What factor has the greatest impact on flexibility?

A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put

olga_2 [115]

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

$F_s = F_g$

$kx = mg$

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

$k = \frac{mg}{x}$

$k = \frac{8*9.8}{0.1}$

$k = 784N/m$

Just before launch the compression is 40cm, then from Potential Elastic Energy definition

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

$PE =\frac{1}{2} 784*0.4^2$

$PE = 63.72J$

Therefore the energy stored in the spring is 63.72J

6 0

2 years ago

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!

8 0

2 years ago

Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for

Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0

2 years ago

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the crit

Dennis_Churaev [7]

Complete Question:

A beam of white light is incident on the surface of a diamond at an angle $\theta_{a}$ , since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. For example, the indices of refraction in diamond are $n_{red} = 2.410$ for red light and $n_{blue} = 2.450$ for blue light. Thus, blue light and red light are refracted at different angles inside the diamond. The surrounding air has $n_{air} = 1.000$ .

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the critical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit. Express your answer in degrees to four significant figures.

Answer:

$\theta_{crit} = 24.09^{0}$

Explanation:

Only the blue refracted ray is related to the critical angle in this question

$n_{air} = 1.000$

$n_{blue} = 2.450$

The relationship between the critical angle( $\theta_{crit}$ ), $n_{air}$ and $n_{blue}$ can be given as $sin \theta_{crit} = \frac{n_{air} }{n_{blue} }$