https://file.aiquickdraw.com/mathbot/file/1726392012270l4937qge.jpg

Let's break down and solve the problems based on the image you've provided.

---

### Problem a)
You are given the equation for acceleration \( a \):

\[
a = \frac{v_2 - v_1}{t_2 - t_1}
\]

Where:
- \( a = 0.01323 \, \text{m/s}^2 \)
- \( v_2 = 5.8 \, \text{km/h} = 5.8 \times \frac{1000}{3600} = 1.611 \, \text{m/s} \)
- \( v_1 = 87 \, \text{cm/s} = 0.87 \, \text{m/s} \)
- \( t_1 = 16 \, \text{s} \)

We need to solve for \( t_2 \).

Rearranging the formula to solve for \( t_2 \):

\[
t_2 = t_1 + \frac{v_2 - v_1}{a}
\]

Substituting the values:

\[
t_2 = 16 + \frac{1.611 - 0.87}{0.01323}
\]

Now let's calculate \( t_2 \).

---

### Problem b)
The power equation is given as:

\[
P = \frac{E_2 - E_1}{\Delta t}
\]

Where:
- \( P = 362 \, \text{kW} = 362 \times 10^3 \, \text{W} \)
- \( E_2 = 0.088 \, \text{kJ} = 0.088 \times 10^3 \, \text{J} \)
- \( \Delta t = 220 \, \mu \text{s} = 220 \times 10^{-6} \, \text{s} \)

We need to solve for \( E_1 \).

Rearranging the formula to solve for \( E_1 \):

\[
E_1 = E_2 - P \times \Delta t
\]

Substituting the values:

\[
E_1 = 0.088 \times 10^3 - (362 \times 10^3) \times (220 \times 10^{-6})
\]

Let's calculate \( E_1 \).

---

### Problem c)
The work equation is given as:

\[
W = mg(y_1 - y_2)
\]

Where:
- \( W = -158 \, \mu J = -158 \times 10^{-6} \, \text{J} \)
- \( m = 16 \, \text{mg} = 16 \times 10^{-3} \, \text{g} = 16 \times 10^{-6} \, \text{kg} \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( y_2 = 17 \, \text{dm} = 1.7 \, \text{m} \)

We need to solve for \( y_1 \).

Rearranging the formula to solve for \( y_1 \):

\[
y_1 = y_2 + \frac{W}{mg}
\]

Substituting the values:

\[
y_1 = 1.7 + \frac{-158 \times 10^{-6}}{16 \times 10^{-6} \times 9.81}
\]

Let's calculate \( y_1 \).

---

### Problem d)
The supporting force \( T \) is given by:

\[
T = mg \sin(\theta) - ma
\]

Where:
- \( m = 55 \, \text{g} = 55 \times 10^{-3} \, \text{kg} \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( \theta = 0.62 \, \text{rad} \)
- \( a = 11 \, \mu \text{m/ms}^2 = 11 \times 10^{-6} \, \text{m/s}^2 \)

Substituting the values:

\[
T = 55 \times 10^{-3} \times 9.81 \times \sin(0.62) - 55 \times 10^{-3} \times 11 \times 10^{-6}
\]

Let's calculate \( T \).

---

I will now compute the answers for the above expressions.Here are the results for the problems:

### Problem a)
The time \( t_2 \) is approximately **72.02 seconds**.

### Problem b)
The initial energy \( E_1 \) is approximately **8.36 Joules**.

### Problem c)
The height \( y_1 \) is approximately **0.693 meters** (69.3 cm).

### Problem d)
The supporting force \( T \) is approximately **0.3135 Newtons**.

Would you like more details on any of the calculations or have any further questions?

Here are 5 related questions you might find interesting:
1. How does the unit conversion impact the precision of the results?
2. Why is \( t_2 \) significantly larger than \( t_1 \) when acceleration is very small?
3. What physical conditions could result in negative work \( W \)?
4. How does the angle \( \theta \) influence the supporting force in Problem d?
5. How could using degrees instead of radians affect the outcome in Problem d?

**Tip:** Always check whether your calculator or software is set to radians or degrees when dealing with trigonometric functions!

Solve the following problems:
a) Solve for t_2 from the equation a = (v_2 - v_1) / (t_2 - t_1), where a = 0.01323 m/s^2, v_2 = 5.8 km/h, v_1 = 87 cm/s, t_1 = 16s.
b) Solve for E_1 from the equation P = (E_2 - E_1) / Δt, where P = 362 kW, E_2 = 0.088 kJ, Δt = 220 μs.
c) Solve for y_1 from the equation W = mg(y_1 - y_2), where W = -158 μJ, m = 16 mg, g = 9.81 m/s^2, y_2 = 17 dm.
d) Calculate the supporting force T = mg sin(θ) - ma, where m = 55 g, g = 9.81 m/s^2, θ = 0.62 rad, and a = 11 μm/ms^2.

Learn to solve kinematic problems involving velocity, acceleration, energy conservation, and force calculations. This problem set covers key concepts such as acceleration, power, work, and force equations. Suitable for college-level physics students.

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