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3 years ago

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A block of an unknown metal has a mass of 227.8 kg and a volume of 0.1189 cubic meters. If the object were a sphere, what would

be its radius? (The value for pi ≈ 3.14159)

1 answer:

otez555 [7]3 years ago

8 0

Explanation:

The given data is as follows.

mass = 227.8 kg, volume = 0.1189 $m^{3}$

It is known that formula for volume of a sphere is $\frac{4}{3} \pi r^{3}$ and density is as follows.

Density = $\frac{mass}{volume}$ ........ (1)

= $\frac{227.8 kg}{0.1189 m^{3}}$

= 1915.895 $kg/m^{3}$

Hence, putting the given values into equation (1) as follows.

Density = $\frac{mass}{volume}$

1915.895 $kg/m^{3}$ = $\frac{227.8 kg}{\frac{4}{3} \pi r^{3}}$

$8021.213 r^{3}$ = 227.8

r = 0.305 $m^{3}$

Thus, we can conclude that radius of the sphere is 0.305 $m^{3}$ .

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Answer:

$\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}$ (for small oscillations)

Explanation:

The total energy of the pendulum is equal to:

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For small oscillations, the equation can be re-arranged into the following form:

$E_{1} \approx m\cdot g \cdot (1-\theta) \cdot L$

Where:

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The factor of change is:

$\frac{E_{2}}{E_{1}} \approx \frac{1 - 4\theta}{1-\theta}$

$\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}$

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3 years ago

Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the

tatuchka [14]

Answer:

7.04 m

Explanation:

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u = Initial velocity

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Initial Velocity of the ball

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s$

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

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The displacement of the ball on the moon is 7.04 m

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Answer:Shifted towards Left by distance of 2.243 m

Explanation:

Given

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Mass of barbara $m_2=55.3 kg$

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