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kotegsom [21]
3 years 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:
marissa [1.9K]3 years ago
7 0

Answer:

maximum static frictional force of the wall on the book (Increasing)

normal force of the wall on the book (Decreasing)

weight of the book (Not changing)

Explanation:

Now according to Newton's third law of motion

"Every action has equal but opposite reaction"

By the data given in question, Albert was pressing the book against the wall.Now, Albert started to reduce his force up against the wall.

First we have to consider all the forces applied on book in this scenario.

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

2. Friction between book and wall acting upward (y-axis)

3. Albert's force on book against wall (x-axis)

4. Normal reaction of wall against Albert's force (x-axis)

Now, when Albert reduced his force, new scenario will be

1. Weight will be remain constant as it is W = mg

Neither mass nor acceleration due to gravity changed, so weight acting upon the book will remain same.

2. When Albert reduced force, normal reaction of wall reduced against it according to Newton's third law of motion

3. Now notice that friction is a force which acts in accordance with the applied force. For example if a box is placed at floor, no friction is applied, but when you drag the box, friction starts to act and increases until its limit comes. So, when Albert reduced his force, weight will try to pull the book and maximum static friction will increase to hinder the movement of book downwards.

Notice that maximum static friction will hinder the book from movement, since Albert reduced his force, but wight didn't pull the book, which means that maximum static friction increased to hinder downward motion.

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3 0
3 years ago
Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. The initial temperature of the gas is 28.0
mel-nik [20]

Answer:

63.5 °C

Explanation:

The expression for the calculation of work done is shown below as:

w=P\times \Delta V

Where, P is the pressure

\Delta V is the change in volume

Also,

Considering the ideal gas equation as:-

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 8.314 J/ K mol

So,

V=\frac{nRT}{P}

Also, for change in volume at constant pressure, the above equation can be written as;-

\Delta V=\frac{nR\times \Delta T}{P}

So, putting in the expression of the work done, we get that:-

w=P\times \frac{nR\times \Delta T}{P}=nR\times \Delta T

Given, initial temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28.0 + 273.15) K = 301.15 K

W=1770 J

n = 6 moles

So,

1770\ J=6 moles\times 8.314\ J/ Kmol \times (T_2-301.15\ K)

Thus,

T_2=301.15\ K+\frac{1770}{6\times 8.314}\ K

T_2=336.63\ K

The temperature in Celsius = 336.63-273.15 °C = 63.5 °C

<u>The final temperature is:- 63.5 °C</u>

7 0
3 years ago
The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current
Komok [63]

Answer:

I=0.047A

Explanation:

Let's use Ohm's law:

V=IR  

or

I=\frac{V}{R}   (1)

Where:

V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

R=\rho*\frac{l}{A}    (2)

Where:

R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}

Keep in mind that the electrical resistivity of the gold is a known constant which is \rho_g_o_l_d=2.35*10^{-8} and the cross sectional area of the conductor is calculated as:

A=\pi *(r^{2})=\pi  *(0.0002m)^{2} =1.256637061*10^{-7} m^{2}

Because we have a wire in this case, so we assume a cylindrical geometry.

Now replacing our data in (2)

R=(2.35*10^{-8})*\frac{80}{1.256637061*10^{-7} }  =14.96056465\Omega

Finally, we know R and V, so replacing these values in (1) we will be able to find the current:

I=\frac{0.7}{14.96056465}\approx0.047A

7 0
3 years ago
NEED HELP NOW. I WILL BRAINLIEST
NeTakaya

Answer:

number of Protons=16

number of Neutrons=16

number of electrons=32

Explanation:

Tip:-

<em><u>Always the number of protons = the number of neutrons.</u></em>

<em><u>Add them and you will get the number of electrons.</u></em>

<u><em>Happy to help</em></u>

ADD ME BRAINLIEST AS YOU SAID

YOUR WELCOME

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