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 ]
What effect man there are a million effects the only one that arent true are fusion or fission and magnetism unless that bowling ball is iron
<h3>
Answer:</h3>
Input work
<h3>
Explanation:</h3>
Concept being tested: Efficiency of machines
Therefore we need to know what is the efficiency of a machine
- Efficiency of a machine is the ratio of work output of machine to the work input expressed as a percentage.
Efficiency = (Work output ÷ Work input) × 100%
- Therefore, if the work input is equal to the work output then the efficiency of the machine will be 100%.
- Most machines are not 100% efficient due to loss of energy in form of heat due to friction of the moving parts of the machine.
Answer:
The correct option is (B).
Explanation:
The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,
It is mentioned that, an asteroid with an orbital period of 8 years. So,
So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.
