A 2.40 kg stone is sliding in the +x-direction on a horizontal, frictionless surface. It collides with a 4.00 kg stone at rest.

#1

The volume of lead measures 100 cm^3

with a density of lead at 11.34 g/cm^3

. Thus, the mass of the lead block equals density multiplied by volume

Therefore, its weight in air is noted as

Next, the buoyant force acting on the lead is defined as

We know that

After solving, we find

V = 11.22 cm^3

(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N

(iii) The buoyant force measures 0.11 N

(iv) The lead block sinks in water due to its density being greater than that of water.

#2

The buoyant force acting on the lead block counterbalances its weight

(ii) This volume of mercury corresponds to the buoyant force weight, confirming that the block floats within mercury, resulting in 11.11 N as its weight.

(iii) The buoyant force is recorded as 11.11 N

(iv) Given that lead's density is less than mercury's, the lead will float in the mercury medium.

#3

Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.

1. τbiceps = +(Positive) 2. τforearm = -(Negative) 3. τball = -(Negative) Explanation: The attached figure illustrates the following: 1. For the biceps, τbiceps indicates that torque is calculated as Torque = r x F, where r and F are vectors. Here, r corresponds to the vector from the elbow to the biceps. In the figure, the force from the biceps is directed upwards. Applying the right-hand rule from r to F results in counterclockwise torque, which is considered positive (+). 2. The torque related to the weight of the forearm, τforearm, uses the same torque formula, with r being the vector from the elbow to the forearm. The weight acts downward, causing a clockwise torque that is negative (-). 3. Similarly, for the weight of the ball, τball, the downward force from the ball's weight generates a clockwise torque, which also registers as negative (-).

Answer:

Explanation:

Provided:

The trolley, with mass M, is allowed to roll freely without friction.

The coefficient of friction between the trolley and mass m is .

A force F is applied to mass m.

The acceleration of the system is

The frictional force will counterbalance the weight of the block.

The frictional force is

The calculation for the horizontal component is performed as follows:
Vhorizontal = V · cos(angle)

For your instance, Vhorizontal = 16 · cos(40) equates to 12.3 m/s

Conclusion: 12.3 m/s