When Carbon -14 decays into Nitrogen -14, what kind of decay is this?

A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting

ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

$T=\dfrac{38}{8}\\\\T=4.75\ s$

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

k = spring stiffness constant

So,

$k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m$

When the cord is in air,

mg=kx

x = the extension in the cord

$x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m$

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

5 0

3 years ago

A 10 kg bowling ball sits at the top of a 10 m hill and then slides down its icy hillside.

Nataly_w [17]

We need more informtion

5 0

3 years ago

Read 2 more answers

An air-filled pipe is found to have successive harmonics at 945 Hz , 1215 Hz , and 1485 Hz . It is unknown whether harmonics bel

Aleonysh [2.5K]

Answer:

L = 0.635m

Explanation:

This problem involves the concept of stationary waves in pipes. For pipes closed at one end,

The frequency f = nv/4L for n = 1,3,5....n

For pipes open at both ends

f = nv/2L for n = 1,2,3,4...n

Assuming the pipe is closed at one end and that velocity of sound is 343m/s in air. If we are right we will obtain a whole number for n.

The film solution can be found in the attachment below.

8 0

3 years ago

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug

telo118 [61]

Answer:

a) $V_{x}=3.72m/s$ , b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

$y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}$

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

$0=y_{0}+\frac{1}{2}a_{y}t^{2}$

so we can solve for t, so we get:

$t=\sqrt{\frac{-(2)y_{0}}{a}}$

so now we can substitute the known values, so we get:

$t=\sqrt{\frac{-(2)(1.42)}{-9.8}}$

which yields:

t=0.538s

So we can use this value to find the velocity in x:

$V_{x}=\frac{x}{t}$

When substituting we get:

$V_{x}=\frac{2m}{0.538s}$

which yields:

$V_{x}=3.72m/s$

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

$\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}$

We know that $V_{0}$ is zero, so we can simplify the expression:

$\Delta y=\frac{V_{yf}^{2}}{2a}$

So we can solve the equation for $V_{yf}^{2}$ so we get:

$V_{yf}=\sqrt{2\Delta y a}$

and when substituting the known values we get:

$V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}$

which yields:

$V_{yf}=-5.28m/s$

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

$tan \theta =\frac{V_{y}}{V_{x}}$

when solving for theta we get:

$\theta = tan^{-1}(\frac{V_{y}}{V_{x}})$

We can substitute so we get:

$\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})$

which yields:

$\theta = -54.83^{o}$

7 0

3 years ago

The Astronomer Carl Sagan is famous. Why?

Zolol [24]

Mainly because he was Johnny Carson's advisor and consultant
on space, astronomy, and science in general, and he appeared
on The Tonight Show Starring Johnny Carson many times.

6 0

3 years ago