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3 years ago

How do the Van de Graaff generator make the tape bend?

1 answer:

7 0

My Van de Graaff generator will create sparks about 10 to 12 inches in length. I like to charge myself on it and point at the aluminum blinds on the window. The charge (electronic wind) will cause the blinds to move. I can do this from about 8 feet away with ease. Soap bubbles are also interesting to play with around the Van de Graaff generator. They initially are attracted to the Van de Graaff generator and float toward it; once they become charged by the Van de Graaff generator, they float away due to repulsion. There are multitudes of fun things you can do with your Van de Graaff generator. Use your imagination!

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Anon25 [30]

Total distance moved by bead is 1.952 cm.

Explanation:

Let first consider all data that are given in question.

1. F = 8 N ...force acting on string

2. f = 2 Hz ...frequency of system

3. β = 4 cm = 0.04 m ...wavelength of wave formed due to vibration 4. A = 1 cm = 0.01 m ...Amplitude of vibration

Under certain conditions, waves can bounce back and forth through a particular region, effectively becoming stationary. These are called standing waves.

Here,it is due to vibration induced in spring due to tension induced in string

Standing wave equation is given by

$y = (x,t) = 2A * sinK x * cos (wt)$ ...(1)

Let first find, value of K, x, w, t

k = 2 * pi / beta ....(2)

Where β is wavelength in meters

K is wave number

k = 2 * pi / 0.01

k = 628.31 $m^{-1}$

now, let us find value of w

W = 2 x pi x f ....(3)

where f is frequency in hertz

W = 2 x pi x 2

W = 4 x pi

W = 0.08 $\frac{m}{s}$

$y = (x,t) = 2A * sinK x * cos (wt)$

now, let us find value of v that is wave speed

Notice that some x-positions of the resultant wave are always zero no matter what the phase relationship is. These positions are called nodes.

Finding the positions where the sine function equals zero provides the positions of the nodes.

In our case, and

K * x = pi

x = 0.04 / 2

x = 0.02

$y = (0.02,1) = 2(0.01) * sin pi * cos (12.5664 * 1)$

$y = (0.02,1) = 2(0.01) * -1 * cos 0.9761$

Y = 1.952 cm

Finally, when bead is at middle of the string, total distance after stretch covered is 1.952 cm.

3 0

3 years ago

Bonnie and Clyde are trying to steal the world's largest diamond from a 10 story

german

Answer:

The speed Clyde will be falling at is 33.72.

6 0

3 years ago

What are the main causes of crack growth in rocks overtime

lana [24]

Ice Wedging and Plant Growth

4 0

3 years ago

Read 2 more answers

A daring stunt woman sitting on a tree limb wishes to drop vertically onto a horse gallop ing under the tree. The constant speed

Shkiper50 [21]

Answer:

0.67 seconds

8.576 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.23=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2.23\times 2}{9.8}}\\\Rightarrow t=0.67\ s$

Time taken by the stunt woman to drop to the saddle is 0.67 seconds which is the time she will stay in the air.

Speed of the horse = 12.8 m/s

Distance = Speed × Time

⇒Distance = 12.8×0.67

⇒Distance = 8.576 m

Hence, the distance between the horse and stunt woman should be 8.576 m when she jumps.

3 0

3 years ago

A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the spee

Naddika [18.5K]

Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

$s = ut +\frac{1}{2}at^{2}$

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

$s = \frac{1}{2}gt'^{2}$

$s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}$ (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

$s = v\times t''$

$s = 343\times t''$ (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t' (3)

Using eqn (2) and (3):

s = 343(3.5 - t') (4)

from eqn (1) and (4):

$4.9t'^{2} = 343(3.5 - t')$

$4.9t'^{2} + 343t' - 1200.5 = 0$

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

3 0

3 years ago