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2 months ago

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A 2.5-kg box, sliding on a rough horizontal surface, has a speed of 1.2 m/s when it makes contact with a spring. The block comes

to a momentary halt when the compression of the spring is 2.0 cm. The thermal energy generated by friction, from the instant the block makes contact with the spring until it comes to a momentary halt, is 0.5 J (b) What is the coefficient of kinetic friction between the box and the rough surface?

1 answer:

ValentinkaMS [3.4K]2 months ago

4 0

Answer:

Explanation:

Given

The box's mass $=2.5 kg$

speed $v=1.2 m/s$

compression in the spring $x=2 cm$

Total Initial energy $K.E.=\frac{mv^2}{2}$

$K.E.=\frac{2.5\times 1.2^2}{2}=1.8 J$

The thermal energy generated is $E=0.5 J$

which is produced by the friction force

The Work done by the Friction Force is $W=f_r\cdot s$

$f_r=\mu _k\times N$

This contradicts the presence of friction since $\mu_k =coefficient\ of\ kinetic\ friction$ suggests no friction, indicating either a frictionless surface or that the compression in the system exceeds 2 cm

$N=mg$

$f_r=\mu _kmg$

$0.5=\mu _k\times 2.5\times 10\times 0.02$

$0.5=\mu _k\cdot 0.5$

$\mu _k=1$

$\mu _k=1$

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3 months ago

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

Softa [3030]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

Projectile Motion

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The height of a projectile can be calculated using

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where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

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3 months ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Sav [3153]

Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

The question indicates that

The wire’s diameter is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

Aluminum's resistivity is $2.75*10^{-8} \ ohm-meters.$

The electric field variation is described as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

The charge is effectively given by the equation

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area expressed as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

Thus,

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

By substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

The question states that t = 5 seconds

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

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3 months ago

What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?

serg [3582]

Response:

U = 12,205.5 J

Clarification:

To determine the internal energy of an ideal gas, use the following equation:

$U=\frac{3}{2}nRT$ (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: the temperature of the gas = 100K

Substituting the parameter values into equation (1):

$U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J$

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1 month ago