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3 years ago

Which can a drop of liquid mercury can be described as?

Physics

2 answers:

Lorico [155]3 years ago

7 0

Answer:

a pure substance and an element

Explanation:

i just did on edgen

Sophie [7]3 years ago

3 0

Option D) A drop of mercury is both a pure substance and and element.
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Which statement best describes the isothermal process? A. the temperature remains constant B. the temperature increases at a con

Veronika [31]

Answer:

The answer would be A. - the temperature remains constant

Explanation:

An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0

5 0

3 years ago

Read 2 more answers

What is Force & how many types of Force express with example <br><br><br><br>

ehidna [41]

Answer:

The force is pull or push acting on the body which tends to change its state of rest or of motion is called force.

There are two types of force:

1.Contact force

2. Non-Contact

7 0

3 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago

What is the distance from axis about which a uniform, balsa-wood sphere will have the same moment of inertia as does a thin-wall

andrey2020 [161]

Answer:

$D_{s}$ ≈ 2.1 R

Explanation:

The moment of inertia of the bodies can be calculated by the equation

I = ∫ r² dm

For bodies with symmetry this tabulated, the moment of inertia of the center of mass

Sphere $Is_{cm}$ = 2/5 M R²

Spherical shell $Ic_{cm}$ = 2/3 M R²

The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass

I = $I_{cm}$ + M D²