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vitfil [10]
3 years ago
6

Bilang nagdadalaga at nagbibinata, sa iyong palagay, ano-ano ang inaasahan sa iyo ng mga magulang, guro at kamag-aral? ​

Physics
1 answer:
dangina [55]3 years ago
8 0

Answer:

Yesyesyesyesyesyesyesyesyes

Explanation:

test

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Since his speed is constant,

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The periodic wave in the diagram below has a frequency of 80. hertz.
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A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected
gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0
3 years ago
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The tow lines for the two skiers must be of the same length. No matter the number of skiers towed, the length of the tow lines must be the same. Towing people on ski requires additional skills and some safety precautions must be taken. Most importantly, it is dangerous and illegal to tow at night. 
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Pressure increases with depth towards the center of earth. In which layer would you expect pressure to be the greatest.
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You could expect to find it in the inner core
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