In a nuclear reactor 1 g of mass is converted into energy. How much energy in Joules is produced?
1 answer:
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1) The law of motion of the projectile is
To find the velocity, we should compute the derivative of h(t):
So now we can calculate the speed at t=2 s and t=4 s:
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:
So we have
And solving this we find
3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
with negative sign, because it is directed downward.
To find the velocity, we should compute the derivative of h(t):
So now we can calculate the speed at t=2 s and t=4 s:
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:
So we have
And solving this we find
3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
with negative sign, because it is directed downward.
Answer:
a. S(t)=350−1t
Explanation:
To determine the equation of motion you take into account the general form of motion with constant velocity:
( 1 )
So is the initial position from a specific reference frame. In this case is 350 m.
v is the speed of the motion, in this case is 1m/s. However, the motion is forward the zero point of the reference frame, hence, the speed is - 1m/s.
You replace the values of So and v in the equation ( 1 ) and you obtain:
Hence, the answer is:
a. S(t)=350−1t
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Para determinar a equação do movimento, você leva em consideração a forma geral do movimento com velocidade constante:
(1)
Assim é a posição inicial de um quadro de referência específico. Neste caso, é de 350 m.
v é a velocidade do movimento, neste caso é de 1m / s. No entanto, o movimento é avançar o ponto zero do quadro de referência, portanto, a velocidade é de - 1m / s.
Você substitui os valores de So ev na equação (1) e obtém:
Portanto, a resposta é:
uma. S (t) = 350-1t, movimento retrógrado