The reaction energy of a reaction is the amount of energy released by the reaction. It is found by determining the difference in

Answer:

Explanation:

The first number is .

The second number is .

We must multiply these two numbers together.

In scientific notation:

Therefore, this is the solution you are looking for.

Answer:

The kinetic energy is higher for the first cart.

Explanation:

For the second cart, its mass is 2kg and the momentum measured is 10kg m/s, which leads to

resulting in

.

Consequently, the kinetic energy for the 3kg cart ends up as

indicating it is less than that of the 1kg cart so it follows that the first cart possesses greater kinetic energy.

Answer: Tension = 47.8N, Δx = 11.5× m.

              Tension = 95.6N, Δx = 15.4× m

Explanation: The speed of a wave on a string under tension can be determined using the following:

denotes tension (N)

μ refers to linear density (kg/m)

Calculating the velocity:

0.0935 m/s

Distance a pulse traveled in 1.23ms:

Δx = 11.5×

With a tension of 47.8N, the distance a pulse will cover is Δx = 11.5×  m.

When tension is doubled:

|v| = 0.1252 m/s

Distance in the same time:

15.4×

With the increased tension, it moves 15.4× m

Response:

The car's acceleration will be

Reasoning:

We are provided with a mass for the ball, m = 1600 kg

The force directed northward is F= 7560 N

The resistance force that counters the car's motion

Thus, the overall force acting on the car

According to Newton's second law, we understand that F=ma

Therefore

Answer:

Part A) Electric fields at the designated point due to charges q₁ and q₂:

E₁ = 33.75 * 10³ N/C (-j), E₂ = (6.48 (-i) + 8.64 (+j)) * 10³ N/C

Part B) The overall electric field at P (Ep)

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C

Explanation:

Conceptual analysis

The electric field at point P caused by a point charge is calculated as:

E = k*q/d²

E: Electric field measured in N/C

q: charge magnitude in Newtons (N)

k: electric constant measured in N*m²/C²

d: distance from the charge q to point P in meters (m)

Equivalence:

1 nC = 10⁻⁹ C

1 cm = 10⁻² m

Data:

k = 9 * 10⁹ N*m²/C²

q₁ = -6.00 nC = -6 * 10⁻⁹ C

q₂ = +3.00 nC = +3 * 10⁻⁹ C

d₁ = 4 cm = 4 * 10⁻² m

d₂ = 5 * 10⁻² m

Part A) Calculation for electric fields at point from q₁ and q₂:

Refer to the attached illustration:

E₁: Electric Field at point P(0,4) cm due to charge q₁. Since q₁ is negative (q₁-), the electric field approaches the charge.

E₂: Electric Field at point P(0,4) cm due to charge q₂. Since q₂ is positive (q₂+), the electric field emanates from the charge.

E₁ = k*q₁/d₁² = 9 * 10⁹ * 6 * 10⁻⁹ / (4 * 10⁻²)² = 33.75 * 10³ N/C

E₂ = k*q₂/d₂²= 9 * 10⁹ * 3 * 10⁻⁹ / (5 * 10⁻²)² = 10.8 * 10³ N/C

E₁ = 33.75 * 10³ N/C (-j)

E₂x = E₂cosβ = 10.8 * (3/5) = 6.48 * 10³ N/C

E₂y = E₂sinβ = 10.8 * (4/5) = 8.64 * 10³ N/C

E₂ = (6.48 (-i) + 8.64 (+j)) * 10³ N/C

Part B) Calculation for net electric field at P (Ep)

The electric field at point P from multiple point charges is the vector sum of the individual electric fields.

Ep = Epx (i) + Epy (j)

Epx = E₂x = 6.48 * 10³ N/C (-i)

Epy = E₁y + E₂y = (33.75 * 10³ (-j) + 8.64 * 10³ (+j)) N/C = 25.11 * 10³ (-j) N/C

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C

Ep = (6.48 * 10³ (-i) + 25.11 * 10³ (-j)) N/C