Your school science club has devised a special event for homecoming. You've attached a rocket to the rear of a small car that ha

Explanation:

• A substance will float if it has a lower density than the liquid it is placed in.
• A substance will sink if its density exceeds that of the liquid.

Density of corn syrup =

1) Density of gasoline =

Gasoline's density is less than that of corn syrup, indicating it will float in corn syrup.

2) Density of water =

Water's density is also less than that of corn syrup, meaning it will float in corn syrup.

3) Density of honey =

Honey's density exceeds that of corn syrup, so it will sink in corn syrup.

4) Density of titanium =

The density of titanium is greater than that of corn syrup, hence it will sink in corn syrup.

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.

The answer is 195 J.

J(r) = Br. We know that the area of a small segment, dA, is represented as 2 π dr. Thus, I = J A and dI = J dA. Plugging in the values gives us dI = B r. 2 π dr which simplifies to dI= 2π Br² dr. Now, integrating the above equation: Given that B= 2.35 x 10⁵ A/m³, with r₁ = 2 mm and r₂ equal to 2 + 0.0115 mm, or 2.0115 mm.

I will assume the girl is on the right while the boy is on the left.
The net force represents the total of all forces acting on an object, factoring in negatives.
Let the force from the boy be denoted as b. We’ll apply the formula F = ma.

b + 3.5 = 0.2(2.5)

This reduces to a straightforward algebraic problem. By solving, we find that the boy is applying a force of -3N to the left.