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Wewaii [24]
3 years ago
8

An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.50 m above the ground. The railcar

has a mass of 38500 kg and is moving to the right at a constant speed of 8.70 m/s on a frictionless rail. The automobile them accelerates to the left, leaving the railcar at a speed of 22.0 m/s with respect to the ground. When the automobile lands, what is the distance D between it and the left end of the railcar?
Physics
1 answer:
igomit [66]3 years ago
7 0
<span>An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 38,500 kg and is moving to the right at a constant speed of 8.7 m/s on a frictionless rail...
</span>
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Answer:

1) λ < 2d,  2)  nfrared imaging technique, 3) each color there is a different index of refraction

Explanation:

We are going to answer the three questions

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5 0
3 years ago
Sort the processes based on the type of energy transfer they involve.
alina1380 [7]
Types of energy transfer involve
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4 years ago
Dante is leading a parade across the main street in front of city hall. Starting at city hall, he marches the parade 4 blocks ea
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Answer:

The correct option is A)

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

Given that Dante is leading a parade across the main street in front of city hall.

Let, Initial location of parade is 0i+0j

One block of city is one units on the XY- graph

Statement 1: Parade marches the parade 4 blocks east, then 3 blocks south

New location of parade is 4i-3j

Statement 2: The parade marches 1 block west and 9 blocks north and finally stops.

Final location of parade is (4i-3j)+(-1i+9j)=3i+6j

Displacement is given by

Displacement = (Final destination)-(Initial destination)

Displacement = (3i+6j)-(0i+0j)=3i+6j

Thus,

Magnitude of displacement = \sqrt{3^{2}+6^{2}}

                                              = 6.71 m

Direction of displacement =  tan^{-1}(\frac{Y}{X} )

                                           =  tan^{-1}(\frac{6}{3} )

                                           = 63.43 NE

Therefore, the correct option is A) Displacement: 6.71 m, Direction: 63.4 degrees north of east

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