The ears are superior and posterior to the shoulders and lateral to the nose.
Answer:
a) 3.37 x 
b) 6.42kg/
Explanation:
a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .
Weight of metal in air = 50N = mg implies the mass of metal is 5kg.
Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x
. So density of metal = mass of metal / volume of metal = 5 / 14 x
= 3.37 x 
b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/
Answer:
Displacement after 5 seconds is 155/2 meters
Explanation:
Let X (t) represent the equation of the position, then you have to d2x / dt2 = 5.
Applying the fundamental theorem of the calculation dx/dt = 5t + vo. The speed equation is V (t) = 5t + vo. Since the initial velocity is 30m/s, V (0) = 5 (0) + vo = 30. Therefore, V (t) = dx/dt = 5t + 30. Applying again the fundamental theorem of the calculation X (t) = 5t^2 / 2 + 30t + xo.
Displacement in 5 seconds is given by X (5) - X (0).
X (5) - X (0) = 5 (5)^2/2 +3 (5) + Xo - 5 (0)^2/2 -3 (0) -Xo = 155/2
Displacement after 5 seconds is 155/2 meters
Answer:
42.9 μF
Explanation:
V = 3.50 V, Q = 150 μC
C = Q/V = 150/3.50 μF = 42.9 μF
Answer:
See below
Explanation:
Energy is lost in the form of friction/heat/sound
you cannot get more work out of a machine than you put into it.