~~~NEED HELP ASAP~~~
Please solve each section and show all work for each section.

~~~NEED HELP ASAP~~~Please Solve Each Section And Show All Work For Each Section.

Answers

Answer 1

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass [tex]m_A[/tex] as

[tex]x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)[/tex]

[tex]y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)[/tex]

Substituting (2) into (1), we get

[tex]\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)[/tex]

where [tex]f_N= \mu_kN[/tex], the frictional force on [tex]m_A.[/tex] Set this aside for now and let's look at the forces on [tex]m_B[/tex]

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on [tex]m_B[/tex] as

[tex]x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)[/tex]

[tex]y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)[/tex]

From (5), we can solve for N as

[tex]N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)[/tex]

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

[tex]T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)[/tex]

Substituting (7) into (4) we get

[tex]m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba[/tex]

Collecting similar terms together, we get

[tex](m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag[/tex]

or

[tex]a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)[/tex]

Putting in the numbers, we find that [tex]a = 1.4\:\text{m/s}[/tex]. To find the tension T, put the value for the acceleration into (7) and we'll get [tex]T = 21.3\:\text{N}[/tex]. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get [tex]N = 50.9\:\text{N}[/tex]


Related Questions

A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.

Answers

Answer:

0.5

Explanation:

because the block is attached to the pulley of the string

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Newtons. What is the frequency of this mode of vibration

Answers

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

[tex]F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz[/tex]

Therefore, the frequency of this mode of vibration is 138.87 Hz


If an electrical component with a resistance of 53 Q is connected to a 128-V source, how much current flows through the component?

Answers

Answer:

the current that flows through the component is 2.42 A

Explanation:

Given;

resistance of the electrical component, r = 53 Ω

the voltage of the source, V = 128 V

The current that flows through the component is calculated using Ohm's Law as demonstrated below;

[tex]V = IR\\\\I = \frac{V}{R} = \frac{128 \ V}{53 \ ohms} = 2.42 \ A[/tex]

Therefore, the current that flows through the component is 2.42 A

A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?

Answers

Answer:

a n c

Explanation:

3. You have a variable-voltage power supply and a capacitor in the form of two metal disks of radius 0.6 m, held a distance of 1 mm apart. What is the largest voltage you can apply to the capacitor without the air becoming highly conductive

Answers

Answer:

The breakdown of air occurs at a maximum voltage of 3kV/mm.

Explanation:

The breakdown of air occurs at a maximum voltage of 3kV/mm.

At this level of voltage the air between the plates become highly ionised and breakdown occurs. Since, the distance held between the plates is 1mm , it can withstand a maximum voltage of 3 kV.

After this voltage the air will become conductive in nature and will form ions in the air between the plates and ultimately breakdown will take place with a flash.

a girl is moving with a uniform velocity of 1.5 m/s then mathematically find her acceleration​

Answers

Answer:

0

Explanation:

a = dv/dt

if v is constant than the slope of the v graph will be 0, so dv/dt is 0

a= 0

2.
Select the correct answer.
Erica is working in the lab. She wants to remove the fine dust particles suspended in a sample of oil. Which method is she most likely to use?

Answers

Answer:

Reverse Osmosis

Explanation:

Reverse osmosis is a type of filtration that involves passing a solvent through a semipermeable membrane in the opposite direction that natural osmosis does. Separation is always enforced through the use of pressure in this process. Ions, fine dust particles, molecules, and larger particles are typically removed from solvents using this method. The technique is particularly popular in the treatment and purification of water.

Answer:

filtration is used to separate

A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frictional force of 2 N, find (a) its acceleration, and (b) its initial velocity.

Answers

B it’s Intail velocity

ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on its circular path is 6.0 m/s. What is its kinetic energy at an instant when the string makes an angle of 50 degree with the vertical

Answers

Answer:

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

Assuming the atmospheric pressure is 1 atm at sea level, determine the atmospheric pressure at Badwater (in Death Valley, California) where the elevation is 86.0 m below sea level.

Answers

Answer:

Atmospheric pressure at Badwater is 1.01022 atm

Explanation:

Data given:

1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa

Elevation (h) = 86m

gravity (g) = 9.8 m/s2

Density of air P = 1.225 kg/m3

Therefore pressure at bad water Pb = Pi + Pgh

Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)

Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa

hence:

Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm

A wave moves in a rope with a certain wavelength. A second wave is made to move in the same rope with twice the wavelength of the first wave. The frequency of the second wave is _______________ the frequency of the first wave.

Answers

Answer:

The frequency of the second wave is half of the frequency of first one.

Explanation:

The wavelength of the second wave is double is the first wave.

As we know that the frequency is inversely proportional to the wavelength of the velocity is same.

velocity = frequency x wavelength

So, the ratio of frequency of second wave to the first wave is

[tex]\frac{f_2}{f_1} =\frac{\lambda _1}{\lambda _2}\\\\\frac{f_2}{f_1} =\frac{\lambda _1}{2\lambda _1}\\\\\frac{f_2}{f_1} =\frac{1}{2}\\\\[/tex]

The frequency of the second wave is half of the frequency of first one.

A car hurtles off a cliff and crashes on the canyon floor below. Identify the system in which the net momentum is zero during the crash.

Answers

Solution :

It is given that a car ran off from a cliff and it crashes on canyon floor. Now the system of a car as well as the earth together have a  [tex]\text{ net momentum of zero}[/tex] when the car crashes on the canyon floor, thus reducing the momentum of the car to zero. The earth also stops its upward motion and it also reduces the momentum to zero.

Traveling waves propagate with a fixed speed usually denoted as v (but sometimes c). The waves are called __________ if their waveform repeats every time interval T.

a. transverse
b. longitudinal
c. periodic
d. sinusoidal

Answers

Answer:

periodic

Explanation:

What are the differences among elements, compounds, and mixtures?

Answers

Answer:

Elements have a characteristic number of electrons and protons.Both Hydrogen(H) and oxygen(O) are two different elements.

••••••••••••••••

Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of .Hydrogen(H) and Oxygen(O) both qr4 naturally gases,but they react to form water(H2O),which is liquid compound.

•••••••••••••••

A mixture is made of atleast two parts》 solid,liquid or gas.The difference is that it's not a chemical substance that's bonded by other elements. 

------------------------------

Hope it helps...

Have a great day!!!

Answer: Elements have a characteristic number of electrons and protons. Both Hydrogen(H) and oxygen(O) are two different elements. Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of.Hydrogen(H) and Oxygen(O) both qr4 naturally gases, but they react to form water(H2O), which is liquid compound. A mixture is made of at least two parts solid, liquid, or gas. The difference is that it's not a chemical substance that's bonded by other elements.

During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race

Answers

Answer:

The maximum speed of Bolt for the 100 m race is 14.66 m/s

Explanation:

Given;

initial distance covered by Bolt, d = 200 m

time of this motion, t = 19.3 s

The second distance covered by Bolt, = 100 m

Assuming Bolt maintained the same acceleration for both races.

His acceleration can be determined from the 200 m race.

d = ut + ¹/₂at²

where;

u is his initial velocity = 0

d =  ¹/₂at²

[tex]at^2 = 2d\\\\a = \frac{2d}{t^2} \\\\a = \frac{2\times 200}{19.3^2} \\\\a = 1.074 \ m/s^2[/tex]

Let the final or maximum velocity for the 100 m race = v

v² = u² + 2ad₂

v² =  2 x 1.074 x 100

v² = 214.8

v = √214.8

v = 14.66 m/s

The maximum speed of Bolt for the 100 m race is 14.66 m/s

A nerve impulse travels along a myelinated neuron at 90.1 m/s.
What is this speed in mi/h?

Answers

Answer:

201.5537 mph

Explanation:

Given the following data;

Speed = 90.1 m/s

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the formula;

Speed = distance/time

To convert this value into miles per hour;

Conversion;

1 meter = 0.000621 mile

90.1 meters = 90.1 * 0.000621 = 0.05595 miles

1 metre per second = 2.237 miles per hour

90.1 meters per seconds = 90.1 * 2.237 = 201.5537 miles per hour

90.1 m/s = 201.5537 mph

No esporte coletivo, um dos principais fatores desenvolvidos é o desenvolvimento social. Qual desses não faz parte das virtudes ensinadas no esporte?

Companheirismo
Humildade
Ser justo (Fair Play)
Vencer independente do que precise ser feito

Answers

Answer:

fair palybtgshsisuehdh

A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?​

Answers

Answer:

The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".

Explanation:

According to the question,

The work will be:

⇒ [tex]Work=-\frac{kQq}{R}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]

              [tex]=\frac{0.3978}{\varepsilon }[/tex]

              [tex]=4.49\times 10^{10} \ joules[/tex]

Thus the above is the correct answer.    

We have that the workdone  is mathematically given as

W=4.49*10e10 J

From the question we are told

A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?​

Workdone

Generally the equation for the workdone   is mathematically given as

W=-kQq/R

Therefore

0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2

Hence

W=4.49*10e10 J

For more information on Charge visit

https://brainly.com/question/9383604

A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity increases to 57 rad/s. Assume that the angular acceleration was constant during this time interval. How many revolutions does the wheel turn through during this time interval

Answers

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]

Where:

[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s

[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]

Hence, the number of revolutions is:

[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]

Therefore, the number of revolutions is 44.6.

       

I hope it helps you!

how can scientific method solve real world problems examples

Answers

The scientific method is nothing more than a process for discovering answers. While the name refers to “science,” this method of problem solving can be used for any type of problem

Two pistons are connected to a fluid-filled reservoir. The first piston has an area of 3.002 cm2, and the second has an area of 315 cm2. If the first cylinder is pressed inward with a force of 50.0 N, what is the force that the fluid in the reservoir exerts on the second cylinder?​

Answers

Answer:

The force on the second piston is 5246.5 N .

Explanation:

Area of first piston, a = 3.002 cm^2

Area of second piston, A = 315 cm^2

Force on first piston, f = 50 N

let the force of the second piston is F.

According to the Pascal's law

[tex]\frac{f}{a} = \frac{F}{A}\\\\\frac{50}{3.002}=\frac{F}{315}\\\\F = 5246.5 N[/tex]

You walk into a room and you see 4 chickens on a bed 2 cows on the floor and 2 cats in a chair. How many legs are on the ground? (I know this answer just a riddle to see who knows it) (:

Answers

Answer:

18

Explanation:

I'm pretty sure I got it right

The correct equation for the x component of a vector named A with an angle measured from the x axis would be which of the following?

Answers

Answer:

Acosθ

Explanation:

The x-component of a vector is defined as :

Magnitude * cosine of the angle

Maginitude * cosθ

The magnitude is represented as A

Hence, horizontal, x - component of the vector is :

Acosθ

Furthermore,

The y-component is taken as the sin of the of the angle multiplied by the magnitude

Vertical, y component : Asinθ

Question 7 of 10
A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s

Answers

Answer:

a

Explanation:

you take 32,000kg ÷2.0m

A cylindrical disk of wood weighing 45.0 N and having a diameter of 30.0 cm floats on a cylinder of oil of density 0.850 g>cm3 (Fig. E12.19). The cylinder of oil is 75.0 cm deep and has a diameter the same as that of the wood. (a) What is the gauge pressure at the top of the oil column

Answers

Answer:

665.25 Pa

Explanation:

Given data :

Weight of the disk, w = 45 N

Diameter, d = 30 cm

                    = 0.30 m

Therefore, radius of the disk,

[tex]$r=\frac{d}{2}$[/tex]

[tex]$r=\frac{0.30}{2}$[/tex]

   = 0.15 m

Now, area of the cylindrical disk,

[tex]$A=\pi r^2$[/tex]

[tex]$A=3.14 \times (0.15)^2$[/tex]

   [tex]$=0.07065 \ m^2$[/tex]

∴ The gauge pressure at the top of the oil column is :

   [tex]$p=\frac{w}{A}$[/tex]

   [tex]$p=\frac{47}{0.07065}$[/tex]

      = 665.25 Pa

Therefore, the gauge pressure is 665.25 Pa.

The definition of pressure allows to find the result for the pressure at the top of the oil cylinder is:

The pressure is: P = 636.6 Pa

The pressure is defined by the relationship between perpendicular force and area.

          [tex]P = \frac{F}{A}[/tex]

where P is pressure, F is force, and A is area.

They indicate that the wooden cylinder weighs W = 45.0 N and has a diameter of d = 30 cm = 0.30 m.

The area is:

        A = π r² = [tex]\pi \frac{d^2}{4}[/tex]  

In the attachment we see a diagram of the forces, where the weight of the cylinder and the thrust are equal.

         B-W = 0

          B = W

The force applied to the liquid is the weights of the cylinder. Let's replace.

          [tex]P= \frac{W}{A} \\P = W \frac{4}{\pi d^2 }[/tex]  

Let's calculate.

          [tex]P = \frac{45 \ 4 }{\pi \ 0.30^2 }[/tex] P = 45 4 / pi 0.30²

          P = 636.6 Pa

In conclusion using the definition of pressure we can find the result for the pressure at the top of the oil cylinder is:

The pressure is: P = 636.6 Pa.

Learn more about pressure here: brainly.com/question/17467912

The slope of a d vs t graph represents velocity. Describe 3 ways you know this to be true.

Answers

Answer:

Look at explanation

Explanation:

I only know 1 way, there is another way you can rephrase this using derivatives but that's pretty much the same thing.

The slope is calculated by Δy/Δx so the slope of distance vs time graph is Δd/Δt which is the velocity

1. A 20.0 N force directed 20.0° above the horizontal is applied to a 6.00 kg crate that is traveling on a horizontal
surface. What is the magnitude of the normal force exerted by the surface on the crate?

Answers

N = 52.0 N

Explanation:

Given: [tex]F_a= 20.0\:\text{N}=\:\text{applied\:force}[/tex]

[tex]m=6.00\:\text{kg}[/tex]

[tex]N = \text{normal force}[/tex]

The net force [tex]F_{net}[/tex] is given by

[tex]F_{net} = N + F_a\sin 20 - mg=0[/tex]

Solving for N, we get

[tex]N = mg - F_a\sin 20[/tex]

[tex]\:\:\:\:\:\:= (6.00\:\text{kg})(9.8\:\text{m/s}^2) - (20.0\:\text{N}\sin 20)[/tex]

[tex]\:\:\:\:\:\:= 52.0\:\text{N}[/tex]

two identical eggs are dropped from the same height. The first eggs lands on a dish and breaks, while the second lands on a pillow and does not break. Which quantities are the same in both situations

Answers

Answer:

The height is the same

Explanation:

Because they were at the same height but they fell at different velocities

It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the spring an additional 0.10 m, does it take 130 J, more than 130 J or less than 130 J? Verify your answer with a calculation.

Answers

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

[tex]W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m[/tex]

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

[tex]W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J[/tex]

So, the new work is more than 130 J.

uppose that 3 J of work is needed to stretch a spring from its natural length of 32 cm to a length of 49 cm. (a) How much work (in J) is needed to stretch the spring from 37 cm to 45 cm

Answers

Answer:

0.113 J

Explanation:

Applying,

w = ke²/2................. Equation 1

Where w = workdone in stretching the spring, k = spring constant, e = extension

make k the subject of the equation

k = 2w/e²................ Equation 2

From the question,

Given: w = 3 J, e = 49-32 = 17 cm = 0.17 m

Substitute these values into equation 2

k = (2×3)/0.17²

k = 6/0.17

k = 35.29 N/m

(a) if the spring from 37 cm to 45 cm,

Then,

w = ke²/2

Given: e = 45-37 = 8 cm = 0.08

w = 35.29(0.08²)/2

w = 0.113 J

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