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3 years ago

Calculate the density of a material that has a mass of 52.457 g and a volume of 13.5 cm3

Physics

1 answer:

Sphinxa [80]3 years ago

8 0

Answer:

Density is 3.90g/cm3

Explanation:

Density = mass ÷ volume

= 52.457g ÷ 13.5cm3

= 3.885703704g/cm3

= 3.90g/cm3

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Answer:

speed are different at different places

Explanation:

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3 years ago

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

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Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

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Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

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A 54 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.3 m/s. The acceleration of

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Answer:

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The previous expression is simplified and required height is found:

$h_{B} = \frac{1}{2\cdot g} \cdot (v_{A}^{2}-v_{B}^{2})$

$h_{B} = \frac{1}{2 \cdot (9.807\, \frac{m}{s^{2}} )} \cdot [(10\, \frac{m}{s} )^{2}-(1.3\, \frac{m}{s} )^{2}]$

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4 years ago

A meteorite has a speed of 90.0 m/s when 850 km above the Earth. It is falling vertically (ignore air resistance) and strikes a

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Before it hits the sand bed, the meteorite is accelerating uniformly with $g=9.80\dfrac{\rm m}{\mathrm s^2}$ , so that its speed $v$ satisfies

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