**Physics Problem: Understanding Free-Fall and Electric Fields on Planet Tehar****Problem Statement:**On planet Tehar, the free-fall acceleration is the same as that on Earth, but there is also a strong downward electric field that is uniform close to the planet's surface. A 2.00 kg ball having a charge of 6.00 μC is thrown upward at a speed of 20.0 m/s and it hits the ground after an interval of 4.50 s. What is the potential difference between the starting point and the top point of the trajectory?**Analysis and Solution:**1. **Free-Fall Acceleration (g):** - Since the free-fall acceleration on planet Tehar is the same as on Earth, we use \( g = 9.8 \, \text{m/s}^2 \).2. **Initial Information:** - Mass of ball, \( m = 2.00 \, \text{kg} \) - Charge of the ball, \( q = 6.00 \, \mu\text{C} = 6.00 \times 10^{-6} \, \text{C} \) - Initial speed, \( v_0 = 20.0 \, \text{m/s} \) - Total time of flight, \( t = 4.50 \, \text{s} \)3. **Trajectory and Time Calculation:** - The ball is thrown upward and returns to the ground, hence the total time for ascending and descending is 4.50 s. - Time to reach the top of the trajectory, \( t_{\text{up}} = \frac{4.50 \, \text{s}}{2} = 2.25 \, \text{s} \)4. **Vertical Displacement Calculation:** - Using the kinematic equation \( v = u + at \): \[ 0 = 20.0 \, \text{m/s} - (9.8 \, \text{m/s}^2) \times 2.25 \, \text{s} \] This confirms the time to the top point as accurate since \( v = 0 \, \text{m/s} \) at the top.5. **Height Calculation (h):** - Using \( h

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On plant Teher, the free-fall acceleration is the same as that on Earth but there is also a strong downward electric field that is unyorm close to the planet's Surface: A 2.ooks Sall having a charge of Charge of 6.00 Mc is thrown upward it a speed of 22.0 m/s. and it hits the ground fter an interval of 4.505. What is the poten- ial difference between the starting point and The top pint of the trajectory 2

On plant Teher, the free-fall acceleration is the same as that on Earth but there is also a strong downward electric field that is unyorm close to the planet's Surface: A 2.ooks Sall having a charge of Charge of 6.00 Mc is thrown upward it a speed of 22.0 m/s. and it hits the ground fter an interval of 4.505. What is the poten- ial difference between the starting point and The top pint of the trajectory 2

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Concept explainers

Magnetic Field Of Coaxial Cable

The word coaxial means same axis. So for a long straight cable to be coaxial, it must have concentric cylindrical shaped conductor around it. Coaxial cable (popularly called ‘coax’) has various applications ranging from current conductors to radiofrequency signal transmitters. This cable has lower emission losses and provides protection from electromagnetic interference which allows signals with less power to transmit over longer distances.For example, currents produced in the pickup of a stereo turntable generally travel to the amplifier through a coaxial cable.The self-inductance of a coaxial cable is another important characteristic, because it influences the propagation of electrical signals in the cable.

Question

**Physics Problem: Understanding Free-Fall and Electric Fields on Planet Tehar**

**Problem Statement:**

On planet Tehar, the free-fall acceleration is the same as that on Earth, but there is also a strong downward electric field that is uniform close to the planet's surface. A 2.00 kg ball having a charge of 6.00 μC is thrown upward at a speed of 20.0 m/s and it hits the ground after an interval of 4.50 s. What is the potential difference between the starting point and the top point of the trajectory?

**Analysis and Solution:**

1. **Free-Fall Acceleration (g):**
- Since the free-fall acceleration on planet Tehar is the same as on Earth, we use \( g = 9.8 \, \text{m/s}^2 \).

2. **Initial Information:**
- Mass of ball, \( m = 2.00 \, \text{kg} \)
- Charge of the ball, \( q = 6.00 \, \mu\text{C} = 6.00 \times 10^{-6} \, \text{C} \)
- Initial speed, \( v_0 = 20.0 \, \text{m/s} \)
- Total time of flight, \( t = 4.50 \, \text{s} \)

3. **Trajectory and Time Calculation:**
- The ball is thrown upward and returns to the ground, hence the total time for ascending and descending is 4.50 s.
- Time to reach the top of the trajectory, \( t_{\text{up}} = \frac{4.50 \, \text{s}}{2} = 2.25 \, \text{s} \)

4. **Vertical Displacement Calculation:**
- Using the kinematic equation \( v = u + at \):
\[
0 = 20.0 \, \text{m/s} - (9.8 \, \text{m/s}^2) \times 2.25 \, \text{s}
\]
This confirms the time to the top point as accurate since \( v = 0 \, \text{m/s} \) at the top.

5. **Height Calculation (h):**
- Using \( h

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