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scoray [572]
3 years ago
7

Một vật dao động điều hòa với phương trình x= 4cos(6πt) cm. Kể từ khi vật bắt đầu dao động, thời điểm vật qua a) vị trí cân bằng

lần thứ 1 là? b) vị trí biên x=4 lần thứ 2 là? c) vị trí x= -2 lần thứ 9 là ? d) vị trí x=2 lần thứ 2020 là?
Physics
1 answer:
dezoksy [38]3 years ago
7 0

Answer:

what are you saying.........

......

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How is urine produced
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How long does it take a plane that is traveling at 350 km/h to travel 1750 km?
olga2289 [7]
S=d/t
T=d/s
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3 years ago
The vibrations along a transverse wave move in a direction _________.
GrogVix [38]

Answer: perpendicular to it oscillations.

Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.

By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.

Examples of transverse waves includes wave in a string, water wave and light.

Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.

If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.

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5 0
3 years ago
True or false the refraction of a wave is how many wavelengths pass a fixed point each second
Ksenya-84 [330]
False. That description fits the wave's 'frequency'. 
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8 0
3 years ago
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A 13500 kg jet airplane is flying through some high winds. At some point in time, the airplane is pointing due north, while the
Mekhanik [1.2K]

Answer

given,

mass of the jet airplane = 13500 kg

Force on the plane = 35700 N due north

force from wind  = 15300 N in direction 80.0° south of west.

Force = 35700 \vec{j} N

force by wind = 15300(-cos \theta \vec{i}-sin \theta \vec{j}) N

                       = 15300(-cos 80^0 \vec{i}-sin 80^0 \vec{j}) N

net force on the jet airplane(ma)

          m a = 35700 \vec{j} + 15300(-cos 80^0 \vec{i}-sin 80^0 \vec{j})

          \vec{a} = \dfrac{35700}{13500} \vec{j} + \dfrac{15300}{13500}(-cos 80^0 \vec{i}-sin 80^0 \vec{j})

          \vec{a} = 2.64\vec{j} -0.197 \vec{i} - 1.116 sin 80^0 \vec{j})

           \vec{a} = -0.197 \vec{i} + 1.524 \vec{j}

a = \sqrt{-0.197^2+1.524^2}

a = 1.54 m/s²

\theta = tan^{-1}(\dfrac{-1.524}{0.197})

\theta = -82.63^0

3 0
3 years ago
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