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:
100,800 Jkg
The heat that is used to change the state of a mater is called latent heat.
In this case it is converting ice to water and it is called latent heat of fusion.
It is given by:
Heat = mc
where m is the mass of ice and l is the specific latent heat of fusion of ice.
l = 0.336 MJ
Heat = 0.3 × 0.336 MJ
= 0.3 × 0.336 × 10⁶
= 100,800 Jkg
Answer:
E = 3600 J
Explanation:
Given that,
Voltage, V = 115 V
Power of electric bulb, P = 60 W
We need to find the electric energy used in 1 minute. The electric energy use is given by :

Hence, the electrical energy is 3600 J.
The cost of developing thermonuclear power with plasmabe defended because D. It can provide an inexpensive power source.
<h3>How did the
cost of developing t
hermonuclear power defended?</h3>
The cost of developing thermonuclear power defended becvause we can see in the paragraph how it was told that the generation of ths power can be donee through the understanding of the occurrence of plasmain nature,
It should be noted that this thermonuclear power with plasmabe posses the characteristics which make it to exist in the ionosphere, and it can be felt in the flames as well; as in the chemical and nuclearexplosions.
In conclusion the power can be seen as an inexpensive source power because the p[roduction of this power cn be found in most of the thing that can be found around us as discused above.
Therefore, option D is correct.
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