All categories

2 years ago

A stone tumbles into a mine shaft strikes bottom after falling for 3.8 seconds. How deep is the mine shaft

Physics

1 answer:

stepladder [879]2 years ago

7 0

Answer:

70.8m

Explanation:

Given parameters:

Time of fall = 3.8s

Unknown:

Depth of the mine = ?

Solution:

To solve this problem, we use the appropriate motion equation as shown below;

S = ut + $\frac{1}{2}$ at²

S is the depth

u is the initial velocity = 0m/s

t is the time taken

a is the acceleration due to gravity

Now insert the parameters and solve;

Since u = 0;

S = $\frac{1}{2}$ at² = $\frac{1}{2}$ x 9.8 x 3.8² = 70.8m

You might be interested in

Ls -2 a solution of 4x +3= -5?.

torisob [31]

Answer:

yes

Explanation:

Let's solve your equation step-by-step.

4x+3=−5

Step 1: Subtract 3 from both sides.

4x+3−3=−5−3

4x=−8

Step 2: Divide both sides by 4.

4x / 4 = −8 / 4

x=−2

Hope it helps,

Please mark me as the brainliest

Thank you

5 0

3 years ago

A positively charged plastic ruler is brought close to a piece paper resting on the desk. The piece of paper was initially neutr

stealth61 [152]

Answer: static electricity

Explanation:

When the plastic ruler is rubbed, friction opposes the motion and causes the transfer of electron from one surface to another such that plastic becomes negatively charged. When ruler is brought nearer to the paper, it induces the positive charge in the piece of paper.

4 0

3 years ago

Name the apparatus used to measure friction force? PLEASE HELP AS SOON AS POSSIBLE

melisa1 [442]

There is no apparatus for it. It is either use something like ruler and table and rub together or rub our hands, and friction force will be showed.

8 0

3 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

$F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

Force on particle A due to particles B & C:

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

Force on particle C due to particles B & A:

$F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

Force on particle B due to particles C & A:

$F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$

$F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$

$F_b=3.49\times 10^{-5}\ N$

3 0

3 years ago

How is aperture measured? What do the measurements mean?

gulaghasi [49]

Aperture is measured in F-stops, in which the f-stops is the amount of light allowed to pass through the aperture, which simply put means that the smaller the aperture, the higher the f-stops. What it does is reduce the amount of light that reaches the film, so the higher the f-stops, the less light reaches the film.

4 0

3 years ago