Answer:
Difference in experimental data.
Explanation:
There is difference of experimental value between the experiment that is performed on the earth and on the international space station because presence of gravity. The result of the experiment on the earth is different due to the presence of gravity that contributes in the result of the experiment as compared to international space station where no gravity is present so there is high difference of the numerical value of the result of both experiments of earth and international space station.
Answer:
The normal stress is 10.7[MPa]
Explanation:
The normal stress can be calculated with the following equation:
![S_{norm} =\frac{F}{A} \\where:\\F= force [Newtons]\\A=area [m^2]\\S_{norm} = Normal stress [\frac{N}{m^{2} }] or [Pa]](https://tex.z-dn.net/?f=S_%7Bnorm%7D%20%3D%5Cfrac%7BF%7D%7BA%7D%20%5C%5Cwhere%3A%5C%5CF%3D%20force%20%5BNewtons%5D%5C%5CA%3Darea%20%5Bm%5E2%5D%5C%5CS_%7Bnorm%7D%20%3D%20Normal%20stress%20%5B%5Cfrac%7BN%7D%7Bm%5E%7B2%7D%20%7D%5D%20or%20%5BPa%5D)
The area of the rod can be calculated using the equation:
![A=\frac{\pi }{4}*d^{2} \\d=8[mm]=0.008[m]\\A=\frac{\pi }{4}*(0.008)^{2} \\A=5.02*10^{-5} [m^{2} ]](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D%2Ad%5E%7B2%7D%20%20%5C%5Cd%3D8%5Bmm%5D%3D0.008%5Bm%5D%5C%5CA%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D%2A%280.008%29%5E%7B2%7D%20%20%5C%5CA%3D5.02%2A10%5E%7B-5%7D%20%5Bm%5E%7B2%7D%20%5D)
The force is the result of the mass multiplied by the gravity.
![F=55[kg]*9.81[m/s^{2} ] = 539.6[N]\\\\S_{norm} = 539.6/5.02*10^{-5} \\S_{norm} = 10.7*10^{6}[Pa] = 10.7[MPa]](https://tex.z-dn.net/?f=F%3D55%5Bkg%5D%2A9.81%5Bm%2Fs%5E%7B2%7D%20%5D%20%3D%20539.6%5BN%5D%5C%5C%5C%5CS_%7Bnorm%7D%20%3D%20539.6%2F5.02%2A10%5E%7B-5%7D%20%5C%5CS_%7Bnorm%7D%20%3D%2010.7%2A10%5E%7B6%7D%5BPa%5D%20%3D%2010.7%5BMPa%5D)
Weight on the Moon = 291 N.
W = g · m, where m stays for the mass and on the Moon g = 1.67 m/s²
291 N = 1.67 m/s² · m
m = 291 kg m / s² : 1.67 m/s²
m = 174.25 kg
Weight on Earth = 9.81 m/s² · 174.25 kg = 1,709.4 N
Answer:
The weight of an astronaut on Earth is 1,709.4 N.
Answer:
a) F = 680 N, b) W = 215 .4 J
, c) F = 1278.4 N
Explanation:
a) Hooke's law is
F = k x
To find the displacement (x) let's use the elastic energy equation
= ½ k x²
k = 2
/ x²
k = 2 85.0 / 0.250²
k = 2720 N / m
We replace and look for elastic force
F = 2720 0.250
F = 680 N
b) The definition of work is
W = ΔEm
W =
- ![K_{eo}](https://tex.z-dn.net/?f=K_%7Beo%7D)
W = ½ k (
² - x₀²)
The final distance
= 0.250 +0.220
= 0.4750 m
We calculate the work
W = ½ 2720 (0.47² - 0.25²)
W = 215 .4 J
We calculate the strength
F = k ![x_{f}](https://tex.z-dn.net/?f=x_%7Bf%7D)
F = 2720 0.470
F = 1278.4 N