Complete question:
A steel component with a tensile strength of 800 MPa and fracture toughness Kic=20 MPa Nm is known to contain internal cracks (also called tunnel cracks) with the maximum length of 1.4 mm. This steel is being considered for use in a cyclic loading application for which the designed stresses vary from 0 to 420 MPa. Would you recommend using this steel in this application?
a. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
b. Yes, this because the tensile strength of steel is much higher than the applied highest stress of 420 MPa.
c. Yes, this because the calculated critical stress to fracture for the cracks is higher than the highest applied stress of 420 MPa and the steel can withstand the stress of 420 MPa.
d. No. Although the calculated critical stress to fracture for the cracks is slightly higher than the highest applied stress of 420 MPa and the steel may withstand the static stress of 420 MPa, the cyclic loading may cause rapid fatigue fracture.
Answer:
A. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
Explanation:
we are not sure if to recommend this alloy for this application given that this material has already been left to experience fatigue degradation. the cyclic load application brings about a growth in the crack. We know that cyclic loading is continuous loading that is useful for the testing of fatigue. Therefore the answer to this question is option a. We cannot make recommendations except fatigue testing has been carried out.
thank you!
Suppose that you have a block of copper which is 300 g. Relevant material parameters follow.
Material atomic weight Density
Copper 63.5 g/mol 8.96 g/cm^3
Using equipartition, compute the molar heat capacity in J/mol K.
Answer:
[tex]C_v= 24.942[/tex] J / mol K
Explanation:
Molar Heat Capacity
using the equipartition, the equation :
[tex]$C_v=\frac{d}{2} R$[/tex]
Here, [tex]C_v[/tex] = molar heat capacity
d = degree of freedom
R = Universal gas constant
Degree of freedom, d is 6
Universal gas constant, R = 8.317 J/ mol K
Therefore, the molar heat capacity is :
[tex]$C_v=\frac{d}{2} R$[/tex]
[tex]$C_v=\frac{6}{2} \times 8.314$[/tex]
[tex]C_v= 24.942[/tex] J / mol K
A power cycle receives QH by heat transfer from a hot reservoir at TH = 1200 K and rejects energy QC by heat transfer to a cold reservoir at TC = 400 K. For each of the following cases, determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
a.QH = 900 kJ, Wcycle= 450 kJ
b. QH = 900 kJ, Qc = 300 kJ
c. Weycle = 600 kJ, Qc= 400 kJ
d. η = 75%
Answer:
a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.
Explanation:
Maximum theoretical efficiency for a power cycle ([tex]\eta_{r}[/tex]), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:
[tex]\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\%[/tex] (1)
Where:
[tex]T_{C}[/tex] - Temperature of the cold reservoir, in Kelvin.
[tex]T_{H}[/tex] - Temperature of the hot reservoir, in Kelvin.
The maximum theoretical efficiency associated with this power cycle is: ([tex]T_{C} = 400\,K[/tex], [tex]T_{H} = 1200\,K[/tex])
[tex]\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%[/tex]
[tex]\eta_{r} = 66.667\,\%[/tex]
In exchange, real efficiency for a power cycle ([tex]\eta[/tex]), no unit, is defined by this expression:
[tex]\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\%[/tex] (2)
Where:
[tex]Q_{C}[/tex] - Heat released to cold reservoir, in kilojoules.
[tex]Q_{H}[/tex] - Heat gained from hot reservoir, in kilojoules.
[tex]W_{C}[/tex] - Power generated within power cycle, in kilojoules.
A power cycle operates irreversibly for [tex]\eta < \eta_{r}[/tex], reversibily for [tex]\eta = \eta_{r}[/tex] and it is impossible for [tex]\eta > \eta_{r}[/tex].
Now we proceed to solve for each case:
a) [tex]Q_{H} = 900\,kJ[/tex], [tex]W_{C} = 450\,kJ[/tex]
[tex]\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 50\,\%[/tex]
Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.
b) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{C} = 300\,kJ[/tex]
[tex]\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 66.667\,\%[/tex]
Since [tex]\eta = \eta_{r}[/tex], the power cycle operates reversibly.
c) [tex]W_{C} = 600\,kJ[/tex], [tex]Q_{C} = 400\,kJ[/tex]
[tex]\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 60\,\%[/tex]
Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.
d) Since [tex]\eta > \eta_{r}[/tex], the power cycle is impossible.
The base of an aluminum block, which is fixed in place, measures 90 cm by 90 cm, and the height of the block is 60 cm. A force, applied to the upper face and parallel to it, produces a shear strain of 0.0060. The shear modulus of aluminum is . What is the displacement of the upper face in the direction of the applied force
The question is incomplete. The complete question is :
The base of an aluminum block, which is fixed in place, measures 90 cm by 90 cm, and the height of the block is 60 cm. A force, applied to the upper face and parallel to it, produces a shear strain of 0.0060. The shear modulus of aluminum is [tex]3.0 \times 10^{10} \ Pa[/tex] . What is the displacement of the upper face in the direction of the applied force?
Solution :
The relation between shear modulus, shear stress and strain,
[tex]$\text{Shear modulus, S =} \frac{\text{Shear stress}}{\text{shear strain}}$[/tex]
Shear stress = shear modulus (S) x shear strain
[tex]$=3 \times 10^{10} \times 0.0060$[/tex]
[tex]$=1.80 \times 10^8$[/tex] Pa
[tex]$=180 \times 10^6$[/tex] Pa
[tex]$=180 \ MPa$[/tex]
The length represents the distance between the fixed in place portion and where the force is being applied.
Therefore,
[tex]$\text{Displacement} = \text{shear strain} \times \text{length}$[/tex]
= 0.006 x 60 cm
= 0.360 cm
= 3.6 mm
Thus, the displacement of the upper face is 3.6 mm in the direction of the applied force.
A 35kg block of mass is subjected to forces F1=100N and F2=75N at agive angle thetha= 20° and 35° respectively.find the distance it slides in 10seconds if the kinetic coefficient is 0.4.
Answer:
21 m
Explanation:
Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.
Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N
The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N
The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N
If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.
Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.
For the block to be in contact with the surface, the vertical forces on the block must balance.
Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,
N = mg + F₃" (since both the weight and the resultant vertical force act downwards)
N = mg + F₃"
Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N
So,
N = mg + F₃"
N = 35 kg × 9.8 m/s² + 8.82 N
N = 343 N + 8.82 N
N = 351.82 N
So, the net horizontal force F = F₃' - f.
F = 155.41 N - 0.4 × 351.82 N
F = 155.41 N - 140.728 N
F = 14.682 N
Since F = ma, where a = acceleration of block,
a = F/m = 14.682 N/35 kg = 0.42 m/s²
To find the distance the block moved, x we use the equation
x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²
Substituting the values of the variables into the equation, we have
x = ut + 1/2at²
x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²
x = 0 m + 1/2 × 0.42 m/s² × 100 s²
x = 0.21 m/s² × 100 s²
x = 21 m
So, the distance moved by the block is 21 m.
what is an OTG USB? how is it useful
Answer:
An OTG or On The Go adapter (sometimes called an OTG cable, or OTG connector) allows you to connect a full sized USB flash drive or USB A cable to your phone or tablet through the Micro USB or USB-C charging port
Explanation:
pls mark brainliest
Write the code using the do-while loop to force the user to enter a number in the range [20,50]
Answer:
Mark as brainlist pls hello
What is protection scheme?
Answer:
The objective of a protection scheme is to keep the power system stable by isolating only the components that are under fault, whilst leaving as much of the network as possible still in operation.
Explanation:
The devices that are used to protect the power systems from faults are called protection devices.
Uses of P-N junction
Answer:
Explanation:
Two that come to mind:
a semiconductor diode is essentially a PN junctiona transistor is made of two pn junctions.beacuse thye want them to hav egoood and thye wn thme tto
Answer:
I don't understand the question
Increasing following distance to
when encountering other motorists who follow too closely
is an example of appropriate implementation of the IPDE defensive driving strategy for the
maintenance of an appropriate Safety Cushion.
Two-seconds
Enree-seconds
Four-seconds
Twenty-seconds
Answer:
Increasing following distance to Four-seconds when encountering other motorists who follow too closely is an example of appropriate implementation of the IPDE defensive driving strategy for the maintenance of an appropriate Safety Cushion.
Explanation:
Maintaining the required safety cushion by utilizing the IPDE defensive driving strategy to manage the nine to fifteen space driving zones involves continuous scanning. Therefore, motorists should be able to identify objects and hazards in the driving scene, line of sight, and path of travel. They should predict points of driving conflicts. They should determine appropriate and safe driving actions to take, when, and where. Finally, action is required to ensure that conflicts are avoided.
Urgent!!!
List the assumptions that can be taken into account in torsion analysis.
Answer:
Explanation:
In the development of a torsion formula for a circular shaft, the following assumptions are made: Material of the shaft is homogeneous throughout the length of the shaft. Shaft is straight and of uniform circular cross section over its length. Torsion is constant along the length of the shaft.
Which statement describes the relay between minerals and rocks ?
Answer:
•○●□■hey hi!■□●○•Explanation:
Minerals and rocks are the same. Aggregates of minerals form rocks. Minerals determine the texture of a rock. Most rocks are made of a single mineral type.
☆♡hope this helps♡☆if a person is injured at the hospital during a natural disaster a correct action to take is
Race cars at the Indianapolis Speedway average speeds of 185 mi/h. After determining the altitude of Indianapolis, find the Mach number of these cars and estimate whether compressibility might affect their aerodynamics.
Answer:
- the Mach number is 0.24.
- Compressibility becomes effective when Mach number is greater than 0.3, the Mach number of the race cars is less than 0.3, hence, compressibility will not affect their aerodynamics.
Explanation:
Given the data in the question;
Average speed V = 185 miles per hour = ( 185 /2.237 ) m/s = 82.7 m/s
From Almanac, we can find that Indianapolis is at 220 m altitude.
So from table, at that altitude, the standard speed of sound will be 339.4 m/s .
Mach number of the race car will be;
Mach Number = Velocity / sound speed
we substitute
Mach Number = ( 82.7 m/s ) / ( 339.4 m/s )
Mach Number = 0.24
Therefore the Mach number is 0.24.
We know that, compressibility becomes effective when the Mach number is greater than 0.3.
Since the Mach number of the race cars is less than 0.3, compressibility will not affect their aerodynamics.
17- The cathodic polarization is ..... *
O
a- activation.
O
b- concentration
O c- both.
What are three types of land reform
Answer:
Abolition of intermediaries (rent collectors under the pre-Independence land revenue system); Tenancy regulation (to improve the contractual terms including the security of tenure); A ceiling on landholdings (to redistributing surplus land to the landless);
Types of Land Reform
Abolition of Intermediaries
The first step taken by the Indian government under land reforms post-independence was passing the Zamindari Abolition Act. The primary reason of a backward agrarian economy was the presence of intermediate entities like, jagirdars and zamindar who primarily focussed on collecting sky-rocketing rents catering to their personal benefits, without paying attention to the disposition of farms and farmers. Abolition of such intermediaries not only improved conditions of farmers by establishing their direct connection with the government but also improved agricultural production.
Regulation of Rents
This was in direct response to the unimaginably high rents which were charged by intermediaries during British rule, which resulted in a never-ending cycle of poverty and misery for tenants. Indian government implemented these regulations to protect farmers and labourers from exploitation by placing a maximum limit on the rent that could be charged for land.
Tenure Security
Legislations were passed in all states of the country to grant tenants with permanent ownership of lands and protection from unlawful evictions on expiry of the lease. This law protects tenants from having to vacate a property immediately after their tenure is over unless ordered by law. Even in that case, ownership can be regained by tenants with the excuse of personal cultivation.
While reflecting on the solutions and the process of concept generation, the development team takes a look at some critical questions such as:________.
1. Is the team developing confidence that the solution space has been fully explored?
2. Are there alternative diagrams and alternative ways to decompose the problem?
3. Have external sources been thoroughly pursued, and everyone’s ideas been accepted and integrated in the process?
4. All of the above
Answer:
While reflecting on the solutions and the process of concept generation, the development team takes a look at some critical questions such as:________.
4. All of the above
Explanation:
The team must explore its solution space, including some external sources. Then, it must integrate its findings with the ideas of team members, ensuring the consideration of all possible ways to decompose the problem. This is because employing a structured process to concept generation enables the team to come up with creative solutions to design concepts.
g A motor driving a 1000-W water pump has a power factor of 0.80 lagging; a second motor driving a 600-W water pump has a power factor of 0.60 lagging assuming the motors are working under 120- Vrms, 60-HZ AC. (a) When both motors working together what is combined power factor
complete Question
A motor driving a 1000-W water pump has a power factor of 0.80 lagging; a second motor driving a 600-W water pump has a power factor of 0.60 lagging assuming the motors are working under 120-Vrms, 60-HZAC. When both motors working together what is the combined power factor? If a 200-μF capacitor is connected to the above system (two motors) what is the new combined power factor?
Answer:
[tex]p.f'=0.960[/tex]
Explanation:
First motor Power [tex]P=1000W[/tex]
First motor Power factor [tex]P.f=0.80[/tex]
Second motor Power [tex]P=600W[/tex]
Second motor Power factor p.f=0.60
Voltage [tex]V=120Vrms[/tex]
Frequency [tex]F=60Hz[/tex]
Capacitor [tex]C=200\mu F[/tex]
Generally power in Var is given as
For First Motor
[tex]Q=\frac{1000}{0.8}\sqrt{1-0.8^2}[/tex]
[tex]Q=750Var[/tex]
For Second Motor
[tex]Q=\frac{600}{0.6}\sqrt{1-0.6^2}[/tex]
[tex]Q=800Var[/tex]
Generally the equation for The Reactive Power is mathematically given by
[tex]Q_c=\frac{V^2}{X_c}[/tex]
Where
[tex]X_c=\frac{1}{2 \pi fc}[/tex]
[tex]X_c=\frac{1}{2 \pi 60*200*10^{-6} }[/tex]
[tex]X_c=13.3[/tex]
Therefore
[tex]Q_c=-\frac{120^2}{13.3}\\\\Q_c=-1085.97j[/tex]
Giving
Total Power Drawn by Supply
[tex]P_t=(1000+j750)+(600+800)-j1085.97[/tex]
[tex]P_t=1600+464.03j[/tex]
Therefore
[tex]p.f'=\frac{1600}{\sqrt{1600^2+464.03^2}}[/tex]
[tex]p.f'=0.960[/tex]
deep invasion lead to???
Answer:
2
Explanation:
usjshshshjeoeieieiiekeiekeiekekekekekkske
Answer:
2
Explanation:
High separation between deep...........
______ is not a type of digital signaling technique
Answer:
Data Rate Signaling
A bird is flying in air. It is stretching its wings all along the sky and is flying in a certain direction Flying a bird is an example of: a. Composition of vector b. Collinear vector c. Addition of vector d. Multiplication of vector
Answer:
a. Composition of Vector.
Explanation:
When a bird flies in the air, it stretches its wings into air and this movement helps it move in a certain direction. This is an example of composition of vector. Air strikes the wings in opposite direction and bird wing movement helps it move against the wind.
8.50 nC of charge is uniformly distributed along a thin rod of length L = 9.20 cm, which is then bent into a semicircle. What is the magnitude of the electric field at the center of the circle
Theo Anh / Chị, để đáp ứng yêu cầu phát triển nền kinh tế thị trường định hướng Xã hội Chủ nghĩa ở Việt Nam trong bối cảnh thời đại hiện nay, cần chú trọng giải quyết những vấn đề gì ?
Plexiglas, polymethylmethacrylate (PMMA), is often used as display windows and cases in art galleries and museums. If its flame imparts 32 kW/m^2 to the surface when it burns estimate the energy release rate of a 3 m^2 square sheet (one side). Assume its vaporization temperature is 350°C. Calculate with units, provide work.
This question is incomplete, the complete question is;
Plexiglas, polymethylmethacrylate (PMMA), is often used as display windows and cases in art galleries and museums. If its flame imparts 32 kW/m^2 to the surface when it burns estimate the energy release rate of a 3 m^2 square sheet (one side). Assume its vaporization temperature is 350°C. Calculate with units, provide work.
a ⇒ [tex]q_{rad[/tex] = σT⁴, where σ = 5.67 × 10⁻¹¹ kW/m².K⁴
b ⇒ [tex]q"[/tex] = [tex]q_{flame[/tex] - [tex]q_{rad[/tex]
c ⇒ [tex]m"[/tex] = [tex]q"[/tex]/L, where L = 1.6 kJ/g
d ⇒ Q = [tex]m"[/tex] × A × ΔHc, where ΔHc is 24.9 kJ/g
Answer:
a) [tex]q_{rad[/tex] = 8.54 kW/m²
b) [tex]q"[/tex] = 23.46 kW/m²
c) [tex]m"[/tex] = 14.6625 g/m²
d) Q = 1095.2888 kJ
Explanation:
Given the data in the question;
[tex]q_{flame[/tex] = 32 kW/m²
Area; A = 3m²
vaporization temperature; T = 350°C = ( 350 + 273 )K = 623 K
Now,
a) ⇒ [tex]q_{rad[/tex] = σT⁴, where σ = 5.67 × 10¹¹ kW/m².K⁴
we substitute
[tex]q_{rad[/tex] = ( 5.67 × 10⁻¹¹ kW/m².K⁴ ) × ( 623 K)⁴
[tex]q_{rad[/tex] = ( 5.67 × 10⁻¹¹ kW/m².K⁴ ) × 150644120641 K⁴
[tex]q_{rad[/tex] = 8.54 kW/m²
b) ⇒ [tex]q"[/tex] = [tex]q_{flame[/tex] - [tex]q_{rad[/tex]
we substitute
[tex]q"[/tex] = 32 kW/m² - 8.54 kW/m²
[tex]q"[/tex] = 23.46 kW/m²
c) ⇒ [tex]m"[/tex] = [tex]q"[/tex]/L, where L = 1.6 kJ/g
we substitute
[tex]m"[/tex] = 23.46 / 1.6
[tex]m"[/tex] = 14.6625 g/m²
d) ⇒ Q = [tex]m"[/tex] × A × ΔHc, where ΔHc is 24.9 kJ/g
we substitute
Q = 14.6625 × 3 × 24.9
Q = 1095.2888 kJ
định khoản nghiệp vụ sau : tạm ứng cho nhân viên A đi công tác bằng tiền mặt 50.000
Answer:
report on a fight you have witnessed
Tech A says that the voltage regulator controls the strength of the rotor s magnetic field. Tech B says that the voltage regulator is installed between the output terminal of the alternator and the positive terminal of the battery. Who is correct?
Answer:
Voltage Regulator
Technician A is correct.
Explanation:
Technician B is not correct. The voltage regulator is not installed between the output terminal of the alternator and the positive terminal of the battery as claimed by Technician B. Technician A's opinion that the voltage regulator controls the strength of the rotor's magnetic field is correct. The computer can also be used to control the output of the alternator by controlling the field current.
You are given a pot of hot tea (350 mL) at a temperature of 85°C and a quantity of ice at -12°C. Determine the absolute minimum amount of ice you can add to the hot tea so that at equilibrium you have iced tea (ie. a very, very small amount of ice and some water). You can assume tea has the same density (1.00 g/mL) and specific heat (4190 J/kgK) as liquid water. The heat of fusion of H2O is 3.33x10^5 J/kg. The specific heat of ice is 2090 J/kgK.
Answer:
348 g
Explanation:
The heat gained by the ice equals the heat lost by the hot tea.
Also, since we require a very, very small amount of ice and some water, to still have some ice, the temperature of the hot tea has to decrease to the melting point of ice which is 0°C. So, the final temperature of the mixture is 0 °C.
So, mc(T - T') + mL = -MC(T - T") where m = mass of ice, c = specific heat of ice = 2090 J/kgK., T = final temperature of mixture = 0 °C, T' = initial temperature of ice = -12 °C, L = heat of fusion of H2O = 3.33 × 10⁵ J/kg, M = mass of hot tea = ρV where ρ = density of tea = 1.00 g/mL and V = volume of hot tea = 350 mL. So, M = ρV = 1.00 g/mL × 350 mL = 350 g = 0.350 kg, C = specific heat of tea = 4190 J/kgK and T" = initial temperature of tea = 85 °C.
Making m subject of the formula, we have
mc(T - T') + mL = -MC(T - T")
m[c(T - T') + L] = -MC(T - T")
m = -MC(T - T")/[c(T - T') + L]
substituting the values of the variables into the equation, we have
m = -MC(T - T")/[c(T - T') + L]
m = -0.350 kg × 4190 J/kgK(0 °C - 85 °C.)/[2090 J/kgK(0 °C. - (-12 °C)) + 3.33 × 10⁵ J/kg]
m = -1466.5 J/kK(- 85 °C)/[2090 J/kgK(0 °C + 12 °C)) + 3.33 × 10⁵ J/kg]
m = 124652.5 J/[2090 J/kgK(12 °C) + 3.33 × 10⁵ J/kg]
m = 124652.5 J/[25080 J/kg + 3.33 × 10⁵ J/kg]
m = 124652.5 J/358080 J/kg
m = 0.3481 kg
m = 348.1 g
m ≅ 348 g
So, the minimum amount of ice to be added is 348 g.
Troy must keep track of the amount of refrigerant he uses from a 50-pound cylinder to ensure that accurate
records are kept. He used 13 pounds on a systein for Ms. Jones and 9 pounds on a system for a commercial
client. How many pounds should he have left in the cylinder?
tof
Troy should have
pounds of refrigerant left in the cylinder.
baon naid Thamar basic
محمود احمد مجد
12
اهداء ما در
Answer:
Amount of gas still in cylinder = 28 pound
Explanation:
Given:
Amount of gas in cylinder = 50 pound
Amount of gas used in Ms. Jones system = 13 pound
Amount of gas used in client system = 9 pound
Find:
Amount of gas still in cylinder
Computation:
Amount of gas still in cylinder = Amount of gas in cylinder - Amount of gas used in Ms. Jones system - Amount of gas used in client system
Amount of gas still in cylinder = 50 - 13 - 9
Amount of gas still in cylinder = 28 pound
what is tracer lathe machine
Answer: The tracer lathe is a roughing operation for the output shaft on rear wheel drive transmissions.
Explanation:
hỗ trợ mình với được không các bạn
Answer:
Explanation:
Be bop
