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Furkat [3]
3 years ago
5

A person pulls a bucket of water up from a well with a rope. Assume the initial and final speeds of the bucket are zero (Vi-Vf-0

), and that the person lifted the bucket a vertical distance h. By looking at energy changes in the bucket-Earth physical system, we can make sense of the force the person must exert to pull the bucket up and the amount of work the person does. a) Think about whether this is an open or closed system. What energy systems undergo a change in energy during the process? Construct a particular Energy-System Diagram for this process and include the algebraic expression of energy conservation (in terms of AE's and if open, any Heat or Work). b) Substitute the algebraic expression we use for the change in gravitational potential energy and solve algebraically for the work done by the person on the rope/bucket. c) Use the definition of work in terms of force and distance (see p.39-40 of Chapter 2 in the Course Notes) to find the average force exerted by the person on the rope and bucket while they are lifting it up.

Physics
1 answer:
n200080 [17]3 years ago
6 0

Answer:

a

This a closed system because the mass of the system is conserved

The energy system that undergoes change is the Potential energy system

The energy system diagram is shown on the first uploaded image

b

Work done = Change in gravitational potential energy

So solving algebraically for work done would be

    Work done   = m*g*h

where m is mass

          g is acceleration due to gravity

          and h is the height

c

Work done in terms of force and distance is = mg

where  m is mass of bucket and

            g is acceleration due to gravity  

Explanation:

a) At the start, potential and kinetic energy were zero. so, energy is zero.

As the person pulls the bucket up, the potential energy becomes mgh.

so,final energy will be consisting of only potential energy.

B) Here work done is equal to change in gravitational potential energy.

W = \Delta P.E

W = m*g*h

where g = 9.9 m/s^2

C) Work = force * distance

mgh = force * h

force = mg

force = weight of bucket

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tigry1 [53]

Explanation:

p  =  \frac{f \times s}{t}

power = Force × distance /time

power = 944N × 12.4m/36secs

power = (944×12.4/36)Nms—¹

power = 390.2Nms—¹ or 390.2Watts or 390.2Js—¹

8 0
3 years ago
There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C
liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

V_{tot} = V_{q1} + V_{q2}

V_q = \dfrac{k \cdot q}{r}

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

V_{tot} =  \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05}  = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0

The electric field at the point exactly midway between the plates, V_{tot} = 0

3) The electric field, 'E', between plates is given as follows;

E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, F_e = E × e

∴ F_e = 1.1294 × 10¹² N/C ×  -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, F_e ≈ 1.807 × 10⁻⁷ N

4 0
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An object with a greater mass will have more inertia as it takes more force to change their motion.

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