50/200×100%=25% is answer the formula is usefull energy output divided by total energy provided into 100%
Answer:
, where the minus indicates the direction is opposite to that of the throw.
Explanation:
a)
Since MKS stands for meter-kilogram-second and we know that:
We can write that:
These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.
So we have that:
b)
Newton's 2nd Law tells us that F=ma, and the definition of acceleration is 
, so we have:
Taking the throw direction as the positive one, for our values we have:
Answer: An equation is missing in your question below is the missing equation
a) ≈ 8396
b) 150 nm/k
Explanation:
<u>A) Determine the number of Oscillators in the black body</u>
number of oscillators = 8395
attached below is the detailed solution
<u>b) determine the peak wavelength of the black body </u>
Black body temperature = 20,000 K
applying Wien's law / formula
λmax = b / T ------ ( 1 )
T = 20,000 K
b = 3 * 10^6 nm
∴ λmax = 150 nm/k
Answer:
<u>B</u>
Explanation:
Planets have different year lengths because it depends how far they revolve from a celestial body. Each planet has its own orbital period. Planets closer to the star will have a lower orbital period compared to the ones that lie far away from it.
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
