Answer:14.14 cm
Explanation:
Given
Spring Compression ![x=12 cm](https://tex.z-dn.net/?f=x%3D12%20cm)
Potential energy Stored in spring![=72 J](https://tex.z-dn.net/?f=%3D72%20J)
Suppose k is the spring constant of spring
Potential Energy of spring is given by ![=\frac{kx^2}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bkx%5E2%7D%7B2%7D)
![\frac{k(0.12)^2}{2}=72](https://tex.z-dn.net/?f=%5Cfrac%7Bk%280.12%29%5E2%7D%7B2%7D%3D72)
![k(0.12)^2=144](https://tex.z-dn.net/?f=k%280.12%29%5E2%3D144)
![k=10,000 N/m](https://tex.z-dn.net/?f=k%3D10%2C000%20N%2Fm)
![k=10 kN/m](https://tex.z-dn.net/?f=k%3D10%20kN%2Fm)
for 100 J energy
![\frac{k(x_0)^2}{2}=100](https://tex.z-dn.net/?f=%5Cfrac%7Bk%28x_0%29%5E2%7D%7B2%7D%3D100)
![10\times 10^3\cdot (x_0)^2=200](https://tex.z-dn.net/?f=10%5Ctimes%2010%5E3%5Ccdot%20%28x_0%29%5E2%3D200)
![(x_0)^2=2\times 10^{-2}](https://tex.z-dn.net/?f=%28x_0%29%5E2%3D2%5Ctimes%2010%5E%7B-2%7D)
![x_0=0.1414](https://tex.z-dn.net/?f=x_0%3D0.1414)
![x_0=14.14 cm](https://tex.z-dn.net/?f=x_0%3D14.14%20cm)
Answer the time you bee spending driving iss 795 because 895-795=100
Explanation:
25 x 10^-5
= 0.00025
25 cm
= 0.00025 km
Ventilation is very important because it helps remove the gas form people’s homes and schools and it redirects the random gas outside so it is less likely to hurt people
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>