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erastovalidia [21]
3 years ago
8

The force pressing two stacked objects together is equal to the weight of the object on top. The friction between a piece of woo

d on a stone surface is 22.5 newtons. The same piece of wood is placed on another surface, and the friction between the wood and the surface is now 46.5 newtons.
What could be the second surface?

A) metal
B) concrete
C) brick
D) skin
Physics
1 answer:
34kurt3 years ago
3 0

Answer: B. Concrete

Explanation:

Let N = reacting force pressing the bodies in context together (units in Newtons),

The question stated that the force pressing the two mounted/stacked objects together is equal to the weight of the object on top.

We need to start by finding the weight of the piece of wood.

friction is given by

f = μN

The value of f is 22.5,

and from the chart reference the coefficient of friction between wood and stone, μ is 0.30.

22.5 = 75. 0.30

Putting the values into the equation: 22.5 = 0.30N.

Divide both sides by 0.30 to find the value of N:

N= 22.5/0.3 = 75

Now that the piece of wood will be placed on another surface, its weight of 75 Newton is the force pressing the two bodies together.

To determine the new surface, you should find the new coefficient of friction by using the new value of the force of friction given 46.5:

46.5 = µ(75).

Divide both sides by 75 to isolate μ.

The refer chart also indicates that the coefficient of friction equals 0.62 between wood and concrete, so the new surface corresponding to 0.62 is the concrete, which is (B).

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A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0
MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

\omega=\frac{2\pi}{T}

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

\omega=\frac{2\pi}{1 s}=6.28 rad/s

Now we can find the linear speed of the ladybug, which is given by

v=\omega r

where:

\omega=6.28 rad/s is the angular speed

r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation

Substituting, we find

v=(6.28)(0.65)=4.1 m/s

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7 0
3 years ago
A beam of red light is made to pass through two slits that are 3.55 E-3 meters apart. On a screen 2.25 meters away from the slit
JulsSmile [24]
I am assuming you know the relation obtained between slit width, distance of screen from slits, distance of interference pattern obtained on the screen from the center and the wavelength of monochromatic light used in Young's Double Slit experiment.
λ =\frac{y*d}{D} =  \frac{3.55*10^{-3}*1.25*10^{-4}  }{2.25} = 1.97*10^{-7} m
λ ~ 1.97 ×10⁻⁷m
7 0
3 years ago
MamaMia's Pizza purchases its pizza delivery boxes from a printing supplier. MamaMia's delivers on-average 200 pizzas each month
Ira Lisetskai [31]

Answer:

a) 138 units

b) 17 units

c) 17 units

d) Total Cost = $353.35

Explanation:

Given:

Average pizzas delivered = 200

Charge of inventory holding = 30% of cost

Lead time = 7 days

Now,

a) Economic Order Quantity =  \sqrt\frac{2\times\textup{Annual Demand}\times\textup{Cost per Order}}{\textup{Carrying cost}}

also,

Annual Demand = 200 × 12 = 2400

Cost per Order = Cost of Box + Processing Costs

= 30 cents + $10

= $10.30

and, Carrying Cost = \frac{\textup{Total Inventory Cost}}{\textup{total annual demand}}

=\frac{\textup{Total Cost per order}\times\textup{Annual demand}\times\frac{25}{100}}{\textup{Annual demand}}

= \frac{\$10.30\times2400}\times\frac{25}{100}}{2400}

= $2.575

Therefore,

Economic Order Quantity =  \sqrt\frac{2\times\textup{2400}\times\textup{10.30}}{\textup{2.575}}

= 138.56 ≈ 138 units

b) Reorder Point

= (average daily unit sales × the lead time in days) + safety stock

= (\frac{200}{30}\times7

= 46.67 ≈ 47 units

c) Number of orders per year = \frac{\textup{Annual Demand}}{\textup{Economic order quantity}}

= \frac{\textup{2400}}{\textup{138}}

= 17.39 ≈ 17 units

d) Total Annual Cost (Total Inventory Cost)

= Ordering Cost + Holding Cost

Now,

The ordering Cost = Cost per Order × Total Number of orders per year

= $10.30 × 17

= $175.1

and,

Holding Cost = Average Inventory Held × Carrying Cost per unit

Average Inventory Held = \frac{\textup{0+138}}{\textup{2}} =  69

Carrying Cost per unit = $2.575

Holding Cost = 69 × $2.575 =  

$177.675

Therefore,

Total Cost = Ordering Cost + carrying cost

= $175.1  + $177.675 = $353.35

5 0
3 years ago
Read 2 more answers
Mercury has a density of 13.56 g/mL. How many kilograms of mercury would you expect to fit in a cylindrical glass cup with a bot
vovangra [49]

Answer:

263.152kg

Explanation:

<em>The density of a substance is related to its mass and volume as follows;</em>

density = mass / volume      

mass = density x volume       -------------(i)

The substance in question here is <em>mercury </em>which has;

density = 13.56g/mL = 13.56g/cm³

Since the mercury is going to be put in the cylindrical glass, the volume of the cylindrical glass is going to be equal to the volume of the mercury that will be put.

And we know that the;

volume of a cylinder = πr²h

<em>Where;</em>

π = 3.142

r = bottom radius of the cylinder = 5.75inches

h = height of the cylinder = 0.950ft

<em>For uniformity, let's convert the radius and height of the cylinder to their corresponding values in cm</em>

r  = 5.75 inches = 5.75 x 2.54 cm = 14.605cm

h = 0.950 ft = 0.950 x 30.48 cm = 28.956cm

<em>Therefore, the volume of the cylinder;</em>

v = 3.142 x (14.605cm)² x 28.956cm = 19406.5cm³

v = 19406.5cm³ [This is also the volume of the mercury necessary to fit the cylinder]

<em>Now the following value has been found;</em>

volume = 19406.5cm³

<em>Substitute the values of density and volume into equation (i)  as follows;</em>

mass = 19406.5cm³ x 13.56g/cm³

mass = 263152.14g

<em>Convert the result to kg by dividing by 1000</em>

mass = 263.152kg

Therefore, 263.152kg kilograms of mercury would fit in the cylindrical glass.

3 0
3 years ago
A 3.0kg weight W is initially at rest on incline AB, which is raised 40° above the horizontal. The effective coefficient is
lakkis [162]

(a) The acceleration of the system is determined as 1.58 m/s².

(b) The relative weight of P is pounds is determined as 0.14 lb.

<h3>Acceleration of the system</h3>

The acceleration of the system is calculated as follows;

W - T = m₂a --- (1)

T = m₁a ----(2)

μmgsinθ - m₁a = m₂a

(0.3 x 3 x 9.8 x sin40) - (0.4 + 0.2)a = 3a

5.67 - 0.6a = 3a

5.67 = 3.6a

a = 5.67/3.6

a = 1.58 m/s²

<h3> Relative Weight of P</h3>

W = ma

W = 0.4 x 1.58

W = 0.632 N = 0.14 lb

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3 0
2 years ago
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