Answer:
F₂= 210 pounds
Explanation:
Conceptual analysis
Hooke's law
Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:
F= K*x Formula (1)
Where;
F is the magnitude of the force applied to the spring in Newtons (Pounds)
K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)
x the elongation of the spring (inch)
Data
The data given is incorrect because if we apply them the answer would be illogical.
The correct data are as follows:
F₁ =80 pounds
x₁= 8 inches
x₂= 21 inches
Problem development
We replace data in formula 1 to calculate K :
F₁= K*x₁
K=( F₁) / (x₁)
K=( 80) / (8) = 10 pounds/ inche
We apply The formula 1 to calculate F₂
F₂= K*x₂
F₂= (10)*(21)
F₂= 210 pounds
<span>436 km
The conversion factor between kilocalorie/hour and watts is 1.163 (1 kcal/hr = 1.163 watt). So let's convert the energy consumption of the bird from watts to kcal/hr
3.7 w / 1.163 w hr/kcal = 3.18 kcal /hr
1 gram of fat has 9 kcal, so the total number of kcals consumed will be 4 * 9 = 36.
So the bird can fly for 36/3.18 = 11.32 hours
The distance traveled will be
11.32 h * 3600 s/h * 10.7 m/s / 1000 m/km = 436 km</span>
<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>
Momentum
= mass x velocity
= 0.2 x 5
= 1 kg m/s
