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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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Answer: Temperature and humidity are the two characteristics used to classify air masses.

Explanation:

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Which mathematical formula would you use to calculate the IMA of a wheel?
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W = F x D

The work you put into a machine -- the input force -- is the force you apply times the distance you apply it. The work done by the machine equals the resisting weight times the distance it moves when you perform the work.

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3 years ago
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A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d
andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x²  - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0
3 years ago
43 kg bear slides, from rest, 15 m down a lodgepole pine tree, moving with a speed of 5.5 m/s just before hitting the ground. (a
olga2289 [7]

Answer:

-6327.45 Joules

650.375 Joules

378.47166 N

Explanation:

h = Height the bear slides from = 15 m

m = Mass of bear = 43 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of bear = 5.5 m/s

f = Frictional force

Potential energy is given by

P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules

Kinetic energy is given by

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J

Kinetic energy of the bear just before hitting the ground is 650.375 Joules

Change in total energy is given by

\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N

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