High School

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**Wire Problem:**

Wire A has length [tex]L_a[/tex] and cross-sectional area [tex]A_a[/tex]. Wire B has length [tex]L_b[/tex] and cross-sectional area [tex]A_b[/tex]. They are made of the same material, but wire A has a smaller spring constant. Which one of the following is not possible?

A. [tex]L_a > L_b[/tex] and [tex]A_a < A_b[/tex]
B. [tex]L_a < L_b[/tex] and [tex]A_a > A_b[/tex]
C. [tex]L_a > L_b[/tex] and [tex]A_a = A_b[/tex]
D. [tex]L_a = L_b[/tex] and [tex]A_a < A_b[/tex]
E. [tex]\frac{A_a}{L_a} < \frac{A_b}{L_b}[/tex]

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**Thermal Contact Problem:**

Solid substance X has mass [tex]m_x[/tex], mass-specific heat capacity [tex]c_x[/tex], and initial temperature 300 K. Solid substance Y has mass [tex]m_y[/tex], mass-specific heat capacity [tex]c_y[/tex], and initial temperature 350 K. When they come into thermal contact, the final equilibrium temperature is 310 K. No change of phase occurs. Which one of the following is not possible?

A. [tex]m_x > m_y[/tex] and [tex]c_x > c_y[/tex]
B. [tex]m_x < m_y[/tex] and [tex]c_x > c_y[/tex]
C. [tex]m_x = m_y[/tex] and [tex]c_x > c_y[/tex]
D. [tex]m_x < m_y[/tex] and [tex]c_x[/tex]
E. [tex]m_x > m_y[/tex] and [tex]c_x = c_y[/tex]

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**Motion Problem:**

An object of mass 2.00 kg is moving horizontally in the positive x direction. The initial position is [tex]x_0 = 55.0 \, \text{m}[/tex], initial velocity is [tex]v_0 = 5.00 \, \text{m/s}[/tex], and final velocity is [tex]v_1 = 3.00 \, \text{m/s}[/tex]. During this time, 4.00 J are transferred to the air by drag, and a constant force [tex]F = -0.200 \, \text{N}[/tex] (i.e., magnitude 0.200 N in the negative x direction) is applied. Find the final position [tex]x_1[/tex].

Answer :

Final answer:

The final position (x1) of the object is 135.0 m.

Explanation:

To find the final position (x1), we need to use the equations of kinematics and consider the work done by the drag force.

Given:

First, let's find the acceleration (a) using the equation:

F = ma

Substituting the values:

-0.200 N = 2.00 kg * a

Solving for a:

a = -0.200 N / 2.00 kg

a = -0.100 m/s^2

Next, let's find the time (t) using the equation:

v1 = v0 + at

Substituting the values:

3.00 m/s = 5.00 m/s + (-0.100 m/s^2) * t

Simplifying the equation:

-2.00 m/s = -0.100 m/s^2 * t

Solving for t:

t = (-2.00 m/s) / (-0.100 m/s^2)

t = 20.0 s

Now, let's find the displacement (s) using the equation:

s = ut + 0.5at^2

Substituting the values:

s = (5.00 m/s) * (20.0 s) + 0.5 * (-0.100 m/s^2) * (20.0 s)^2

Simplifying the equation:

s = 100.0 m - 20.0 m

s = 80.0 m

Finally, let's find the final position (x1) using the equation:

x1 = x0 + s

Substituting the values:

x1 = 55.0 m + 80.0 m

x1 = 135.0 m

Learn more about kinematics here:

https://brainly.com/question/28037202

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