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34kurt
8 days ago
6

Albert presses a book against a wall with his hand. As Albert gets tired, he exerts less force, but the book remains in the same

spot on the wall Classify each force on the book as increasing, decreasing, or not changing in magnitude when Albert reduces the force he is applying to the book. Not every classification must be used
Increasing Decreasing Not changing



Answer Bank

frictional force of the wall on the book
maximum static frictional force of the wall on the bookweight of the book
normal force of the wall on the book
Physics
1 answer:
Maru [1K]8 days ago
7 0

Answer:

the maximum static friction force of the wall acting on the book (Increasing)

the normal force of the wall acting on the book (Decreasing)

the weight of the book (Constant)

Explanation:

According to Newton's third law of motion:

"Every action has an equal and opposite reaction"

In the scenario provided, Albert is pressing the book against the wall and subsequently decreases the force applied against the wall.

Let's evaluate all forces influencing the book in this situation.

1. Weight of the book acting downwards (y-axis)

2. Friction from the book against the wall acting upwards (y-axis)

3. Albert’s force exerted on the book against the wall (x-axis)

4. Normal force of the wall reacting to Albert’s applied force (x-axis)

As Albert eases off his force, the new scenario reads:

1. The weight remains constant as represented by W = mg

Since neither mass nor gravitational acceleration has changed, the weight exerted on the book remains the same.

2. As Albert reduces his force, the wall’s normal reaction force decreases correspondingly, following Newton's third law of motion.

3. Friction operates in response to the force applied to it. With a box resting on the floor, no friction acts upon it until it is dragged, at which point friction begins to manifest and rise until it reaches its maximum. Therefore, when Albert diminishes his force, the weight's pull will influence the book and the maximum static friction will rise to resist the book’s downward movement.

It should be noted that the maximum static friction is working to prevent movement of the book. With Albert's force reduced, but the weight of the book unchanged, maximum static friction increases to prevent downward movement.

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Answer:

a) v_{o} =16m/s

b) v=9.8m/s

c) \beta =-35.46º

Explanation:

According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.

With this information, we can compute the ball’s starting speed.

a) Let's first assess the horizontal trajectory.

x=v_{ox}t

x=v_{o}cos(60)t

v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)} (1)

This gives us our initial equation.

Next, we need to examine the vertical trajectory.

y=y_{o}+v_{oy}t+\frac{1}{2}gt^2

y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2

Utilizing v_{o} in our first equation (1)

8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2

\frac{1}{2}(9.8)t^2=16tan(60)-8

Now let’s solve for t.

t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s

The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.

v_{o}=\frac{16m}{(2s)cos(60)}=16m/s

b) To determine the velocity magnitude just before impact, we must calculate both x and y components.

v_{x}=v_{ox}+at=16cos(60)=8m/s

v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s

The computed velocity magnitude is:

v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s

c) The ball's angle is:

\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º

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Answer:

Maximum emf = 5.32 V

Explanation:

Provided data includes:

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions each second

Magnetic field, B = 0.5 T

We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:

\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

For the maximum emf, \sin\omega t=1

Therefore,

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Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

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k(r) = 0.5 W/m.K

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k(h) = 25 W/m².K

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Applying the energy balance equation,

Energy In = Energy Out

This equates to q, which is 754L

From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C

Additionally, the outer surface temperature records as 51° C

Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C

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