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 ]
Answer:
A title
Explanation:
Because this is middle school.
Answer:
The metal will melt but their will be no change in temperature.
Explanation:
The metal is at its melting temperature which means it is still in solid phase but have to cross the enthalpy of its condensation at this same temperature to convert into liquid phase.
<u>On supplying heat, the metal's temperature will not change as the heat will be required as enthalpy of condensation to melt the solid to liquid at the melting temperature.</u>
Answer:
q = 400 nC
the correct answer is b
Explanation:
The expression for the electric potential of a point charge is
V = k q / r
they ask us for the electrical charge
q = V r / k
let's calculate
Q = 600 6.0 / 9 10⁹
Q = 4 10⁻⁷ C
let's reduce to nC
Q = 4 10⁻⁷ C (10⁹ nC / 1C)
q = 4 10² nC = 400 nC
the correct answer is b
Traslate
La expresión para el potencial eléctrico de una carga puntual es
V = k q/r
nos piden la carga eléctrica
q= V r /k
calculemos
Q= 600 6,0 / 9 10⁹
Q= 4 10⁻⁷ C
reduzcamos a nC
Q = 4 10⁻⁷ C(10⁹ nC/1C )
q = 4 10² nC = 400 nC
la respuesta correcta es b
Answer:
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