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Umnica
19 days ago
7

A hiker walks due east for a distance of 25.5 km from her base camp. On the second day, she walks 41.0 km northwest till she dis

covers the cave she wanted to see. Determine the magnitude and direction of her resultant displacement between the base camp and the cave​
Physics
1 answer:
ValentinkaMS [3.3K]19 days ago
3 0
The resultant displacement amounts to 29.2 km at north of west.
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A man makes a 27.0 km trip in 16 minutes. (a.) How far was the trip in miles? (b.) If the speed limit was 55 miles per hour, wa
Maru [3272]

Answer:

(a) 16.777 miles

(b) Yes, he exceeded the speed limit

Explanation:

(a)

We need to perform the necessary calculations to convert kilometers to miles:

27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi

Thus, the distance of the trip in miles is:

d=16.77706389mi

(b)

Next, we will compute the man's speed during the journey:

v=\frac{d}{t}

Before that, we must convert minutes to hours:

16min*\frac{1h}{60min} =2.666666667h

The resulting speed is:

v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}

Consequently:

62.91398959\frac{mi}{h}>55\frac{mi}{h}

Thus, it can be concluded that the driver was speeding

8 0
1 month ago
A wire, of length L = 3.8 mm, on a circuit board carries a current of I = 2.54 μA in the j direction. A nearby circuit element g
Keith_Richards [3153]

It equals 5z

Explanation:

5z

5 0
18 days ago
How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. th
Softa [2959]

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:

                               n*lambda = \frac{sin(theta)}{N}

- 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:

                            y_n = n*lamda*L*N

- 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.

4 0
1 day ago
Which of the following statements about ycarrier(x,t) is correct?
serg [3485]

Answer: Option D: indicates rapid travel with slow oscillation.

Clarification:

ycarrier(x,t) is traveling quickly but has slow oscillations.


8 0
16 days ago
A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?
ValentinkaMS [3368]
The formula is W = 1/2k*x^2.

Here, k represents the spring constant which is 2500 n/m.
x signifies the distance as 4 cm, converted to 0.04m.

Therefore, W = 1/2(2500)(0.04)^2 = 2J.
5 0
1 month ago
Read 2 more answers
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