A 29-mm-diameter copper rod is 1.1 m long with a yield strength of 73 MPa. Determine the axial force necessary to cause the diameter of the rod to reduce by 0.01 percent, assuming elastic deformation. Check that the elastic deformation assumption is valid by comparing the axial stress to the yield strength. The axial force necessary to cause the diameter of the rod to reduce by 0.01 percent is ____kN.

Answer:

a) 0 mi/s^2

b) 52 mi/s

Explanation:

Assuming the crossing is 1/2 mile past point A and that point B is near point A (it isn't clear in the problem)

The train was running at 70 mi/h at point A and with constant deceleration reachesn the crossing 1/2 mile away with a speed of 52 mi/h

The equation for position under constant acceleration is:

X(t) = X0 + V0 * t + 1/2 * a * t^2

I set my reference system so that the train passes point A at t=0 and point A is X = 0, so X0 = 0.

Also the equation for speed under constant acceleration is:

V(t) = V0 + a * t

Replacing

52 = 70 + a * t

Rearranging

a * t = 52 - 70

a = -18/t

I can then calculate the time it will take it to reach the crossing

1/2 * a * t^2 + V0 * t  - X(t) = 0

Replacing

1/2 (-18/t) * t^ + 70 * t - 1/2 = 0

-9 * t + 70 * t = 1/2

61 * t = 1/2

t = (1/2)/61 = 0.0082 h = 29.5 s

And the acceleration is:

a = -18/0.0082 = -2195 mi/(h^2)

To beath the train the car must reach the crossing in 29.5 - 4.3 = 25.2 s

X(t) = X0 + V0 * t + 1/2 * a * t^2

52 mi/h = 0.0144 mi/s

1/2 = 0 + 0.0144 * 25.2 + 1/2 * a * 25.2^2

1/2 = 0.363 + 317.5 * a

317.5 * a = 0.5 - 0.363

a = 0.137/317.5 = 0.00043 mi/s^2 (its almost zero)

The car should remain at about constant speed.

It will be running at the same speed.

Answer:

specific volume v = 0.0010385 m³/kg

and internal energy u = 416.23 KJ/kg

specific volume vf = 0.001043 m³/kg

and internal energy uf = 419.06 KJ/kg

% error in specific volume = 0.43 %

% error in internal energy = 0.679 %

Explanation:

given data

pressure = 10MPa

temperature = 100 C

to find out

specific volume and internal energy and  What percent error do you get

solution

we know pressure 10 MPa and temperature 100 C

so from compressed liquid water table we get

specific volume v = 0.0010385 m³/kg

and internal energy u = 416.23 KJ/kg

and

by using saturated liquid approx

from saturated water table of 100 C

specific volume vf = 0.001043 m³/kg

and internal energy uf = 419.06 KJ/kg

so

% error in specific volume is

% error in specific volume = [tex]\frac{0.001043-0.0010385}{0.0010385}[/tex] × 100

% error in specific volume = 0.43 %

and

% error in internal energy is

% error in internal energy = [tex]\frac{419.06-416.23}{416.23}[/tex] × 100

% error in internal energy = 0.679 %

Answer:

0.023 Pa*s

Explanation:

The surface area of the side of the inner cylinder is:

A = π*d*l

A = π*0.15*0.75 = 0.35 m^2

At 200 rpm the inner cylinder has a tangential speed of:

u = w * r

u = w * d/2

w = 200 rpm * 2π / 60 = 20.9 rad/s

u = 20.9 * 0.15 / 2 = 1.57 m/s

The torque is of 0.8 N*m, this means that the force is:

T = F * r

F = T / r

F = 2*T / d

For Newtoninan fluids with two plates moving respect of each other with a  fluid between the viscous friction force would be:

F = μ*A*u / y

Where

μ: viscocity

y: separation between pates

A: surface area of the plates

Then:

2*T / d = μ*A*u/y

Rearranging:

μ = 2*T*y / (d*A*u)

μ = 2*0.8*0.0012 / (0.15*0.35*1.57) = 0.023 Pa*s

Answer:

470 rad/s

Explanation:

The acceleration of the motor shaft is:

γ1 = 4*w1^(3/4)

When connected by a belt the pulleys have the same tangential speed

vt = w * r

vt1 = vt2

w1 * r1 = w2 * r2

w2 = w1 * r1/r2

Therefore:

γ2 = 4 * (w1 * r1/r2)^(3/4)

d(w1 * r1/r2)/dt = 4 * (w1 * r1/r2)^(3/4)

(r1/r2) * dw1/dt = 4 * (r1/r2)^(3/4) * (w1 * r1/r2)^(3/4)

dw1/dt = 4 * (r1/r2)^(-1/4) * (w1)^(3/4)

This is a differential equation.

Solving it through Wolfram Alpha:

w1(t) = (1 / 256) * (4 * (r1/r2)^(-1/4) * t - 4)^4

w1(4) = (1 / 256) * (4 * (0.25 / 1)^(-1/4) * 4 - 4)^4 = 470 rad/s

The angular velocity of the beater bar at t=4s is approximately 6.37 rad/s, based on the given angular acceleration equation and initial angular velocity.

The angular velocity of the beater bar can be found using the relationship between angular acceleration and angular velocity. The given equation states that the angular acceleration is four times the angular velocity raised to the 3/4 power. Therefore, we can write:

α = 4 * ω^(3/4)

To find the angular velocity at t=4s, we can integrate the equation to get:

ω = 4/7 * t^7/4 + C

When t = 0, ω = ω_0 = 1 rad/s. Substituting these values, we can solve for C:

1 = 0 + C

Therefore, C = 1. Finally, we can substitute t = 4s into the equation to get the angular velocity:

ω = 4/7 * 4^7/4 + 1

Calculating this expression, we find that the angular velocity of the beater bar at t=4s is approximately 6.37 rad/s.

Answer:

The air mass in the tank is 23.78 kg

Solution:

As per the question:

Volume of the tank, [tex]V_{t} = 200 l = 2\times 10^{- 3} m^{3}[/tex]

Pressure, P = 1 MPa = [tex]1\times 10^{6} Pa[/tex]

Temperature, T = [tex]20^{\circ}C[/tex] = 273 + 20 = 293 K

Pressure, P' = 0.5 MPa = [tex]0.5\times 10^{6} Pa[/tex]

Now,

To calculate the air mass, [tex]m_{a}[/tex] we use:

[tex]PV_{t} = m_{a}RT[/tex]

where

R = Rydberg constant = 0.287 J/kg.K

[tex]1\times 10^{6}\times 2\times 10^{- 3} = m_{a}0.287\times 293[/tex]

[tex]m_{a} = 23.78 kg[/tex]

Answer:

a) on mars W=1454.7N

b)on earth W=3825.9N

Explanation:

The weight of any body with mass is given by the following equation

W=mg

where

m=mass

g=gravity

W=weight

Remember that the weight is expresed in Newton and the units are kgm/s^2

A)weight on the surface of mars

W=(390kg)(3.73m/s^2)=1454.7N

b) on earth

W=(390kg)(9.81m/s^2)=3825.9N

Answer:

Pressure of pressure head will be [tex]863.6pound/ft^2[/tex]

Explanation:

We have given pressure head = 14 feet

Specific weight of water [tex](\rho g)_{water}=62.4pound/ft^3[/tex]

Pressure is given by [tex]P=(\rho g)_{water}h[/tex], here h is pressure head.

So pressure [tex]P=(\rho g)_{water}h=62.4pound/ft^3\times 14ft=863.6pound/ft^2[/tex]

So the pressure of pressure head will be [tex]863.6pound/ft^2[/tex]

Answer:

The force over the plane windows are 764 lbf in the EE unit system and 3398 N in the international unit system.

Explanation:

The net force over the window is calculated by multiplying the difference in pressure by the area of the window:

F = Δp*A

The pressure inside the plane is around 1 atm, hence the difference in pressure is:

Δp = 1atm - 0.35 atm = 0.65 atm

Expressing in the EE unit system:

Δp = 0.65 atm * 14.69 lbf/in^2 = 9.55 lbf/in^2

Replacing in the force:

F = 9.55 lbf/in^2 * 80 in^2  = 764 lbf

For the international unit system, we re-calculate the window's area and the difference in pressure:

A = 80 in^2 * (0.0254 m/in)^2 = 0.0516 m^2

Δp =  0.65 atm * 101325 Pa  = 65861 Pa  = 65861 N/m^2

Replacing in the force:

F = 65861 N/m^2  *0.0516 m^2  = 3398 N

Answer:

The pressure and volume of the tank are 246.878 Kpa and 9.449 m³ respectively.

Explanation:

Volume is constant as the tank is rigid. Take the saturation condition of water from the steam table for pressure at 127°C.  

Given:  

Total mass of water is 22 kg.  

Mass of liquid water is 9 kg.  

Temperature of water is 127°C.  

From steam table at 127°C:  

The pressure in the tank is 246.878 Kpa.  

Specific volume of saturated water is 0.00106683 m³/kg.  

Specific volume of saturated steam is 0.72721 m³/kg.  

Calculation:  

Step1  

From steam table at 127°C:  

The pressure in the tank is 246.878 Kpa.  

Step2  

Dryness fraction is calculated as follows:  

[tex]x=\frac{m_{v}}{m_{t}}[/tex]

Here, dyness fraction is x, mass of vapor is [tex]m_{v}[/tex]and total mass is [tex]m_{t}[/tex].  

Substitute the values in the above equation as follows:  

[tex]x=\frac{m_{v}}{m_{t}}[/tex]

[tex]x=\frac{22-9}{22}[/tex]

x = 0.59  

Step3  

Specific volume of tank is calculated as follows:  

[tex]v=v_{f}+x(v_{g}-v_{f})[/tex]

[tex]v=0.00106683+0.59(0.72721-0.00106683)[/tex]

[tex]v=0.00106683+0.42842447[/tex]

v=0.4295 m³/kg.  

Step4  

Volume is calculated as follows:  

[tex]V=v\times m_{t}[/tex]

[tex]V=0.4295 \times22[/tex]

V=9.449 m³.  

Thus, the pressure and volume of the tank are 246.878 Kpa and 9.449 m³ respectively.

Answer:

The correct answer is option 'a': Black

Explanation:

As we know that for an object which is black in color it absorbs all the electromagnetic radiation's that are incident on it. Thus if we need to transfer energy to an object by radiation the most suitable color for the process  is black.

In contrast to black color white color is an excellent reflector, reflecting all the incident radiation that may be incident on it hence is the least suitable material for radiative heat transfer.

Answer:

Hooke's law is a principle of physics that states that the force F needed to extend or compress a spring by some distance X is proportional to that distance. That is: F = kX, where k is a constant factor characteristic of the spring: its stiffness, and X is small compared to the total possible deformation of the spring.

Explanation:

Answer:3142 Pound force

Explanation:

Given

Length(L)=16 in.

diameter=0.5 in.

Modulus of elasticity (E)[tex]=6.4\times 10^6 psi.[/tex]

Allowable stress [tex]\left ( \sigma _{allowable}\right )=17000 psi[/tex]

Max elongation of the bar =0.04 in.

Also we Know

Elongation is given

[tex]\Delta L=\frac{PL}{AE}[/tex]

where

P=Force applied

L=length of bar

A=Cross-section

E=Modulus of Elasticity

[tex]A=\frac{\pi d^2}{4}=\frac{\pi}{16} in.^2[/tex]

[tex]0.04=\frac{P\times 16\times 16}{\pi \times 6.4\times 10^6}[/tex]

[tex]P=3142 pound\ force\approx 13.976 kN[/tex]

corresponding stress is [tex]\frac{3142}{\frac{\pi }{16}}=16,000 psi[/tex]

Which is less than Allowable stress

thus allowable value of Force P is 3142 Pound force or 13.976 kN

Answer:

The thermal efficiency of cycle is 42.6%.

Explanation:

Given that

[tex]T_1=300 K[/tex]

[tex]P_1=100KPa[/tex]

mass flow rate = 6 kg/s

Compression ratio = 7

Turbine inlet temperature = 1200 K

γ=1.4

We know that thermal efficiency of Brayton cycle given as

[tex]\eta=1-\dfrac{1}{r_p^{\frac{\gamma-1}{\gamma}}}[/tex]

Now by putting the values

[tex]\eta=1-\dfrac{1}{r_p^{\frac{\gamma-1}{\gamma}}}[/tex]

[tex]\eta=1-\dfrac{1}{7^{\frac{1.4-1}{1.4}}}[/tex]

η=0.426

So the thermal efficiency of cycle is 42.6%.

Answer:

total weight of liquid = 4788.25 N or 488.09 kg

Explanation:

given data

total volume = 0.5 m³

volume of ethanol = 10 %  of volume = 0.10 × 0.5 = 0.05 m³

volume of water = 90 % at 25 °C of volume = 0.90 × 0.5 = 0.45 m³

to find out

weight of the liquid

solution

we know that density of water at 25  is 997 kg/m³

and density of ethanol is 789 kg/m³

so weight of water is = density × volume × g

put here value and we take g = 9.81

weight of water is = 997 × 0.45 × 9.81

weight of water = 4401.25 N     ......................1

weight of ethanol is = density × volume × g

put here value and we take g = 9.81

weight of ethanol is = 789 × 0.05 × 9.81

weight of ethanol = 387.00 N       ...............2

so total weight of liquid = sum of equation 1 add 2

total weight of liquid =  4401.25 + 387

total weight of liquid = 4788.25 N or 488.09 kg

Answer:

The number of inputs processed by the new machine is 64

Solution:

As per the question:

The time complexity is given by:

[tex]T(n) = 10\times 2n[/tex]

where

n = number of inputs

T = Time taken by the machine for 'n' inputs

Also

The new machine is 65 times faster than the one currently in use.

Let us assume that the new machine takes the same time to solve k operations.

Then

T(k) = 64 T(n)

[tex]\frac{T(k)}{T(n)} = 64[/tex]

[tex]\frac{20k}{20n} = 64[/tex]

k = 64n

Thus the new machine will process 64 inputs in the time duration T

Answer:

In the steel: 815 kPa

In the aluminum: 270 kPa

Explanation:

The steel pipe will have a section of:

A1 = π/4 * (D^2 - d^2)

A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2

The aluminum core:

A2 = π/4 * d^2

A2 = π/4 * 0.7^2 = 0.3848 m^2

The parts will have a certain stiffness:

k = E * A/l

We don't know their length, so we can consider this as stiffness per unit of length

k = E * A

For the steel pipe:

E = 210 GPa (for steel)

k1 = 210*10^9 * 0.1178 = 2.47*10^10 N

For the aluminum:

E = 70 GPa

k2 = 70*10^9 * 0.3848 = 2.69*10^10 N

Hooke's law:

Δd = f / k

Since we are using stiffness per unit of length we use stretching per unit of length:

ε = f / k

When the force is distributed between both materials will stretch the same length:

f = f1 + f2

f1 / k1 = f2/ k2

Replacing:

f1 = f - f2

(f - f2) / k1 = f2 / k2

f/k1 - f2/k1 = f2/k2

f/k1 = f2 * (1/k2 + 1/k1)

f2 = (f/k1) / (1/k2 + 1/k1)

f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN

f1 = 200 - 104 = 96 kN

Then we calculate the stresses:

σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa

σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa

The answer is: within the steel: 815 kPa

When within the aluminum: 270 kPa

When The steel pipe will have a piece of:Then A1 = π/4 * (D^2 - d^2)After that A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2

The aluminum core is: Now A2 = π/4 * d^2Then A2 = π/4 * 0.7^2 = 0.3848 m^2

After that The parts will have a particular stiffness:k = E * A/l

We don't know their length, so we are able to consider this as stiffness per unit of length k = E * A

For the steel pipe: E = 210 GPa (for steel) k1 = 210*10^9 * 0.1178 = 2.47*10^10 N

For the aluminum: E = 70 GPak2 = 70*10^9 * 0.3848 = 2.69*10^10 NHooke's law:Δd = f / k

Since we are using stiffness per unit of length we use stretching per unit of length:ε = f / k

When the force is distributed between both materials will stretch the identical length:f = f1 + f2f1 / k1 = f2/ k2

Replacing: f1 = f - f2(f - f2) / k1 = f2 / k2f/k1 - f2/k1 = f2/k2f/k1 = f2 * (1/k2 + 1/k1)f2 = (f/k1) / (1/k2 + 1/k1)f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KNf1 = 200 - 104 = 96 kN

Then we calculate the stresses:σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPaσ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa

Find out more information about Aluminum here:

brainly.com/question/4229590

Answer:

option D

Explanation:

The correct answer is option D.

For jet aircraft engine, all the system that is nozzle, turbine and Compressor are used.

Nozzle is used to provide thrust to the aircraft. It is attached at the downstream of the turbine.

Turbine is used to release hot gas which is produced in air due to burning of gas.

Engine suck air from the fan attached in front and compressor increase the pressure of the air.

Answer

given,

P₁ = 8 bar     T₁  = 500 K       V₁ = 150 m/s

P₂ = 1 bar     T₂  = 320 K        V₂ = 10 m/s

writing energy equation

h₁ + (KE)₁ + (PE)₁  + Q m = h₂ + (KE)₂ + (PE)₂ + W

[tex]W = (h_1 - h_2 ) + \dfrac{V_1^2-V_2^2}{2000}[/tex]

ideal gas property of air

T₁  = 500 K       h₁ = 503.02 KJ/kg  S₁ = 2.21952 kJ/kgK

T₂  = 320 K       h₂ = 320.29 KJ/kg  S₂ = 1.7679 kJ/kgK

[tex]W = (503.02-320.29) + \dfrac{150^2-10^2}{2000}[/tex]

W = 193.93 KJ/Kg

calculation of energy destruction

= [tex]T_0(S_2-S_1-Rln(\dfrac{P_2}{P_1}))[/tex]

= [tex]T_0(S_2-S_1+Rln(\dfrac{P_1}{P_2}))[/tex]

= [tex]300(1.7679-2.21952-\dfrac{8.314}{28.97}ln(\dfrac{8}{1}))[/tex]

=[tex]300 \times 0.145152[/tex]

=43.54 KJ/Kg

Answer:

The correct answer is option 'c':Convection.

Explanation:

When we ignite a campfire the heat produced by combustion heats the air above the fire. As we know that if a gases gains heat it expands thus it's density decreases and hence it rises, if we hold our hands directly above the fire this rising hot air comes in contact with our hands thus warming them.

The situation is different if we are at some distance from the campfire laterally. Since the rising air cannot move laterally the only means the heat of the fire reaches our body is radiation.

But in the given situation the correct answer is convection.  

Answer:

Q(h=200)=0.35W

Q(h=3000)=5.25W

Explanation:

first part h=200W/Km^2

we must use the convection heat transfer equation for the chip

Q=hA(Ts-T∞)

h= convective coefficient=200W/m2 K

A=Base*Leght=5mmx5mm=25mm^2

Ts=temperature of the chip=85C

T∞=temperature of coolant=15C

Q=200x2.5x10^-5(85-15)=0.35W

Second part h=3000W/Km^2

Q=3000x2.5x10^-5(85-15)=5.25W

Answer:

The velocity of the fluid is 1.1012 m/s

Solution:

As per the question, for the fluid:

Diameter of the capillary tube, d = 1.0 mm = [tex]1.0\times 10^{- 3} m[/tex]

Reynolds No., R  = 1000

Kinematic viscosity,  [tex]\mu_{k} = 1.1012\times 10^{- 6} m^{2}/s[/tex]

Now, for the fluid velocity, we use the relation:

[tex]R = \frac{v_{f}\times d}{\mu_{k}}[/tex]

where

[tex]v_{f}[/tex] = velocity of fluid

[tex]v_{f} = \frac{R\times \mu_{k}}{d}[/tex]

[tex]v_{f} = \frac{1000\times 1.1012\times 10^{- 6}}{1.0\times 10^{- 3}} = 1.1012 m/s[/tex]

Answer:

(1) [tex]\Delta E = 4845.43 kW[/tex]

(2) [tex]\Delta E_{m} = 5.7319 kW[/tex]

(3) [tex]\Delta E_{t} = 4839.69 kW[/tex]

(4) q = 4839.69 kW[/tex]

Solution:

Using Saturated water-pressure table corresponding to pressure, P = 10 bar:

At saturated temperature, Specific enthalpy of water, [tex]h_{ws} = h_{f} = 762.5 kJ/kg[/tex]

At inlet:

Saturated temperature of water, [tex]T_{sw} = 179.88^{\circ}C[/tex]

Specific volume of water, [tex]V_{wi} = V_{f} = 0.00127 m^{3}/kg[/tex]

Using super heated water table corresponding to a temperature of [tex]600^{\circ}C[/tex] and at  7 bar:

At outlet:

Specific volume of water, [tex]V_{wso} = 0.5738 m^{3}/kg[/tex]

Specific enthalpy of water, [tex]h_{wo} = 3700.2 kJ/kg[/tex]

Now, at inlet, water's specific enthalpy is given by:

[tex]h_{i} = C_{p}(T - T_{sw}) + h_{ws}[/tex]

[tex]h_{i} = 4.187(110^{\circ} - 179.88^{\circ}) + 762.5[/tex]

[tex]h_{i} = -292.587 + 762.5= 469.912 kJ/kg[/tex]

(1) Now, the change in combined thermal energy and work flow is given by:

[tex]\Delta E = E_{o} - E_{i}[/tex]

[tex]\Delta E = m(h_{wo} - h_{i})[/tex]

[tex]\Delta E = 1.5(3700.2 - 469.912) = 4845.43 kW[/tex]

(2) The mechanical energy can be calculated as:

velocity at inlet, [tex]v_{i} = \rho A V_{wi}[/tex]

[tex]v_{i} = \frac{mV_{wi}}{frac{\pi d^{2}}{4}}[/tex]

[tex]v_{i} = \frac{mV_{wi}}{frac{\pi d^{2}}{4}}[/tex]

[tex]v_{i} = \frac{1.5\times 0.00127}{frac{\pi (63\times 10^{- 3})^{2}}{4}}[/tex]

[tex]v_{i} = 0.542 m/s[/tex]

Similarly,, the velocity at the outlet,

[tex]v_{o} =  \frac{1.5\times 0.57378}{frac{\pi (63\times 10^{- 3})^{2}}{4}}[/tex]

[tex]v_{o} =  276.099 m/s[/tex]

Now, change in mechanical energy:

[tex]\Delta E_{m} = E_{mo} - E_{mi}[/tex]

[tex]\Delta E_{m} = m[(\frac{v_{o}^{2}}{2} + gz_{o}) - (\frac{v_{i}^{2}}{2} + gz_{i})][/tex]

[tex]\Delta E_{m} = 1.5[(\frac{276.099^{2}}{2} + 9.8(z_{o} - z_{i}) - (\frac{0.542^{2}}{2}][/tex]

[tex]\Delta E_{m} = 57319 J = 5.7319 kW[/tex]

(3) The total energy of water is given by:

[tex]\Delta E_{t} = E - E_{m} = 4845.43 - 5.7319 = 4839.69 kW[/tex]

(4) The rate of heat transfer:

q = [tex]\Delta E_{t} = 4839.69 kW[/tex]

Answer:

e)v=120 m/s

Explanation:

Given that

Scale ratio = 1/4

Speed of car =30 m/s

lets wind tunnel speed is v

We know that Reynolds number given as

[tex]Re=\dfrac{\rho\ L\ V}{\mu }[/tex]

If all conditions taken as similar then

[tex](L\ V)_c=(L\ V)_w[/tex]

Given that

[tex]\dfrac{L_w}{L_c}=\dfrac{1}{4}[/tex]

So we can say that

4 x 30 = v x 1

v=120 m/s

Answer:

101.42235 kPa

Explanation:

The unit inHg means "inches of mercury", Its a pressure unit commonly used  by the US aviators.

The conversion value to KPa (kilopascal) is

1 inHg= 3.386389 kPa

So now we only have to multiply:

29.95 inHg * 3.386389 kPa/in Hg =101.42235 kPa

Have a nice day and Good Luck!

Answer with Explanation:

The general equation of simple harmonic motion is

[tex]x(t)=Asin(\omega t+\phi)[/tex]

where,

A is the amplitude of motion

[tex]\omega [/tex] is the angular frequency of the motion

[tex]\phi [/tex] is known as initial phase

part 1)

Now by definition of velocity we have

[tex]v=\frac{dx}{dt}\\\\\therefore v(t)=\frac{d}{dt}(Asin(\omega t+\phi )\\\\v(t)=A\omega cos(\omega t+\phi )[/tex]

part 2)

Now by definition of acceleration we have

[tex]a=\frac{dv}{dt}\\\\\therefore a(t)=\frac{d}{dt}(A\omega cos(\omega t+\phi )\\\\a(t)=-A\omega ^{2}sin(\omega t+\phi )[/tex]

part 3)

The angular frequency is related to Time period 'T' as[tex]T =\frac{2\pi }{\omega }[/tex]

where

[tex]\omega [/tex] is the angular frequency of the motion of the particle.

Part 4) The acceleration and velocities are plotted below

since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is [tex]A\omega ^{2}[/tex] and [tex]A\omega [/tex] respectively.

Answer:

1)  75Mpa

2) 1.125 MPa

Explanation:

given data:

gauge length = 4 mm

displacement [tex]= 1.5\mu m = 1.5\times 10^{-3} m[/tex]

a) structural steel

Young modulus for steel is [tex]200 GPa = 200\times 10^3 MPa[/tex]

we know that

[tex]E  =\frac{stress}{strain}[/tex]

[tex]Stress = 200\times 10^3 \times \frac{1.5\times 10^{-3}}{4}[/tex]

          = 75Mpa

b) PMMA

Young's modulus [tex]= 3GPa = 3\times10^3 MPa[/tex]

[tex]stress = 3\times \frac{1.5\times10^{-3}}{4}[/tex]

stress = 1.125 MPa

Answer:

b)false

Explanation:

false, missing two variables, the current and the material.

To find the volume of the tube we must know the transverse area and the length.

It is possible to calculate the length of the tube knowing the resistance, since this is a value that depends on the material, the area and the length.

It is possible to calculate the electrical resistance using the ohm equation.

Resitence = Voltage / current.

Explanation:

Given T = 10 °C

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

T(K) = T( °C) + 273.15  

So,  

T = (10 + 273.15) K = 283.15 K

T = 283.15 K

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

T (°F) = (T (°C) × 9/5) + 32  

So,

T (°F) = (10 × 9/5) + 32 = 50 °F

T = 50 °F

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

T (R) = (T (°C) × 9/5) + 491.67

So,

T (R) = (10 × 9/5) + 491.67 = 509.67 R

T = 509.67 R

A 10°C rise in temperature is equivalent to a 10 K increase in the Kelvin scale and an 18°F increase in the Fahrenheit scale. For Rankine, it's an 18 R increase.

The temperature of a system rising by 10 °C can be expressed in different units of measurement. Firstly, since the Celsius and Kelvin scales are offset by 273.15, an increase by 10 °C is equivalent to an increase by 10 K. For the Fahrenheit scale, we use the conversion formula (°C × 9/5) + 32, which means an increase by 10 °C corresponds to an increase by 18 °F. Therefore, a rise by 10 °C equates to a rise by 10 K in the Kelvin scale and by 18 °F in the Fahrenheit scale. In the Rankine scale, which is Fahrenheit-based, the increase would be 18 R since it also uses the 9/5 scaling factor.  

Answer:

mg = 30.415 lbf

Explanation:

from figure body of size 3ft*2ft is tend to move down side

weight is divided into two component

vertical component =  mgcos15 and

horizontal component = mg sin15

considering horizontal component equal to shear force

mgsin15 = \tau A

mgsin15 =\mu \frac{dv}{dh} A

       mg =\frac{ \mu v A}{h*sin15}

             =\frac{8.2*10^{-2}*0.2*3*2}{0.0125*sin15}

        mg = 30.415 lbf

Answer:

Final Velocity (Vf)= 139.864 ft/s

Time (t)= 4,34 s

Explanation:

This is a free fall problem, to solve it we will apply free  fall concepts:

In a free fall the acceletarion is gravity (g) = 9,81 m/s2, if we convert it to ft/s^2 = g= 32.174 ft/s^2

where h is height (304 ft in this case).

Vo =0 since the hammer wasn't moving when it stared to fall

Then Vf^2= 0 + 2* 32.174 ft/s^2 *304 ft

          Vf^2= 19,561.8224  ft^2/s^2

          Vf=[sqrt{19561.8224 ft^2/s^2}

          Vf=139.864 ft/s

Time t= (Vf-Vo)/g => (139.864 ft/s-0)/32.174 ft/s^2 = 4.34 sec

Good luck!