Question 4 (point)

What causes a liquid to freeze? A. When particles speed up and get closer together. B. When particles stop moving. C. When parti

Fittoniya [83]

Answer:The answer is C

Explanation:

3 0

3 years ago

What is the likely identity of a metal if a sample has a mass of 63.5 g when measured in air and an apparent mass of 60.2 g when

Gre4nikov [31]

Answer:

Gold

Explanation:

Given:

Mass of sample = 63.5 g

Mass of water = 60.2 g

Find:

Object

Computation:

Mass of water displaced = 63.5 g - 60.2 g

Mass of water displaced = 3.3 g

So, volume in water = 3.3 cm³

Density = Mass / Volume

Density = 63.5 g / 3.3

Density = 19.24

So,

Object ,must be gold.

7 0

3 years ago

A scientist observes rock masses that have moved past each other in opposite horizontal directions. Which feature

Reil [10]

Answer:

C. strike-slip fault

Explanation:

The scientist must have observed a strike- slip fault.

A fault is an evidence of brittle deformation of the crust in the presence of applied stress on earth materials. Here, the earth material is the rock subjected to tension.

Where a fault occurs, there must have been movement between two blocks of rocks. The direction of movement helps us to delineate the fault type.

When two blocks moves past each other horizontally, it is a strike-slip fault like rubbing your palms together.
When a block moves in the direction of the dip, it forms a dip-slip fault which results in a fault-block mountain characterized by graben and horst systems.

Option A, Plateau is a table landform usually a mountain with flat peak.

Option B is a bowl shaped stratigraphic pattern in which the youngest sequence is at the core of the strata or a fold.

So, the most fitting option is C, a strike-slip fault.

8 0

3 years ago

To become positively charged an átom must

Ivanshal [37]

Lose electrons because they have a negative charge.

7 0

3 years ago

Read 2 more answers

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