Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Case A:

A.75 kg 65 N/m 1.2 m

m = weight of the car = 0.75 kg

k = spring's stiffness = 65 N/m

h = elevation of the hill = 1.2 m

x = spring's compression = 0.25 m

Applying the principle of energy conservation from the Top of the hill to the Bottom of the hill

Energy at the Top of the hill equals Energy at the Bottom of the hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s

Case B:

B.60 kg 35 N/m.9 m

m = weight of the car = 0.60 kg

k = spring's stiffness = 35 N/m

h = height of the hill = 0.9 m

x = spring's compression = 0.25 m

Using energy conservation from the Top of the hill to the Bottom of the hill

Top hill energy = Bottom hill energy

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s

Case C:

C.55 kg 40 N/m 1.1 m

m = weight of the car = 0.55 kg

k = spring's stiffness = 40 N/m

h = height of the hill = 1.1 m

x = spring's compression = 0.25 m

Using conservation of energy from the Top of the hill to the Bottom of the hill

Energy at the Top of the hill = Energy at the Bottom of the hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s

Case D:

D.84 kg 32 N/m.95 m

m = weight of the car = 0.84 kg

k = spring's stiffness = 32 N/m

h = height of the hill = 0.95 m

x = spring's compression = 0.25 m

Using energy conservation from the Top of the hill to the Bottom of the hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s

thus, the closest result is from case C at 5.1 m/s

Answer:

A flea can attain a maximum elevation of 51 mm.

Explanation:

Hello!

The following equations describe the height and velocity of the flea:

During the jump:

h = h0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

In free fall:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = flea's height at time t.

h0 = initial height.

v0 = starting velocity.

t = time interval.

a = flea's acceleration while jumping.

v = flea's velocity at that specific time.

g = gravitational acceleration.

Initially, we need to determine the time taken for the flea to attain a height of 0.0005 m. This will help us calculate the flea's velocity during the jump:

h = h0 + v0 · t + 1/2 · a · t²

If we assume the ground as the origin, thus h0 = 0. Since the flea starts stationary, v0 = 0. Therefore:

h = 1/2 · a · t²

We need to find the value of t when h = 0.0005 m:

0.0005 m = 1/2 · 1000 m/s² · t²

0.0005 m / 500 m/s² = t²

t = 0.001 s

Next, we calculate the velocity achieved during that time:

v = v0 + a · t (v0 = 0)

v = a · t

v = 1000 m/s² · 0.001 s

v = 1.00 m/s

At a height of 0.50 mm, the flea's velocity stands at 1.00 m/s. This initial speed will reduce due to gravity's downward pull. When the speed reaches zero, the flea will have reached its peak height. Using the velocity equation, let's determine the time taken to reach maximum height (v = 0):

v = v0 + g · t

At peak height, v = 0:

0 m/s = 1.00 m/s - 9.81 m/s² · t

-1.00 m/s / -9.81 m/s² = t

t = 0.102 s

Now, we can compute the height attained by the flea during this time:

h = h0 + v0 · t + 1/2 · g · t²

h = 0.0005 m + 1.00 m/s · 0.102 s - 1/2 · 9.81 m/s² · (0.102 s)²

h = 0.051 m

A flea reaches a maximum height of 51 mm.

Response:

Explanation:

Torque is defined as:

where

F is the force's magnitude,

d represents the distance from where the force is applied to the center of rotation,

and is the angle between the force's direction and d.

In this question, we have:

, the force

, the distance of applying the force from the center (0,0)

, and the angle between the force's direction and a.

Thus, the torque is

Response:

Explanation:

The solution can be found in the attached file.