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Karolina [17]
3 years ago
12

I need huge help understanding this type of questions :( Please and thank you

Physics
1 answer:
Mrrafil [7]3 years ago
5 0

Answer:

67.28 N

Explanation:

Given that,

The mass of an box, m = 15.2 kg

The coefficients of static and kinetic frictions for plastic on wood are 0.55 and 0.36, respectively.

The force of static friction,

F_s=\mu_smg\\\\F_s=0.55\times 15.2\times 9.8\\\\F_s=81.92\ N

The force of kinetic friction,

F_k=\mu_kmg\\\\F_k=0.36\times 15.2\times 9.8\\\\F_k=53.62\ N

Net force acting on the object is :

F = 120.9 -53.62

= 67.28 N

Hence, this is the required solution.

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From what height must an oxygen molecule fall in a vacuum so that its kinetic energy at the bottom equals the average energy of
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Answer:

The  value is  h  = 11930 \ m

Explanation:

From the question we are told that

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Generally the root mean square speed of the  oxygen molecules is mathematically represented as

        v  =  \sqrt{\frac{3 *  R  *  T }{M} }  =  \sqrt{ 2 *  g  *  h }

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An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo
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Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

Where:

v_{a}, v_{b} - Initial and final velocities, measured in meters per second.

t_{a}, t_{b} - Initial and final times, measured in seconds.

a(t) - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

v_{4} = 2\,\frac{m}{s}  +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt

v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)

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Region II (t = 4 s to t = 6 s)

v_{6} = -6\,\frac{m}{s}  +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt

v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)

v_{6} = -4\,\frac{m}{s}

Region III (t = 6 s to t = 10 s)

v_{10} = -4\,\frac{m}{s}  +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt

v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)

v_{10} = 4\,\frac{m}{s}

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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