A 5.74 kg rock is thrown upwards with a force of 317 N at a location where the local gravitational acceleration is 9.81 m/s^2. What is the net acceleration of the rock?

Answer:

Both are correct.

Explanation:

Graphing multi meter is used to verify signals that move from electrical components. Digital oscilloscope is an equipment which stores and analyzes input signals with digital technique.

Complete Question:

Hans, a scheduler at Mangel-Wurzel Transport, gets a call from a regular customer needing to move 19.8m³ of rock and soil, which Hans knows from previous experience has an average density of 650/kgm³. Hans has available a dump truck with a capacity of 4m³ and a maximum safe load of 3700kg.  Required: calculate the number of trips the dump truck will have to make to haul the customer's load away.

Answer:

Mangel-Wurzel Transport

The number of trips that the dump truck will have to make to haul the customer's load away is:

= 5 trips.

Explanation:

a) Data and Calculations:

Volume of customer's load (rock and soil) = 19.8m³

Density of load = 650 kg/m³

Mass of load = Volume of load * Density of load

= 19.8m³ × 650 kg/m³

= 12,870 kg

The maximum safe load (mass) of the dump truck = 3,700 kg

Volume of the dump truck = 4m³  

Assuming the truck is to carry 4m³ of the load.

The mass of load that the 4m³ capacity truck can carry = 4m³ × 650kg/m³

= 2,600kg

Quick Check:

Mass = 2,600kg < 3,700 kg, satisfying required conditions.  

The number of trips that the truck would make to haul the customer's load away is, therefore, calculated as follows:

Number of trips = N

N = total volume of load/ volume per trip

N = 19.8/4

N = 4.95

N = 5 trips approx.

Solution :

The spring is expanded by 2 times of the block when it moves down an inclined by x times.

Here, [tex]$x_1$[/tex] = 39 mm

        [tex]x_2[/tex] = 225 mm

a). From the work energy principal,

   Work forces = kinetic energy

[tex]$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$[/tex]

[tex]$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$[/tex]

[tex]$9.75= 7.5(V^2_2-0.08^2)$[/tex]

[tex]$1.3= V^2_2-0.08^2$[/tex]

[tex]$V_2=1.14\ m/s$[/tex]

b). calculating the distance travelled by the block before it comes to rest.

Substitute the value of [tex]V_2[/tex] in (1),

[tex]$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$[/tex]

[tex]$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$[/tex]

[tex]$98.43x - 100(4x^2+0.156x)+0.048=0$[/tex]

[tex]$98.43x - 400x^2-15.6x+0.048=0$[/tex]

[tex]$82.83x - 400x^2+0.048=0$[/tex]

[tex]$ 400x^2- 82.83x-0.048=0$[/tex]

x = 0.20 m

Answer:

International Film Festival

Judging the best best film:

a. The probability that out of five judges (random sample),  three are in favor of film FA is:

= 33%.

b. The demerits of classical probability are:

1. Classical probability can only be used with events that have definite numbers of possible outcomes.  

2. Classical probability can only handle events where each outcome is equally likely.

3. Classical probability is based on the assumption of linear relationship (which is not always true in real life) between the latent variable and observed scores.

Explanation:

a) Number of judges = 11

Number of judges in favor of FA film = 6

Number of judges in favor of FB film = 5

Probability of judges in favor of FA film = 6/11

Probability of judges in favor of FB film = 5/11

Random sample of judges = 5

Probability that out of five judges, three are in favor of film FA = 3/5 * 6/11

= 18/55

= 33%

b) Classical probability is the simple probability showing that each event has equal chance of happening.  It can be contrasted with empirical probability that is obtained from experiments.

Answer:

Interval:  x∈ ( 0, ∞ )

There are no transient terms

Explanation:

x (dy/dx) – y= x^2sinx

Attached below is the detailed solution of the Given problem

There are no transient terms found in the general solution

Interval:  x∈ ( 0, ∞ )

Answer:

A. The biceps muscle contracts while the triceps muscle relaxes. Only then can the forearm move up to lift the mug.

Explanation:

Skeletal muscle is also known as voluntary muscle and it can be defined as a type of muscle connected with the skeleton by tendons, so as to form a mechanical system that enables the movement of the limbs and other body parts with respect to another.

Generally, skeletal muscles are only found in vertebrates.

The muscles in the arm work to lift a mug of coffee to your lips when the biceps muscle contracts while the triceps muscle relaxes.

Basically, the only way the forearm move up to lift a mug for example, is through the contraction of the biceps muscle and the relaxation of the triceps muscle.

This ultimately implies that, the elbow joint is straightened or bent due to the actions of the biceps muscle and triceps muscle against each other.

Answer:A

Explanation:

got it right on plato

Answer:

Explanation:

[tex]r_2=[/tex]∞

[tex]q=4\pi kT_1(T_2-T_1)\\[/tex]

[tex]q=2\pi kD.[/tex]ΔT--------(1)

[tex]q=hA[/tex] ΔT[tex]=4\pi r_1^2(T_2_s-T_1_s)\\[/tex]

[tex]N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)[/tex]

By equation (1) and (2)

[tex]2\pi kD.[/tex]ΔT=h.4[tex]\pi r_1^2[/tex]ΔT

[tex]2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\[/tex]-------(4)

From equation (3) and (4)

So for sphere [tex]R_a[/tex]→0

Answer:

H(s) =  10 / [ 1 + s / (200*10^3π ) ]^2

Explanation:

Characteristics of the high pass filter

Cutoff frequency = 100 kHz

stopband attenuation rate = 40 dB/decade

nominal passband gain = 20dB  = 20logK = 20

Formula for H(s) satisfying the requirements above

given that the stopband attenuation = 40 dB/decade the formula for H(s) that will satisfy the requirements is a second order filter

H(s) = K / ( 1 + s/Wo ) ^2  ----- ( 1 )

Wo = 2πf = 2π ( 100 * 10^3 ) =  200 * 10^3 π

 K = 10

back to equation ( 1 )

H(s) =  10 / [ 1 + s / (200*10^3π ) ]^2

The following should be done by the team member:

Learn more: brainly.com/question/17429689

Answer:

[tex]h_{1} = h_2} = 293.45 KJ/kg[/tex].

The quality of the refrigerant exiting the expansion valve is

[tex]x_{2}=0.193596[/tex].

Explanation:

Fluid given Ammonia.

Inlet 1:-

Temperature [tex]T_{1}[/tex] = [tex]24^{o} C[/tex].

Pressure [tex]P_{1}[/tex] = 10 bar.

Exit 2:-

Pressure [tex]P_{2}[/tex] = 1 bar.

Solution:-

Answer:

H(s) = 20 / [ 1 + s / 10^5 ]^2

Explanation:

Given data:

cutoff frequency = 100 kHz

stopband attenuation rate = 40 dB/decade

nominal passband gain = 20 dB

new nominal passband gain at cutoff = 14 dB

Represent the transfer function H(s)

The attenuation rate show that there are two(2) poles

H(s) = k / [ 1 + s/Wc ]^2  ----- ( 1 )

where : Wc = 100 kHz = 10^5 Hz , K = 20 log k = 20 dB ∴ k = 20

Input values into equation 1

H(s) = 20 / [ 1 + s / 10^5 ]^2

Answer:

Input power = 465.63 W

current = 1.94 A

Explanation:

we have the following data to answer this question

V = 9130

i = 0.051

the input power = VI

I = 51.0 mA = 0.051

= 9130 * 0.051

= 465.63 watts

the current = 465.63/240

= 1.94A

therefore the input power is 465.63 wwatts

while the current is 1.94A

the input power is the same thing as the output power.

Answer:

[tex]\eta_{turbine} = 0.603 = 60.3\%[/tex]

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = [tex]h_{g\ at\ 125KPa}[/tex] = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than [tex]s_g[/tex] and greater than [tex]s_f[/tex] at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

[tex]x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88[/tex]

Now, we will find [tex]h_{2s}[/tex](enthalpy at the outlet for the isentropic process):

[tex]h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg[/tex]

Now, the isentropic efficiency of the turbine can be given as follows:

[tex]\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%[/tex]

Answer:

True

Explanation:

Yes it is true that the Temperature gradient would also decrease with magnitude just as the distances rise from the centre line.

We have this cylinder equation as

[T1-T2 / ln(r1-r2)]2πKL

The radial distance is r2-r1

The gradient of temperature is T1-T2

From the equation,

The temperature gradient has a direct and proportional relationship to radial distance

T1-T2 ∝ ln(r2-r1)

1/T1-T2 = k(r2-r1)

This inverse relationship above confirms that the statement is true

Answer:

The code is as follows:

<form name = "myForm">

       <div>

           <input type="radio" name="vehicle" value="D0" id="D0"/>

           <label for="D0">Car</label>

       </div>

       <div>

           <input type="radio" name="vehicle" value="D1" id="D1"/>

           <label for="D1">Bike</label>

       </div>

   </form>

Explanation:

This defines the first button

           <input type="radio" name="vehicle" value="D0" id="D0"/>

           <label for="D0">Car</label>

This defines the second button

           <input type="radio" name="vehicle" value="D1" id="D1"/>

           <label for="D1">Bike</label>

The code is self-explanatory, as it follows all the required details in the question

Answer:

35 miles

Explanation:

When you are driving a truck that has an air brake system you have to keep in mind that when driving down a steep downgrade, the truck will automatically accelerate due to the inclination of the road, so in order to keep the speed to a controllable situation, you need to activate the service brake until you've reached the 35 miles per hour mark.

Answer:

The answer is "15 N".

Explanation:

Please find the complete question in the attached file.

In frame B:

For just slipping:

[tex]\to \frac{P}{2} \cos \theta =mg \sin \theta\\\\\to P=2 mg \tan \theta \\\\[/tex]

        [tex]=2 \times 1 \times g \times \tan 37^{\circ}\\\\ =2 \times 10 \times \frac{3}{4}\\\\ =15 \ N[/tex]

Solution :

Characteristic length  = thickness / 2

                                    [tex]$=\frac{0.04}{2}$[/tex]

                                    = 0.02 m

Thermal conductivity for steel is 42.5 W/m.K

[tex]$\text{Biot number} = \frac{\text{convective heat transfer coefficient} \times \text{characteristic length}}{\text{thermal conductivity}}$[/tex]

                  [tex]$=\frac{30 \times 0.02}{42.5}$[/tex]

                  = 0.014

Since the Biot number is less than 0.01, the lumped system analysis is applicable.

[tex]$\frac{T-T_{\infty}}{T_0-T_{\infty}} = e^{-b\times t}$[/tex]

Where,

T = temperature after t time

[tex]$T_{\infty}$[/tex] = surrounding temperature

[tex]$T_0$[/tex] = initial temperature

[tex]$b=\frac{\text{heat transfer coefficient}}{\text{density} \times {\text{specific heat } \times \text{characteristic length }}}$[/tex]

t = time

We calculate B:

[tex]$b=\frac{30}{7833 \times 460 \times 0.02}$[/tex]

  = 0.000416

Thus, [tex]$\frac{100-50}{500-50}=e^{-0.00416 \times t}$[/tex]

t = 5281.78 second

  = 88.02 minutes

Thus the time taken for reaching 100 degree Celsius is 88.02 minutes.

Answer:

a) -505.229 kJ/Kg

b) -1.724 kJ/kg

Explanation:

T1 = 400°C

P1 = 3 MPa

P2 = 125 kPa

work output   = 530 kJ/kg

surrounding temperature = 20°C = 293 k

A) Calculate heat transfer from Turbine to surroundings

Q = h2 + w - h1

h ( enthalpy )

h1 = 3231.229 kj/kg

enthalpy at P2

h2 = hg = 2676 kj/kg

back to equation 1

Q = 2676 + 50 - 3231.229  = -505.229 kJ/Kg  ( i.e.  heat is lost )

b) Entropy generation

entropy generation = Δs ( surrounding )  + Δs(system)

                                =  - 505.229 / 293   + 0

                                = -1.724 kJ/kg  

Actions violated:

Actions to be taken:

Edit: Use these to write your paragraph.

Answer:

a. unit of length: [L]

b. unit of volume: [[tex]L^3[/tex]]

c. unit of pressure:[tex]P=\frac{F}{A} \equiv\frac{[MLT^{-2}]}{[L^2]}[/tex] [tex][ML^{-1}T^{-2}][/tex]

d. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]

e. unit of force: [tex][kg.m/s^2]\equiv [MLT^{-2}][/tex]

f. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]

Force: [tex]F=m.a=m.\frac{v}{t}=m.\frac{x}{t}\div t[/tex]

Power: [tex]P=\frac{W}{t}=\frac{F.x}{t}[/tex]

where:

F = force

A = area

W = work

t = time

a = acceleration

v = velocity

x = displacement

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Answer:

1. Drawing grid.

2. Orthogonal.

3. Geometries.

4. CTRL+N.

5. Thirteen (13).

Explanation:

CAD is an acronym for computer aided design and it is typically used for designing the graphical representation of a building plan. An example of a computer aided design (CAD) software is AutoCAD.

Some of the features of an AutoCAD software are;

1. Drawing grid: is a regular pattern of dots displayed on the screen of an AutoCAD software, which acts as a visual aid and it's also used to define the extent of a drawing.

2. Ortho is short or an abbreviation for orthogonal, which means either vertical or horizontal.

3. Tangent is a point where two geometries meet at just a single point.

4. If you want to create a new drawing, simply press CTRL+N for the short cut key.

5. There are thirteen object snaps (Osnap) that can help you perform your task on AutoCAD easily. The 13 object snaps (Osnap) are; Endpoint, Midpoint, Apparent intersect, Intersection, Quadrant, Extension, Tangent, Center, Insert, Perpendicular, Node, Parallel, and Nearest.

Answer:

7.9 billion people

Explanation:

Explanation:

mass=19kg

density=800kg/m³

volume=?

as we know that

density=mass/volume

density×volume=mass

volume=mass/density

putting the values

volume=19kg/800kg/m³

so volume=0.02375≈0.02m³

The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

Reducing such a need to move in between multiple tabs, the split-screen has been valuable for increasing wages. In the several instances running a two or more desktop system will allow different programs to run throughout multiple devices. That works with the same process on both PC and laptop monitors.

Just display them side by side, instead of the switching among both the apps that has been used frequently. In this phase, an app that the snap to either left or right occupies a third of the display, and yet another app holds the two-thirds remaining. It refers to Split-Screen Mode.

Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization.

Therefore, The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

Learn more about average power on:

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#SPJ2

Answer: Can these nuts fit in your mouth?

Explanation:

im just here for the points >:)