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MrRa
1 month ago
11

A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft

must hold a 730 kg steel cube with edges of length 45.0 cm without sinking. Assume the density of water in the pool is 1000 kg/m3.
The students use a crane to gently place the cube on the raft but accidentally place it off center. The cube remains on the raft for a few moments and then the raft tilts, causing the cube to slide off and sink to the bottom of the pool. The raft remains floating in the pool. In a coherent paragraph-length response, indicate whether the water level in the pool when the cube is on the bottom of the pool is higher than, lower than, or the same as when the cube is on the raft, and explain your reasoning. For both cases, assume that there is no motion of the water.
Physics
1 answer:
Sav [3.1K]1 month ago
7 0

Answer:

The water level increases more when the cube is above the raft before it sinks.

Explanation:

The principle involved is based on Archimedes' theory, which states that immersing an object in water will raise the initial water level. This means that the height of the water in the container increases. The increase in water volume corresponds to the volume of the submerged object.

We can consider two scenarios: when the steel cube rests on the raft and when it settles at the pool's bottom.

When the cube rests at the pool’s bottom, the volume will indeed rise, and we can ascertain this using the cube's volume.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

When an object floats, it's because the densities of the object and water are in equilibrium.

Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]

The formula for density is:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyant force can be calculated with this equation:

F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]

Vs > Vc indicates that the combined volume of the raft and the cube exceeds that of the cube alone resting at the bottom of the pool. Hence, when the cube is positioned above the raft, the water level rises more before it becomes submerged.

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A window washer on a hanging platform cleans the outside of windows on a skyscraper. The washer hoists the platform up the side
kicyunya [3294]

Answer:

The acceleration of the platform is - 1.8 m/s²

Explanation:

The net force on a body causes that body to accelerate in the direction of the resultant force applied.

Setting up the force equilibrium for the configuration:

ma = 800 - mg

100a = 800 - 100×9.8

100a = - 180

100a = - 180

a = - 1.8 m/s²

This indicates that the body is falling downward.

6 0
1 month ago
A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of
Sav [3153]

Answer:

A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.

Explanation:

τchild=τrock  

We will utilize the formula for torque:

(F)child(d)child)=(F)rock(d)rock)

The gravitational force acts equally on both objects.

(m)childg(d)child)=(m)rockg(d)rock)

We can eliminate gravity from both sides of the equation for simplification.

 (m)child(d)child)=(m)rock(d)rock)  

Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.

(25kg)(1m)=(50kg)drock

Solve for the distance where the rock should be positioned in relation to the seesaw's center.

drock=25kg⋅m50kg

drock=0.5m

6 0
1 month ago
A child on a 2.4 kg scooter at rest throws a 2.2 kg ball. The ball is given a speed of 3.1 m/s and the child and scooter move in
Maru [3345]

Answer:

The mass of the child is 14.133 kg

Explanation:

According to the principle of linear momentum conservation, we get;

(m₁ + m₂) × v₁ + m₃ × v₂ = (m₁ + m₂) × v₃ - m₃ × v₄

The negative sign signifies that their velocities were directed oppositely

Considering both the child and the ball are initially at rest, we have;

v₁ = 0 m/s and v₂= 0 m/s

Thus;

0 = m₁ × v₃ - m₂ × v₄

(m₁ + m₂) × v₃ = m₃ × v₄

Where:

m₁ = Mass of the child

m₂ = Mass of the scooter = 2.4 kg

v₃ = Final velocity of the child and scooter = 0.45 m/s

m₃ = Mass of the ball = 2.4 kg

v₄ = Final velocity of the ball = 3.1 m/s

Substituting the known values gives us;

(m₁ + 2.4) × 0.45 = 2.4 × 3.1

(m₁ + 2.4) = 16.533

This implies m₁ + 2.4 = 16.533

Thus, m₁ = 16.533 - 2.4 = 14.133 kg

The mass of the child equals 14.133 kg.

3 0
13 days ago
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6 0
22 days ago
a force of 6lbs acts on an object with a weight of 35 lbs on earth. determine the objects acceleration. final answer must be 5.5
kicyunya [3294]
Weight of the object = 35 lbs
F = ma
m = F/a = 35/32 (with acceleration of 32 ft/s²)
m= 1.09

Again applying the same formula,
a = F/m
a= 6/1.09
a= 5.489

Thus, the acceleration is approximately 5.5 ft/s²!!
5 0
1 month ago
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