The reservoir stores 6 500 000 m3 of water. The density of the water is 998 kg/m3. Calculate the mass of water in the reservoir.
1 answer:
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Answer:
(a) 2.5 m/s
(b) 37.5 KJ
Explanation:
(a)
From the law of conservation of momentum, Initial momentum=Final momentum
and making
the subject then
and since
is initial velocity of car, value given as 4 m/s,
is the initial velocity of the three cars stuck together, value given as 2 m/s and
is the final velocity which is unknown. By substitution
(b)
Initial kinetic energy is given by
Final kinetic energy is given by
The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence
Energy lost=350-312.5=37.5 KJ
Answer:
A) 16.6 m/s i -17.2 m/s j B) 23.9 m/s c) 46º below horizontal.
Explanation:
A) Once released, the football is not under the influence of any external force in the horizontal direction, so it continues moving at a constant speed equal to the initial velocity, i.e., 16.6 m/s.
If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:
vₓ = 16.6 m/s i
In the vertical direction, the football, once released, is in free fall, starting from rest.
So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:
vy = a*t = -g*t = -9.81 m/s²*1.75 s = -17.2 m/s
Assuming that the upward direction is the positive for the y-axis (perpendicular to the chosen x-axis), we can write the vertical component of the velocity vector, at t=1.75 s, as follows:
vy = -17.2 m/s j
So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:
v = 16.6 m/s i -17.2 m/s j
b) The magnitude of this vector can be found applying trigonometry, as the magnitude is the hypotenuse of a triangle with sides equal to vx and vy, as follows:
v = 23.9 m/s
c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:
tg θ =
⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.