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kow [346]
1 year ago
14

A visitor at a Michigan lighthouse is trying to measure the height of the lighthouse. She has a big spool of string, but she doe

sn't have a measuring tape. She has a wrist watch with a timer function on it. She attaches a weight to the end of the string and she makes a simple pendulum. She hangs the pendulum down from the top of the spiral staircase and she measures the period of the pendulum. The period of the pendulum is 8.48 s. What is the height of the light house?
Physics
1 answer:
Pie1 year ago
4 0

The height of the light house is 17.86m.

To find the answer, we have to know about the simple pendulum.

<h3>How to find the height of the light house?</h3>
  • We have the expression for time period of the pendulum as,

                       T=2\pi \sqrt{\frac{l}{g} } \\

where, g is acceleration due to gravity, and l is the length of the pendulum.

  • For the pendulum with time period T kept at some height under the influence of gravity, then the height will be equal to,

              h=(\frac{T}{2\pi }) ^2g=(\frac{8.48}{2*3.14}) ^2*9.8=17.86m

Thus, we can conclude that, the height of the light house is 17.86m.

Learn more about the simple pendulum here:

brainly.com/question/23269083

#SPJ1

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What is the final speed of a 60 kg boulder dropped from a 111 meter cliff
saveliy_v [14]

After rolling off the edge of the cliff and falling ' M ' meters down,
the speed of the boulder is

       Square root of ( 19.6 M ) .

If M=111 meters, then the speed is <em>46.64 meters per second</em>.

We have known for roughly 500 years that if there's no air resistance,
the mass of the falling object makes no difference, and all objects fall
with the same acceleration, speed, time to splat, etc.



3 0
3 years ago
The half-life of the radioactive isotope Carbon-14 is about 5730 years. Find N in terms of t. The amount of a radioactive elemen
Fudgin [204]

The complete queston is The amount of a radioactive element A at time t is given by the formula

A(t) = A₀e^kt

Answer:  A(t) =N e^( -1.2 X 10^-4t)

Explanation:

Given

Half life =  5730 years.

A(t) =A₀e ^kt

such that

A₀/ 2 =A₀e ^kt

Dividing both sides by A₀

1/2 = e ^kt

1/2 = e ^k(5730)

1/2 = e^5730K

In 1/2 =  5730K

k = 1n1/2 / 5730

k = 1n0.5 / 5730

K= -0.00012 = 1.2 X 10^-4

So that expressing   N in terms of t, we have

A(t) =A₀e ^kt

A₀ = N

A(t) =N e^ -1.2 X 10^-4t

7 0
3 years ago
Which example best illustrates that light behaves like particles?
AlekseyPX
I would say B. Because actual mass would ricochet off the sidewalk.
8 0
3 years ago
Read 2 more answers
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
3 years ago
32. Which type of electromagnetic wave is used for nuclear power and medical treatment?
adelina 88 [10]

Answer:

Answers at the bottom!!

Explanation:

For the first answer: Gamma ray

Gamma rays and x-rays consist of high-energy waves that can travel great distances at the speed of light and generally have a great ability to penetrate other materials. For that reason, gamma rays (such as from cobalt-60) are often used in medical applications to treat cancer and sterilize medical instruments.

For the second answer: They both have technological uses.

In my opinion I'm pretty sure it's A because we do use microwaves and x-rays as technological uses.

Hope this helps!!

Happy Holidays and Season Greeting!!

ii Feliz Navidad a todos !!

3 0
3 years ago
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