A 6.00 kg object is released from rest while fully submerged in a liquid. The liquid displaced by the submerged object has a mass of 3.45 kg. How far and in what direction does the object move in 0.200 s, assuming that it moves freely and that the drag force on it from the liquid is negligible

The object moves 0.0833 m downwards.

The object moves 8.33 cm downwards.

Explanation:

Weight of object = W = 6 × 9.8 = 58.8 N

Mass of liquid displaced = 3.45 kg

Meaning that buoyant force = Fb = 3.45 × 9.8 = 33.81 N

drag force on the object from the liquid is negligible.

Net force acting on the object = ma

Force balance on the object gives

Net force = W - Fb + D

W = 58.8 N

Fb = 33.81 N

D = 0

Net force = 58.8 - 33.81 = 24.99 N

ma = 24.99

6a = 24.99

a = (24.99/6) = 4.165 m/s²

Now, using one of the equations of motion

u = 0 m/s

a = 4.165 m/s²

y = ?

t = 0.200 s

y = ut + ½at²

y = 0 + (0.5 × 4.165 × 0.2²)

y = 0.0833 m

Since the Weight of the body is more than the buoyant force exerted by the liquid on the object, the object moves downwards.

Hope this Helps!!!

0.4 m

Explanation:

Given

Speed of ambulance, vs = 61.9 m/s

Speed of car = 28.5 m/s

Frequency of ambulance siren, f = 694 Hz

Velocity of sound in air, v = 343 m/s

With speed of ambulance being (61.9 m/s) -> We solve using

fd = f(v + vr) / (v - vs), where vr = 0

fd = 694 * (343 + 0) / (343 - 61.9)

fd = 694 * (343 / 281.1)

fd = 694 * 1.22

fd = 847 Hz

Recall,

λ = v/f

λ = 343/847

λ = 0.4 m

Therefore, the wavelength of the sound of the ambulance’s siren if you are standing at the position of the car is 0.4 m