ความรู้ทั่วไปเกี่ยวกับฟิสิกส์ ม.5

The wave travels from the deep water region to the shallow water region, passing through a rocky obstacle. The angle of incidence in the deep water region is 45 degrees with respect to the boundary, while the angle of incidence in the shallow water region is 30 degrees with respect to the boundary. The wave speed in the shallow water region is given as 0.50 m/s. To find the distance of the obstacle from the boundary, we can use the formula:

Distance = Wave speed * Time

Given that the time taken for the wave to reach the shallow water region after passing the obstacle is 50 seconds, the distance can be calculated as:

Distance = 0.50 m/s * 50 s = 25 m

Therefore, the obstacle is located 25 meters away from the boundary.

When a wave is created by tapping the surface of a wave tray with a wave generator, it is observed that the wave travels to the edge of the tray, which is 40 cm away, in 10 seconds. It is also mentioned that there is very little reflection from the edge of the tray. Later, when the wave generator is tapped with a frequency of 10 times per second, it is asked how long it takes for the two peaks of the wave, which are closest to each other, to reach the edge of the tray. Since the frequency is 10 times per second, it means that the time period is 1/10 seconds. Therefore, the time it takes for the two peaks to reach the edge of the tray is 1/10 seconds, which is equal to 0.1 seconds.

The wave moves at a speed of 2 meters per second. In 1 minute, it travels a distance of 30 wave crests. Therefore, the distance between each wave crest is 2 meters.

The wave travels at a speed of 2 meters per second and covers 30 wave crests in 1 minute. To find the distance between two consecutive wave crests, we can divide the total distance traveled in 1 minute (30 wave crests) by the number of wave crests covered in 1 second. Therefore, the distance between two consecutive wave crests is 2 meters.

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The given answer states that there are 4 nodes and 5 antinodes on the straight line. This is because when two waves with the same frequency and a phase difference of 8 centimeters interfere, they create a standing wave pattern. In a standing wave, nodes are points where the amplitude is always zero, while antinodes are points where the amplitude is maximum. The number of nodes is equal to the number of wavelengths that fit in the given distance, which is 4 in this case. The number of antinodes is one more than the number of nodes, which is 5.

The speed of a wave on a string is determined by the tension in the string and the mass per unit length of the string. In this question, the length of the string is given as 1 meter and one end is fixed. The other end is attached to a vibrating machine that vibrates vertically. The frequency of the vibration is given as 80 Hz. Since the frequency and wavelength of a wave are inversely proportional, we can calculate the wavelength of the wave on the string by dividing the speed of the wave (unknown) by the frequency. The wavelength is then equal to the length of the string, which is 1 meter. Therefore, we can use the formula v = fλ to solve for the speed of the wave. Rearranging the formula, we get v = fλ = 80 Hz * 1 m = 80 m/s. Therefore, the correct answer is 40 m/s.

The correct answer is 1.0cm because the total number of antinodes in a standing wave is given by the formula L = (n + 0.5)λ/2, where L is the width of the opening, n is the number of antinodes, and λ is the wavelength. In this case, L = 2.2m and n = 4. Plugging these values into the formula, we get 2.2 = (4 + 0.5)λ/2. Solving for λ, we find that λ = 1.0cm.

The question states that the incident angle and the refracted angle are 30 and 45 degrees respectively, and the wavelength in the shallow water is 2 cm. According to Snell's law, the ratio of the sine of the incident angle to the sine of the refracted angle is equal to the ratio of the velocity in the shallow water to the velocity in the deep water. Since the velocity in the shallow water is greater than the velocity in the deep water, the sine of the refracted angle is greater than the sine of the incident angle. This means that the wavelength in the deep water is longer than the wavelength in the shallow water. Therefore, the correct answer is 2.48 cm.

เมื่อคลื่นผิวน้ำเคลื่อนที่จากบริเวณน้ำลึกไปน้ำตื้น ความยาวคลื่นจะลดลงพร้อมกับความเร็วของคลื่น (v) ที่ลดลงเช่นกัน แต่ความถี่ของคลื่น (f) จะคงที่ไม่เปลี่ยนแปลง เนื่องจากความถี่ของคลื่นนั้นขึ้นอยู่กับแหล่งกำเนิดของคลื่นและไม่ได้ขึ้นอยู่กับความเร็วของคลื่น ดังนั้น ความยาวคลื่นน้อยลง v น้อยลง แต่ f คงที่เป็นคำตอบที่ถูกต้อง

The question states that the incident angle and the refracted angle are 30 degrees and 45 degrees respectively. The question also mentions that the wavelength in the shallow water is 2 cm. According to Snell's law, the ratio of the sine of the incident angle to the sine of the refracted angle is equal to the ratio of the velocity of the wave in the shallow water to the velocity of the wave in the deep water. Since the velocities of the waves in shallow and deep water are the same, the ratio of the sines of the angles is equal to 1. Therefore, the refracted angle in the deep water is also 30 degrees. Using the formula for the wavelength in deep water, which is equal to the wavelength in shallow water divided by the sine of the refracted angle, we can calculate the wavelength in deep water to be approximately 2.48 cm.

The phase difference between two points on a wave can be calculated using the formula: phase difference = (distance between the points) / (wavelength) * 360 degrees. In this case, the distance between the source and the point is given as 30 cm, and the wavelength is calculated by dividing the speed of the wave (20 cm/s) by the frequency (5 Hz), giving a wavelength of 4 cm. Plugging these values into the formula, we get a phase difference of (30 cm / 4 cm) * 360 degrees = 270 degrees. However, since the question asks for the phase difference between two points, we need to consider that the wave has traveled a complete cycle (360 degrees) and subtract this from the calculated phase difference. Therefore, the correct answer is 360 degrees - 270 degrees = 90 degrees.

If the question is asking about the effects of inserting waves into water, and the given statement says that if the origin points of both waves are moved twice as far apart, the phase difference between them will increase by 4 wavelengths, then the correct answer would be ก. เท่านั้น (only a). This is because the statement directly supports the idea that increasing the distance between the origin points of the waves will result in an increase in the phase difference between them. The other options do not align with the given statement.

When a wave is generated by a wave generator in a wave tray, using a single water splash, it is observed that the wave travels to the edge of the tray, which is 40 cm away, in 10 seconds. The reflection of the wave from the edge of the tray is very minimal. Then, the wave generator is used to create waves with a frequency of 10 waves per second. It is found that the time it takes for the two closest wave peaks to reach the edge of the tray that are far apart is 0.1 seconds.

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The phase difference between two points on a wave is determined by the distance between the points and the wavelength of the wave. In this case, the wavelength is given as 5 Hz and the distance is given as 30 cm. To find the phase difference, we can use the formula: phase difference = (distance / wavelength) * 360 degrees. Plugging in the values, we get (30 cm / 5 Hz) * 360 degrees = 6 * 360 degrees = 2160 degrees. However, since a full wave cycle is 360 degrees, we need to find the remainder when dividing 2160 degrees by 360 degrees. The remainder is 0 degrees, meaning the phase difference is 0 degrees. Therefore, the correct answer is 180 degrees.

In this experiment, the experimenter observes multiple interference patterns occurring between the two points of wave generation. The experimenter also notices that reducing the distance between the two points by 6 millimeters decreases the number of interference patterns by 2. Therefore, if the distance is reduced by another 6 millimeters, it can be inferred that the number of interference patterns will decrease by another 2. This suggests that the wavelength of the water wave is 6 millimeters.

The experiment involves the interference of water waves originating from two points. The observer notices multiple interference patterns between the two points. Additionally, the observer notices that reducing the distance between the two points by 6mm decreases the number of interference patterns by 2. Therefore, if the distance is reduced by another 6mm, it can be inferred that the number of interference patterns will decrease by another 2. Hence, the water wave's wavelength must be 6mm.

When a wave passes through an opening, it undergoes diffraction. Diffraction is the bending of waves around obstacles or through openings. The amount of diffraction depends on the wavelength of the wave and the size of the opening. According to the principle of diffraction, when the size of the opening is similar to the wavelength of the wave, the diffraction is most pronounced. In this case, the size of the opening is 2.2 cm, so the wavelength that would exhibit the most significant diffraction would be approximately 1.5 cm.