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Andreyy89
3 years ago
12

You get pulled over for running a red light. You explain to the police officer that the light appeared green to you due to the D

oppler effect. The police officer writes you a ticket for speeding instead. The fine is $100 for every 10km/hr over the speed limit of 60km/hr you were traveling. How much is the speeding ticket?
Physics
2 answers:
Ludmilka [50]3 years ago
8 0

Answer:

$ 3085713685.71

Explanation:

\lambda_0 = Actual wavelength = 700 nm

\lambda = Changed wavelength = 500 nm

Let the wavelength of red color be 700 nm and green be 500 nm

Change in wavelength is

\Delta \lambda=700-500\\\Rightarrow \theta\lambda=200\ nm

We have the relation

\dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}\\\Rightarrow v=\dfrac{\Delta\lambda}{\lambda_0}c\\\Rightarrow v=\dfrac{200}{700}\times 3\times 10^8\\\Rightarrow v=85714285.7143\ m/s

The speed of the vehicle is 85714285.7143 m/s

\dfrac{85714285.7143\times 3600}{1000}=308571428.571\ km/h

By how much was the car speeding

308571428.571-60=308571368.571\ km/h

The number of 10 km/h in the above speed

308571368.571\times\dfrac{1}{10}=30857136.8571

Cost of the ticket

30857136.8571\times 100=\$ 3085713685.71

The cost of the ticket is $ 3085713685.71

mr Goodwill [35]3 years ago
5 0

Answer:

not enough info

Explanation:

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A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then di
jeyben [28]

Answer:

3.6μF

Explanation:

The charge on the capacitor is defined by the formula

q = CV

because the charge will be conserved

q₁ = C₁V₂

q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor  and the voltage drop across the two capacitor will be the same

q = q₁ + q₂ = C₁V₂ + C₂V₂

CV = CV₂ + C₂V₂

CV - CV₂ = C₂V₂

C ( V - V₂) = C₂V₂

C ( V/ V₂ - V₂ /V₂) = C₂

C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF

7 0
3 years ago
DO NOT ANSWER IF YOU DON'T KNOW
Andrew [12]
The last one, the soil will become weak & unable to support plant growth
4 0
3 years ago
A(n) 131 g ball is dropped from a height
larisa [96]

Answer:

26.59 N/m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 131 g

Extention (e) = 4.82755 cm

Acceleration due to gravity (g) = 9.8 m/s²

Spring constant (K) =?

Next, we shall convert 131 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

131 g = 131 g × 1 Kg / 1000 g

131 g = 0.131 Kg

Thus, 131 g is equivalent to 0.131 Kg.

Next, we shall the force exerted by the ball on the spring. This can be obtained as follow:

Mass (m) = 0.131 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = ma

F = 0.131 × 9.8

F = 1.2838 N

Next, we shall convert 4.82755 cm to metre (m)

This can be obtained as follow:

100 cm = 1 m

Therefore,

4.82755 cm = 4.82755 cm × 1 m / 100 cm

4.82755 cm = 0.0482755 m

Thus, 4.82755 cm is equivalent to 0.0482755 m

Finally, we shall determine the spring constant as follow:

Force (F) = 1.2838 N

Extention (e) = 0.0482755 m

Spring constant (K) =?

F = Ke

1.2838 = K × 0.0482755

Divide both side by 0.0482755

K = 1.2838 / 0.0482755

K = 26.59 N/m

Thus the spring constant is 26.59 N/m

7 0
3 years ago
On his way off to college, Russell drags his suitcase 19 m from the door of his house to the car at a constant speed with a hori
Mashcka [7]

Answer:

The work done on the suitcase is, W = 1691 J

Explanation:

Given data,

The force on the suitcase is, F  = 89 N

The distance Russell dragged the suitcase, S = 19 m

The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction

The work done on the suitcase by Russell is given by the formula

                          W = F · S

Substituting the given values,

                           W = 89 N x 19 m

                           W = 1691 J

Hence, the work done on the suitcase is, W = 1691 J

8 0
3 years ago
Which statement is true according to Newton's first law of motion?
katrin2010 [14]
C) In the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.

hope this helps and have a great day :)







8 0
3 years ago
Read 2 more answers
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