Answer:
(a) 1.257 x 10^5 J
(b) 1.456 Watt
Explanation:
Volume of blood, v = 7500 L = 7.5 m^3
Height, h = 1.63 m
density of blood, d = 1.05 x 10^3 kg/m^3
(a) work done = m x g x h
W = v x d x g x h = 7.5 x 1.05 x 1000 x 9.8 x 1.63 = 1.257 x 10^5 J
(b) time = 1 day = 24 x 60 x 60 s = 86400 seconds
Power = Work / time = 1.257 x 10^5 / 86400 = 1.456 Watt
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
Answer:
Explanation:
Given
mass of steel ball 
initial speed of ball 
Final speed of ball 
(in upward direction)
Impulse imparted is given by change in the momentum of object
therefore impulse J is given by
so magnitude of Impulse =4 N-s
Answer:
θ = 28.9
Explanation:
For this exercise let's use the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where we use index 1 for air and index 2 for water where the fish is
sin θ₂ = n₁ / n₂ sin θ₁
in this case the air repair index is 1 and the water 1.33
we substitute
sin θ₂ = 1 / 1.33 sin t 40
sin θ = 0.4833
θ = sin⁻¹ 0.4833
θ = 28.9
