[ÔN TẬP GIỮA KÌ 1] BUỔI SỐ 06 - THẦY VNA

Ok, Hi Tien Dat, Hi Nguyen Hoang, Hi Trung Kien, Hi Khanh An, Hi Trong Viet Ok, so this morning is nothing special. Today is nothing special, yesterday was just a special day. Today is the day we go back to our daily work. That is to study and practice for the university entrance exam. We will start a new day with a live stream of me studying for the 1st semester of the 6th grade. This Saturday is the 7th day, the last day. Next week, I will have new strategies, such as learning topics such as mechanical waves and rotating electricity. I will only study for this semester until the end of Saturday. Next week, we will have a new strategy. Ok, everyone, please share with me. Ok, let's start the live stream today. It's been a long time since we came back to study in the morning. Many of you wake up early to study, right? Today, I will give you 35 questions. It is a topic that is close to the standard of the mid-term exam. Ok, let's start this Saturday. Remember to share with me. Ok, let's start. Have you done this topic yet? Comment for me to know the answer. Let's start. The first question. 1. There is a single wave with a length of L. How long is the magnetic field? It is T by 2P multiplied by L over G. So what is the answer to our first question? The answer is A. Next, the second question. The frequency of the object. The equation is x by 4 cos Pt plus P over 3. The frequency of the object is the magnitude outside the cosine. The answer is 4 cm. Our x is equal to cm. So our second question is C. Next, the third question. 1. A spring with a small object and a light spring with a hardness of K. It is oscillating evenly. What is the characteristic of a spring when it has a magnitude of x? What is the answer? It is Kx divided by 2. Kx divided by 2 is 0.5 Kx. So our answer to this question is D. Next, the fourth question. 1. An object oscillating evenly on the orbit is 10 cm long. The frequency of the object is 5 cm. L is the orbit of 2A. So the answer to the fourth question is B. The fifth question. 1. An object oscillating on a very long wire with a speed of 1 meter per second and a magnitude of 0.5 seconds. This object has a magnitude of λ. It is equal to V times T. V times T or V times C or F can be. 1 times 0.5 is 0.5 meters per second. 0.5 meters is 50 cm. So our answer to the fifth question is C. Next, the sixth question. 1. An object oscillating evenly on a very long wire with a speed of 1 meter per second and a magnitude of 0.5 seconds. The distance between the two points closest to each other, and the two particles there, oscillate evenly. The distance between the two points closest to each other, is λ divided by 4. Thank you, Nguyen Khang. The distance between the two points closest to each other, is λ divided by 4. Because it is closest to each other. Or 3λ divided by 4 is also fine. But it is further away. λ divided by 4 is still 0.5 meters. The answer is A. The seventh question. 2. An object oscillating evenly on a very long wire with a speed of 1 meter per second and a magnitude of 0.5 seconds. The distance between the two points closest to each other, and the two particles there, oscillate evenly. The distance between the two points closest to each other, oscillate evenly. The answer is A.