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denis-greek [22]

2 years ago

9

A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate

r at the bottom of the well 5.00 seconds after being dropped from the tower. Calculate the depth of the well. Given: g = -9.81 meters/second2. 22.5 meters 50.7 meters 100 meters 110 meters 152.45 meters

1 answer:

mestny [16]2 years ago

3 0

22.5 meters. I attached all my work below. I hope this helped. If anything is still unclear please feel free to comment back with any questions you still have! It is helpful to draw pictures for problems like these, but if you know what you're doing you don't have to.

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A physics department has a Foucault pendulum, a long-period pendulum suspended from the ceiling. The pendulum has an electric ci

antoniya [11.8K]

Answer:

t=37 mins -> 2220sec

We want "T" which is the pendulum time constant

Using this equation

.5A=Ae^(-t/T)

The .5A is half the amplitude

Take ln of both sides to get ride of Ae

=ln(.5)=-2220/T

Now rearrange to = T

T=-2220/ln(.5) = 3202.78sec / 60 secs = 53.38 mins -> first part of the answer.

The second part is really easy. It took 37 mins to decay half way. meaning to decay another half of 50% which equals 25% it will take an additional 37 mins!

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2 years ago

A flute player hears four beats per second when she compares her note to an 880 Hz tuning fork (note A). She can match the frequ

ludmilkaskok [199]

Answer:

884Hz

Explanation:

Beats is the absolute difference between two frequencies therefore

Beats = f1-f2

4=f1-880

F1=880+4

F1=884Hz

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2 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

2 years ago

A 4kg block is sliding on a horizontal frictionless floor at a speed of 2.5ms and runs into a horizontal spring. The spring has

alexgriva [62]

Answer:

Explanation:

Given

mas of block $m=4\ kg$

speed of block $v=2.5\ m/s$

spring constant $k=30\ N-m$

As the mass collides with the spring its kinetic energy is converted to the Elastic Potential energy of the spring

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$

$x=v\sqrt{\frac{m}{k}}$

$x=2.5\times \sqrt{\frac{4}{30}}$

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2 years ago

Read 2 more answers

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init

Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

$h=\frac{gt^2}{2}$

For second stone

$h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$ ---2

Equating 1 &2 we get

$\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$

$\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$26.46t-35.721-52.92t+142.884=0$

$t=4.05 s$

4 0

2 years ago