Answer:
C.
Explanation:
Option C
The information given in the question tells us about the number of electrons in an atom and also the number of shells in the atom. So, we will come to know about the atomic number, size and chemical properties of the atom. But we cannot determine atomic mass. Atomic mass is a function of number of neutrons and protons.
<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>
Answer:
The equation of motion is 
Explanation:
Lets calculate
The weight attached to the spring is 24 pounds
Acceleration due to gravity is 
Assume x , is spring stretched length is ,4 inches
Converting the length inches into feet 
The weight (W=mg) is balanced by restoring force ks at equilibrium position
mg=kx
⇒ 
The spring constant , 
= 72
If the mass is displaced from its equilibrium position by an amount x, then the differential equation is
Auxiliary equation is, 
=
Thus , the solution is 
The mass is released from the rest x'(0) = 0
=0
Therefore , 
Since , the mass is released from the rest from 4 inches
inches
feet
feet
Therefore , the equation of motion is 
a = 7.8 m/s^2
Explanation:
Let Fnet = net force = ma
m = mass of the skydiver
a = acceleration caused by Fnet
W = weight = mg
f(air) = frictional force due to air resistance
Fnet = W - f(air)
= (100 kg)(9.8 m/s^2) - (200 N)
= 780 N
Therefore, the acceleration of the skydiver due to Fnet is
a = Fnet/m
= (780 N)/(100 kg)
= 7.8 m/s^2
