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dezoksy [38]
2 years ago
8

The angular width of localizers varies between ______ and ______ degrees in order to provide a signal width of approximately 700

feet at the runway thresholds.
Physics
1 answer:
natali 33 [55]2 years ago
5 0

Answer: 10 and 35 degrees

Explanation: Localizers width below 10 degree and 35 degree signal arc is unreliable and considered unusable for navigation and as a result, aircrafts may loose alignment

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If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
yulyashka [42]

Answer:

178200 g mile pounds

Explanation:

Work= Force * Distance= Fh

F=ma=mg where m is mass and g is acceleration due to gravity

Work= 165 pounds *g* 1080 m=  178200 g mile pounds

5 0
2 years ago
A snowball is melting at a rate of 324π mm3/s. At what rate is the radius decreasing when the volume of the snowball is 972π mm3
Oduvanchick [21]

Answer:

The radius is decreasing at 4 mm/s

Explanation:

The volume of a sphere is:

V = 4/3*\pi *r^3   So, when the volume is 972π mm^3 the radius r is:

r = 9mm

Now, the change rate is given by the derivative:

dV/dt = 4/3*\pi *3*r^2*dr/dt  

Where: dV/dt = -324π mm^2/s

            r = 9mm

Solving for dr/dt:

dr/dt = -4mm/s

5 0
2 years ago
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea
madreJ [45]
In order to answer this exercise you need to use the formulas

 S = Vo*t + (1/2)*a*t^2

Vf = Vo + at

The data will be given as

Vf = final velocity = ?

Vo = initial velocity = 1.4 m/s

a = acceleration = 0.20 m/s^2

s = displacement = 100m

And now you do the following:

100 = 1.4t + (1/2)*0.2*t^2

t = 25.388s

and

Vf = 1.4 + 0.2(25.388)

Vf = 6.5 m/s

So the answer you are looking for is 6.5 m/s
7 0
2 years ago
Read 2 more answers
Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward
KATRIN_1 [288]

Answer:

Tension in the cable is T = 16653.32 N

Explanation:

Give data:

Cross section Area A = 1.3 m^2

Drag coefficient CD = 1.2

Velocity V = 4.3 m/s

Angle made by cable with horizontal  =30 degree

Density \rho \ of\  water= 1000 kg/m3

 Drag force FD is given as

F_{D} = \fracP{1}{2} \rho v^{2} C_{D} A

        = 0.5\times 1000\times 4.32\times  1.2\times 1.3

Drag force = 14422.2 N acting opposite to the motion

As cable made angle  of 30 degree with horizontal  thus horizontal component is take into action to calculate drag force

TCos30 = F_D

T = \frac{F_D}{cos30}

T =\frac{ 14422.2}{cos 30}

T = 16653.32 N

7 0
2 years ago
Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. Wha
devlian [24]

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

K_{max}=\frac{hc}{\lambda}-W

Here h is the Planck's constant, c is the speed of light, \lambda is the wavelength of the light and W the work function of the element:

W=\frac{hc}{\lambda}-K_{max}\\W=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{495*10^{-9}m}-0.5eV\\W=2.01eV

Now, we calculate the wavelength for the new maximum kinetic energy:

W+K_{max}=\frac{hc}{\lambda}\\\lambda=\frac{hc}{W+K_{max}}\\\lambda=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{2.01eV+1.5eV}\\\lambda=3.54*10^{-7}m=354*10^{-9}m=354nm

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

8 0
2 years ago
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