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nikklg [1K]
3 years ago
11

Technician A says that daytime running lights (DRLs) light the headlights, usually at reduced intensity, whenever the engine is

running. Technician B says high-intensity discharge (HID) headlights are brighter and have a yellow tint. Who is right?
Physics
1 answer:
miskamm [114]3 years ago
3 0

Answer:

technician A is right

Explanation:

Daytime running lights are auxiliary lights that allow other cars to see us better. They are not intended to illuminate the road, so their intensity is low.

Consequently technician A is right

HID lights are high intensity lights, generally bluish-white in color, yellow lights are halogen lights with lower intensity than HID lights.

Consequently, technician B is wrong

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Yes because once the accident has happen the air bag will then push out
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Answer:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

Explanation:

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Please help me find the fraction
Ira Lisetskai [31]

Answer:

C. 1/2

Explanation:

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3 years ago
Air at 30°C and 2MPa flows at steady state in a horizontal pipeline with a velocity of 25 m/s. It passes through a throttle valv
erastovalidia [21]

Answer:

V_2=159.9\ m/s

T_2=290.6K

Explanation:

At initial condition

P=2 MPa

T=30°C

V=25 m/s

At final condition

P=0.3 MPa

Now from first law for open system

h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}

We know that for air

h= 1.010 x T  KJ/kg

1.01\times 303+\dfrac{25^2}{2000}=1.01\times T+\dfrac{V_2^2}{2000}                               -----1

Now from mass balance

\rho_1A_1V_1=\rho_2A_2V_2

We also know that

\rho=\dfrac{P}{RT}

\dfrac{P_1}{RT_1}V_1=\dfrac{P_2}{RT_2}V_2

\dfrac{2}{R\times 303}\times 25=\dfrac{0.3}{RT_2}V_2

T_2=1.81V_2                  ----------2                                                                                                

Now from equation 1 and 2

V_2^2+3673.749V-612916.49=0

So we can say that

V_2=159.9\ m/s

This is the outlet velocity.

Now by putting the values in equation 2

T_2=1.81\times 159.9  

T_2=290.6K

This is the outlet temperature.

4 0
4 years ago
A grasshopper jumps straight up at 5.33 m/s. How much time does it take for it to return to earth? If u know please help.
sattari [20]
1) Time to get the highest point of it pathway>

Vf = Vo - g*t

Vf = 0 (condition of maximum height)
Vo = 5.33 m/s
g = 9.81 m/s

=> t = Vo / g = 5.33 m/s / 9.81 m/s = 0.543 s

2) Time to return from the highest point = time to get to the highest point = 0.543 s

Total time = 0.543s + 0.543 s = 1.086 s ≈ 1.09 s

Answer: 1.09 s
5 0
3 years ago
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