4 years ago

How was the earliest version of the periodic table organized before Mendeleev?

Physics

1 answer:

astraxan [27]4 years ago

7 0

This is how the periodic table is organized.

Hope this helps!

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A steel ball with mass m=5.21 g is moving horizontally with speed ????=412 m/s when it strikes a block of hardened steel with ma

docker41 [41]

Answer:

a) The speed of the block immediately after the collision is $v_{2f}=(0.289\±0.002)m/s$ .

b) The impulse exerted on the block is $p_{2}=(4.2772\±0.0296)kg*m/s$ .

Explanation:

a) As this is a perfectly elastic collision, we can use the formula $v_{2f}=(\frac{2m_{1} }{m_{1}+m_{2}} ) v_{1i}+ (\frac{m_{2}-m_{1}}{m_{1}+m_{2}} )v_{2i}$ , due $v_{2i}=0m/s$ , we obtain $v_{2f}=(\frac{2m_{1} }{m_{1}+m_{2}} ) v_{1i}$ . Then with the data that we know $m_{1}=5.21g=0.00521kg, m_{2}=14.8kg$ and $v_{1i}=412m/s$ , therefore $v_{2f}=(\frac{2(0.00521kg) }{0.00521kg+14.8kg} ) 412m/s=0.289m/s$ or $v_{2f}=(0.289\±0.002)m/s$ adding uncertainty.

b) Now that we know the speed we can use $p_{2}=m_{2}*v_{f2} =14.8kg*(0.289\±0.002)m/s=(4.2772\±0.0296)kg*m/s.$

8 0

4 years ago

Mimas, a moon of Saturn, has an orbital radius of 1.62 × 108 m and an orbital period of about 23.21 h. Use Newton’s version of K

Drupady [299]

Answer:

$3.60432\times 10^{26}\ kg$

Explanation:

a = Orbital radius = $1.62\times 10^8\ m$

T = Orbital period = 23.21 hours

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

From Kepler's third law we get

$M=\frac{4\pi^2a^3}{GT^2}\\\Rightarrow M=\frac{4\pi^2\times (1.62\times 10^8)^3}{6.67\times 10^{-11}\times (23.21\times 3600)^2}\\\Rightarrow M=3.60432\times 10^{26}\ kg$