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3 months ago

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For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir

st hill, where a horizontal spring that is initially compressed 0.25 m will push a small car forward. Each group of students will choose a car and a spring to push the car and then build a track. The assignment is to make the car go 5.0 m/s when it reaches the bottom of the first hill. Four groups of students choose springs and build tracks as described in the table. Which group’s roller coaster will most likely make the car travel closest to 5.0 m/s when it is at the bottom of the first hill?
Group Car Mass (Kg) Spring Constant (N/m) Hill Height (m)
A .75 kg 65 N/m 1.2 m
B .60 kg 35 N/m .9 m
C .55 kg 40 N/m 1.1 m
D .84 kg 32 N/m .95 m

2 answers:

Keith_Richards [3.2K]3 months ago

7 0

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

inna [3.1K]3 months ago

5 0

The correct choice is C. I recently completed the assessment on edge.

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3 months ago

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

Maru [3345]

1) For x = 6.6 cm, $E_x=3.47\cdot 10^6 N/C$

2) For x = 6.6 cm, $E_y=0$

3) For x = 1.45 cm, $E_x=-3.76\cdot 10^6N/C$

4) For x = 1.45 cm, $E_y=0$

5) Surface charge density at b = 4 cm: $+62.75 \mu C/m^2$

6) At x = 3.34 cm, the x-component of the electric field equals zero

7) Surface charge density at a = 2.9 cm: $+65.25 \mu C/m^2$

8) None of these regions

Explanation:

1)

The electric field from an infinite charge sheet is perpendicular to it:

$E=\frac{\sigma}{2\epsilon_0}$

where

$\sigma$ is the surface charge density

$\epsilon_0=8.85\cdot 10^{-12}F/m$ represents vacuum permittivity

Outside the slab, the electric field behaves like that of an infinite sheet.

Consequently, the electric field at x = 6.6 cm (situated to the right of both the slab and sheet) results from the combination of the fields from both:

$E=E_1+E_2=\frac{\sigma_1}{2\epsilon_0}+\frac{\sigma_2}{2\epsilon_0}$

where

$\sigma_1=-2.5\mu C/m^2 = -2.5\cdot 10^{-6}C/m^2\\\sigma_2=64 \muC/m^2 = 64\cdot 10^{-6}C/m^2$

The field from the sheet points left (negative, inward), and the slab’s field points right (positive, outward).

Thus,

$E=\frac{1}{2\epsilon_0}(\sigma_1+\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}+64\cdot 10^{-6})=3.47\cdot 10^6 N/C$

and the negative sign indicates a rightward direction.

2)

Both the sheet’s and slab’s fields are perpendicular to their surfaces, directing along the x-axis, hence there's no y-component for the total field.

<pThus, the y-component totals zero.

This happens because both the sheet and slab stretch infinitely along the y-axis. Choosing any x-axis point reveals that the y-component of the field, generated by a surface element dS of either the sheet or slab, $dE_y$ , will be equal and opposite to the corresponding component from the opposite side, $-dE_y$ . Thus, the combined y-direction field is always zero.

3)

This scenario resembles part 1), but the point here is

x = 1.45 cm

which lies between the sheet and the slab. The fields from both contribute leftward as the slab has a negative charge (resulting in an outward field). Thus, the total field computes to

$E=E_1-E_2$

Replacing with expressions from part 1), we get

$E=\frac{1}{2\epsilon_0}(\sigma_1-\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}-64\cdot 10^{-6})=-3.76\cdot 10^6N/C$

where the negative illustrates a leftward direction.

4)

This portion parallels part 2). Since both fields remain perpendicular to the slab and sheet, no component exists along the y-axis, thus the electric field's y-component is zero.

5)

Notably, the slab behaves as a conductor, signifying charge mobility within it.

The net charge on the slab is positive, indicating a surplus of positive charge. With the negatively charged sheet on the left of the slab, positive charges shift towards the left slab edge (at a = 2.9 cm), while negative charges move to the right edge (at b = 4 cm).

The surface charge density per unit area of the slab is

$\sigma=+64\mu C/m^2$

This average denotes the surface charge density on both slab sides at points a and b:

$\sigma=\frac{\sigma_a+\sigma_b}{2}$ (1)

Additionally, the infinite sheet at x = 0 negatively charged $\sigma_1=-2.5\mu C/m^2$ , induces an opposite net charge on the slab's left surface, thus

$\sigma_a-\sigma_b = +2.5 \mu C/m^2$ (2)

Having equations (1) and (2) allows for solving the surface charge densities at a and b, yielding:

$\sigma_a = +65.25 \mu C/m^2\\\sigma_b = +62.75 \mu C/m^2$

6)

We aim to compute the x-component of the electric field at

x = 3.34 cm

This point lies inside the slab, bounded at

a = 2.9 cm

b = 4.0 cm

In a conducting slab, the electric field remains at zero owing to charge equilibrium; thus, the x-component thereof in the slab is zero

7)

From part 5), we determined the surface charge density at x = a = 2.9 cm is $\sigma_a = +65.25 \mu C/m^2$

8)

As mentioned in part 6), conductors have zero electric fields internally. Since the slab is conductive, the electric field inside remains zero; therefore, the regions where the electric field is null are

2.9 cm < x < 4 cm

Thus, the suitable answer is

"none of these regions"

Learn more about electric fields:

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Answer:

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$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

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3 months ago

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