as

d = 2r

2.15 = 2 r

r = 1.075 m

Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C

Potential near the surface is given as

V = \frac{kQ}{r}

V = \frac{(9\times 10^{9})(5.05\times 10^{-3})}{1.075}

V = 4.23 x 10⁷ volts

V = 42.3 MV

5 0
4 years ago
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