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Dahasolnce [82]
3 years ago
12

Student A states that when she sits down on a chair, she is exerting a force on the chair and that is all that happens. Student

B states that when she sits on a chair, the chair is actually exerting a force back on her in reaction to her force exerted upon the chair. Which student is correct and why?
A.
Student A is correct because Newton’s 1st Law of Motion states that if the chair exerted a force on the student, her motion would change.

B.
Student A is correct because Newton’s 3rd Law of Motion states that for every action there is an equal and opposite action and the chair does not show any action.

C.
Student B is correct because Newton’s 1st Law of Motion states that if the chair were to be exerting a force on the student, her motion would change.

D.
Student B is correct because Newton’s 3rd Law of Motion states that for every action there is an equal and opposite action and the chair’s action counteracts the student’s so that neither move.
Physics
1 answer:
zvonat [6]3 years ago
7 0

Answer:

C

Explanation:

I got it right on the test !!

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An object elongates from a length of 45 cm to a length of 55 cm. The percent strain is
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22%

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10/45 simplifies to 2/9

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Explain where most of the mass of an atom is located. Also, explain why some particles that make up the atom do not contribute m
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Read 2 more answers
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
timama [110]

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)

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

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.

3 0
4 years ago
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