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notka56 [123]
2 years ago
14

When driving at night switch to low-beams whenever you come within ___ ft of an oncoming vehicle.

Physics
2 answers:
Goshia [24]2 years ago
8 0

Answer:

500 ft

Explanation: One must dim their high-beam lights when the oncoming vehicle is  500 ft away so that the oncoming driver can look clearly and drive safely. High beam light should be used at night to drive safely when one is not able to see ahead of the road clearly. Even for the most experienced drivers low visibility is an issue.

Ainat [17]2 years ago
3 0
B. 500ft.  If you are driving at night with high beams on you must dim them within 500 feet of an oncoming vehicle.
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you need at least two out of the three to get any aenser

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Is the process of introducing a non blood fluid into the blood​
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The lymphatic system
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Three basic electrical effects occur when an electric current flows in a conductor: A magnetic field is set up around the conduc
Alex787 [66]

Answer:

A drop in voltage occurs  

Explanation:

When electric current flows through a conduct, there are three basic electrical effects that occur to the conductor;

1. A magnetic field is set up around the conductor,

      A magnetic field is formed around a conductor when current flows through it which makes it acts like a magnet. Application is used in electric bells.

2. Heat is generated, and

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                          H = I²Rt

3. A drop in voltage occurs  

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7 0
3 years ago
As a pole of a 2nd-order discrete-time system moves away from the origin in the z-plane, while its phase remains constant, the d
Rudik [331]

Answer:

False

Explanation:

When the location of the poles changes in the z-plane, the natural or resonant frequency (ω₀) changes which in turn changes the damped frequency (ωd) of the system.

As the poles of a 2nd-order discrete-time system moves away from the origin then natural frequency (ω₀) increases, which in turn increases damped oscillation frequency (ωd) of the system.

ωd = ω₀√(1 - ζ)

Where ζ is called damping ratio.

For small value of ζ

ωd ≈ ω₀

4 0
2 years ago
At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak
Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

P = AI

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

A = \frac{\pi d^2}{4}

Now the intensity is inversely proportional to the square of the distance from the source, then

I \propto \frac{1}{r^2}

The expression for the intensity at different distance is

\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}

Here,

I_1 = Intensity at distance 1

I_2 = Intensity at distance 2

r_1 = Distance 1 from light source

r_2 = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

I_2 = I_1 (\frac{r_1^2}{r_2^2})

If we replace with our values at this equation we have,

I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})

I_2 = 1.11*10^{-4} W/m^2

Now using the equation to find the area we have that

A = \frac{\pi (8.4*10^{-3}m)^2}{4}

A = 5.5*10^{-5}m^2

Finally with the intensity and the area we can find the sound power, which is

P = AI

P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)

P = 6.1*10^{-9}J/s

Power is defined as the quantity of Energy per second, then

E = 6.1*10^{-9}J

8 0
3 years ago
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