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3 years ago

The electric eye that prevents an elevator door from closing when you are in the doorway relies upon which physics principle?

2 answers:

5 0

Just took the test as well that dude is right.. lol I rested my grade in your hands and I got an A.. for more help with that test all answers on here are 100% correct: https://quizlet.com/212791759/1007-unit-10-test-part-1-flash-cards/

goblinko [34]3 years ago

3 0

It's Photoelectric Effect, I just a test with this same question. I am not good for explaining exactly how, but I was right.

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A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w

Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s$

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

$\theta= \omega t$

and substituting t = 75 seconds, we find

$\theta= (0.20 rad/s)(75 s)=15 rad$

In degrees, it is

$15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}$

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

$v=\omega r$

where we have

$\omega=0.20 rad/s$

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

$v=(0.20 rad/s)(13.5 m)=2.7 m/s$

6 0

3 years ago

Help pleaseeeeeeeeeeeee

Tasya [4]

A it’s not that hard tbh

7 0

3 years ago

You're an electrical engineer designing an alternator (the generator that charges a car's battery). Mechanical engineers specify

Julli [10]

Answer:

13.78 mT

Explanation:

The peak voltage ε = ωNAB where ω = angular speed of coil = 1500 rpm = 1500 × 2π/60 rad/s = 50π rad/s = 157.08 rad/s, N = number of turns of coil = 250, A = area of coil = πr² where r = radius of coil = 10 cm = 0.10 m,

A = π(0.1 m)² = 0.03142 m² and B = magnetic field strength

So,

B = ε/ωNA

substituting the values of the variables into the equation given that ε = 17 V

So, B = ε/ωNA

B = 17 V/(157.08 rad/s × 250 turns × 0.03142 m²)

B = 17 V/(1233.8634 rad-turns-m²/s)

B = 0.01378 T

B = 13.78 mT

8 0

2 years ago

What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from

ElenaW [278]

When is at the end of the runway the velocity of the plane is given by the equation $vf^{2}=0+2*a*s$ where s=1800 m is the runway length. Thus
$vf^{2}=2*5*1800=18000 (m/s)^{2}$
$vf =134.164 (m/s)$

At half runway the velocity of the plane is
$v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
$
$v= \sqrt{9000}=94.87 ( \frac{m}{s})$

Therefore at midpoint of runway the percentage of takeoff velocity is
‰ $P= \frac{v}{vf}= \frac{94.87}{134.164}=0.707$

6 0

3 years ago

Wagon wheel. While working on your latest novel about settlers crossing the Great Plains in a wagon train, you get into an argum

OverLord2011 [107]

Answer:

I = 16.7kgm²

Explanation:

Since, Torque is given by,

\tau = F*r = I*\alpha

here, I = Moment of inertia = ??

\alpha = angular acceleration of wheel = a/r

F = tangential tension acting on the wheel = T

a = acceleration of bag of sand = 2.95 m/s^2

r = radius of wheel = d/2 = 120/2 = 60 cm = 0.60 m

from force balance on sand bag,

mg - T = m*a

T = m*(g-a)

m = mass of sand bag = 20 kg

So, I = T*r/\alpha = m*(g-a)*r/(a/r)

Using known values:

I = 20*(9.81 - 2.95)*0.60/(2.95/0.60) = 16.74

I = 16.7 kgm² = Moment of inertia of wheel experimentally

also, Moment of inertia of wheel theoretically(I') = M*r²

given, M = mass of wheel = 70 kg

I' = 70*0.60²= 25.2 kgm² = Moment of inertia of wheel theoretically

7 0

3 years ago