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natali 33 [55]
3 years ago
11

The density of air at room temperature is about 1.2 g/L. This is the same as

Physics
1 answer:
Verdich [7]3 years ago
3 0
The correct answer is
<span>C) 1200 g/m3. 

Let's see why. The relationship between liters and cube decimeters is
</span>1 L = 1 dm^3
Therefore, 
1 g/L= 1 g/dm^3
However, we also know that
1 dm^3 = 10^{-3} m^3
Therefore, 
1 L = 10^{-3} m^3
and 
1  \frac{g}{L}= 1 \frac{g}{10^{-3} m^3} =1 \cdot 10^3  \frac{g}{m^3} =1000  \frac{g}{m^3}

Therefore, the density of the problem 1.2 g/L becomes
d=1.2 g/L=1200 g/m^3
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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:

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Mass of object = 5 kg

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Mass of stationary object = 3 kg

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Stationary object deflected = 31°

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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.

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The correct option is B.
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