During a car accident, a 125 kg driver is moving at 31 m/s and in 1.5 seconds is brought to rest by an inflating air bag. What i
2 answers:
The momentum of a moving object can be expressed as:
where
m is the object's mass
and v stands for velocity.
Prior to the vehicular mishap, the driver (with a mass of m=125 kg) was traveling at a rate of v=31 m/s, indicating his initial momentum as:
However, after the accident, the driver is at rest: this implies his final velocity is nil, resulting in a final momentum of zero:
. This indicates the change in momentum equals to his initial momentum:
thus the accurate response is D.
where
m is the object's mass
and v stands for velocity.
Prior to the vehicular mishap, the driver (with a mass of m=125 kg) was traveling at a rate of v=31 m/s, indicating his initial momentum as:
However, after the accident, the driver is at rest: this implies his final velocity is nil, resulting in a final momentum of zero:
thus the accurate response is D.
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Answer:
Refer to the explanation
Explanation:
Solution:-
- For this problem, we will assume the grating has a defined line density ( N ) representing the lines per mm.
- The angle formed by each fringe on the screen is given by ( θ ).
- The order of bright/dark spots is characterized by an integer ( n )
- The incident light's wavelength is ( λ )
- Using the diffraction grating relationship from the Young's Experiment is as follows:
- The provided formula corresponds to constructive interference.
- We will examine how increasing the distance between the screen and the grating affects the pattern. Let ( L ) be the distance from the grating center to the screen center. The distance ( yn ) indicates the vertical spacing between each fringe on the screen.
- For small angles ( θ ), we use the approximation of sin ( θ ) ≈ tan ( θ ). Thus,
sin ( θ ) ≈ tan ( θ ) = [ yn / L ]
- Replacing this approximation in the diffraction grating relation results in:
- To double the distance from the screen to the grating, we use the relation with ( 2L ):
yn ∝ L
Result: The separation between each order of bright and dark fringe doubles. The interference pattern will spread further! Consequently, there will be fewer bright spots visible on the screen since a larger surface area is required to accommodate the entire pattern. The increased distance also diminishes the intensity contrast between bright and dark fringes due to the extended path traveled by light rays. Intensity is inversely related to the square of the travel distance.
- If the line density of the grating ( N ) were doubled, it follows that:
yn ∝ N
Result: The spacing between bright and dark fringes would also double. The interference pattern expands further, leading to the necessity of a bigger screen to show the entire pattern.