Apply the impulse-momentum relation and the work-energy theorem to calculate the maximum value of t if the cake is not to end up

Answer:

2.0

Explanation:

because I'm a geek and ik

Since the roundabout operates at a constant angular velocity, the input power equals the frictional power. Given that the frictional power is 2.5 kW, we can express this as frictional torque multiplied by angular velocity: frictional torque x 0.47 = 2.5 kW. Therefore, solving for frictional torque gives us 2.5 / 0.47 kN.m, which amounts to approximately 5.32 kN.m, leading to a rounded value of 5 kN.m. When the power supply is interrupted, the roundabout experiences deceleration due to the influence of the frictional torque.

Answer:

Q = 12.5 kJ

Explanation:

The formula used to compute heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

ΔH° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

ΔH° = 35.0 kJ/mol

Using these two reactions, we can determine the heat change.

Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

2A + B + 2A2B -------> 2AB + A2

So, for the enthalpy, the values are summed:

H°3 = -25 + 35 = 10 kJ/mol

Now we can calculate the heat:

Q = 10 * 2.5 = 25 kJ

However, as we have 2A in the reaction, it does not maintain a 1:1 mole ratio, instead, it is 1:2, which requires us to adjust; thus:

Q = 25 / 2 = 12.5 kJ

Answer:

20 cm

Explanation:

The electric potential energy U is calculated with the formula U = kq₁q₂/r, where q₁ = 5 nC (5 × 10⁻⁹ C) and q₂ = -2 nC (-2 × 10⁻⁹ C) and r is determined as √(x - 2)² + (0 - 0)² + (0 - 0)² = x - 2. This leads to U = -0.5 µJ (-0.5 × 10⁻⁶ J), where k = 9 × 10⁹ Nm²/C².

Thus, solving for r gives us r = kq₁q₂/U

which leads to x - 2 = kq₁q₂/U

Then, rearranging gives x = 0.02 + kq₁q₂/U m

So, x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

Resulting in x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

This simplifies to x = 0.02 + 0.18 = 0.2 m, or 20 cm

The answer is 9938.8 km. Explanation: 1 pound-force = 4.48 N. Hence, 30.0 pounds-force = 134.4 N. The gravitational force between Earth and an object on its surface is defined by: Where M denotes Earth’s mass, m is the object's mass, and R represents the Earth's radius (6371 km). To determine height (h) above Earth's surface, we compare ratios. Ultimately, Pete's weight would be 30 pounds at a height of 9938.8 km from the Earth's surface.