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
The aluminumWhen 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
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You want a potof water to boil at 105 celcius. How heavy a
lid should youput on the 15 cm diameter pot when Patm =
101kPa?
Answer:
35.7 kg lid we put
Explanation:
given data
temperature = 105 celcius
diameter = 15 cm
Patm = 101 kPa
to find out
How heavy a lid should you put
solution
we know Psaturated from table for temperature is 105 celcius is
Psat = 120.8 kPa
so
area will be here
area = [tex]\frac{\pi }{4} d^2[/tex] ..................1
here d is diameter
put the value in equation 1
area = [tex]\frac{\pi }{4} 0.15^2[/tex]
area = 0.01767 m²
so net force is
Fnet = ( Psat - Patm ) × area
Fnet = ( 120.8 - 101 ) × 0.01767
Fnet = 0.3498 KN = 350 N
we know
Fnet = mg
mass = [tex]\frac{Fnet}{g}[/tex]
mass = [tex]\frac{350}{9.8}[/tex]
mass = 35.7 kg
so 35.7 kg lid we put
A carpenter uses a hammer to strike a nail. Approximate the hammer's weight of 1.8lbs, as being concentrated at the head, and assume that at impact the head is traveling in the -j direction. If the hammer contacts the nail at 50mph and the impact occurs over 0.023 seconds, what is the magnitude of the average force exerted by the nail of the hammer?
Answer:
The average force F exherted by the nail over the hammer is 178.4 lbf.
Explanation:
The force F exherted by the nail over the hammer is defined as:
F = |I|/Δt
Where I and Δt are the magnitude of the impact and the period of time respectively. We know that the impact can be calculated as the difference in momentum:
I = ΔP = Pf - Pi
Where Pf and Pi are the momentum after and before the impact. Recalling for the definition for momentum:
P = m.v
Where m and v are the mass and the velocity of the body respectively. Notice that final hummer's momentum is zero due to the hammers de-acelerate to zero velocity. Then the momentum variation will be expressed as:
ΔP = - Pi = -m.vi
The initial velocity is given as 50 mph and we will expressed in ft/s:
vi = 50 mph * 1.47 ft/s/mph = 73.3 ft/s
By multiplyng by the mass of 1.8 lbs, we obtain the impulse I:
|I|= |ΔP|= |-m.vi| = 1.8 lb * 73.3 ft/s = 132 lb.ft/s
Dividing the impulse by a duration of 0.023 seconds, we finally find the force F:
F = 132 lb.ft/s / 0.023 s = 5740 lb.ft/s^2
Expressing in lbf:
F = 5740 lb.ft/s^2 * 0.031 lbf/lb.ft/s^2 = 178.4 lbf
A heating cable is embedded in a concrete slab for snow melting. The heating cable is heated electrically with joule heating to provide the concrete slab with a uniform heat of 1200 W/ m2. The concrete has a thermal conductivity of 1.4 W/m⋅K. To minimize thermal stress in the concrete, the temperature difference between the heater surface (T1) and the slab surface (T2) should not exceed 16°C (2015 ASHRAE Handbook—HVAC Applications, Chapter. 51). Formulate the temperature profile in the concrete slab, and determine the thickness of the concrete slab (L) so that T1 − T2 ≤ 16°C.
Answer:
18.7 mm
Explanation:
Fourier's law for heat conduction on a plate is:
q = -k/t * ΔT
Where
q: heat conducted per unit of time and surface
k: thermal conductivity
t: thickness of the plate
ΔT: temperature difference
Rearranging:
t = -k/q * ΔT
t = -1.4/(-1200) * 16 = 0.0187 m = 18.7 mm (q is negative because it is heat leaving the plate)
The maximum thickness of the concrete slab is 18.7 mm.
When all network cables connect to one central point the network topology is typically referred to as a(n)_______
Answer:
Star
Explanation:
When all hosts are connected to a central hub it is known as a star topology.
The advantages of the star network is that it is very easy to add new devices, a failure in a host will not cause problems on the rest of the network, it is appropriate for large networks and works well under heavy loads,
The disadvantages are that a failure of the central hub brings the whole networks down and it is expensive due to the amount of cables needed.
A car is accelerated 5.5 ft/s^2. Calculate the initial velocity v, the car must have if it is to attain a final velocity v of 45 mph in a distance of 352 ft. Compute also the time t required to attain that final velocity.
Answer:
1) The initial velocity must be 22 feet/sec.
2) The time required to change the velocity is 8 seconds.
Explanation:
This problem can be solved using third equation of kinematics
According to the third equation of kinematics we have
[tex]v^{2}=u^{2}+2as[/tex]
where
'v' is the final velocity of object
'u' is the initial velocity of object
'a' is acceleration of the object
's' is the distance in which this velocity change is brought
Since the final velocity is 45 mph converting this to feet per seconds we get
45 mph =66 feet/sec
Applying Values in the above equation and solving for 'u' we get
[tex]66^{2}=u^{2}+2\times 5.5\times 352\\\\u^{2}=484\\\\\therefore u=22feet/sec[/tex]
2)
The time required to attain this velocity can be found using first equation of kinematics as
[tex]v=u+at[/tex]
where
't' is the time in which the velocity change occurs
'v', 'u', 'a' have the usual meaning
Thus applying given values we get
[tex]66=22+5.5\times t\\\\\therefore t=\frac{66-22}{5.5}=8seconds[/tex]
Applying the given values we get
A Carnot heat engine operates between 1000 deg F and 50 deg F, producing 120 BTU of work. What is the heat input to the engine?
Answer:
184.6 BTU
Explanation:
The thermal efficiency for a Carnot cycle follows this equation:
η = 1 - T2/T1
Where
η: thermal efficiency
T1: temperature of the heat source
T2: temperature of the heat sink
These temperatures must be in absolute scale:
1000 F = 1460 R
50 F = 510 R
Then
η = 1 - 510/1460 = 0.65
We also know that for any heat engine:
η = L / Q1
Where
L: useful work
Q1: heat taken from the source
Rearranging:
Q1 = L / η
Q1 = 120 / 0.65 = 184.6 BTU
In this exercise we will deal with the Carnot cycle and calculate how much heat was initially placed, so we have that:
The heat input to the engine was 184.6 BTU
How does the Carnot Cycle work?The theoretical Carnot cycle is formed by two isothermal transformations and two adiabatic transformations. One of the isothermal transformations is used for the temperature of the hot source, where the expansion process takes place, and the other for the cold source, where the compression process takes place.
The thermal efficiency for a Carnot cycle follows this equation:
[tex]\eta= 1 - T_2/T_1[/tex]
Where:
η: thermal efficiencyT1: temperature of the heat sourceT2: temperature of the heat sinkThese temperatures must be in absolute scale:
[tex]\eta = 1 - 510/1460 = 0.65[/tex]
Rearranging:
[tex]Q_1 = L / \eta\\Q_1 = 120 / 0.65 = 184.6 BTU[/tex]
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Domestic units have power input ratings of between (a) 35 W and 375 W (b) 3.5 and 37.5 W (c) 350W and 375 W
Answer:
The answer is "b" 3.5W and 37.5w.
Explanation:
Domestic use normally refers to refrigeration equipment that is not subject to high thermal loads, for example, refrigerators should only extract heat from food, and domestic air conditioners should only extract heat from small spaces and a number reduced people.
A spherical tank is being designed to hold 10 moles of carbon dioxide gas at an absolute pressure of 5 bar and a temperature of 80°F. What diameter spherical tank should be used? The molecular weight of methane is 44 g/mole.
Answer:
r=0.228m
Explanation:
The equation that defines the states of a gas according to its thermodynamic properties is given by the general equation of ideal gases
PV=nRT
where
P=pressure =5bar=500.000Pa
V=volume
n=moles=10
R = universal constant for ideal gases = 8.31J / (K.mol)
T=temperature=80F=299.8K
solvig For V
V=(nRT)/P
[tex]V=(\frac{(10)(8.31)(299.8)}{500000} )\\V=0.0498m^3[/tex]
we know that the volume of a sphere is
[tex]V=\frac{4\pi r^3}{3} \\[/tex]
solving for r
[tex]r=\sqrt[3]{ \frac{3 V}{4\pi } }[/tex]
solving
[tex]r=\sqrt[3]{ \frac{3 (0.049)}{4\pi } }\\r=0.228m[/tex]
An electric motor supplies 200 N·m of torque to a load. What is the mechanical power supplied to the load if the shaft speed is 1000 rpm? Express the result in watts and horsepower.
Answer:
power = 20943.95 watts
power = 28.086 horsepower
Explanation:
given data
torque = 200 N
speed = 1000 rpm
to find out
What is the mechanical power in watts and horse power
solution
we know that mechanical power formula that is
power = torque × speed ...................1
here we have given both torque and speed
we know speed = 1000 rpm = [tex]\frac{2* \pi *1000}{60}[/tex] = 104.66 rad/s
so put here value in equation 1
power = 200 × 104.719
power = 20943.95 watts
and
power = [tex]\frac{20943.95}{745.7}[/tex]
power = 28.086 horsepower
Answer:
Part 1) Power required for motor = 20944 watts.
Part 2) Power required for motor in Horsepower equals = 28.075H.P
Explanation:
Power is defined as the rate of consumption of energy. For rotational motion power is calculated as
[tex]Power=Torque\times \omega[/tex]
where,
[tex]\omega [/tex] is the angular speed of the motor.
Since the rotational speed of the motor is given as 1000 rpm, the angular speed is calculated as
[tex]\omega =\frac{N}{60}\times 2\pi[/tex]
where,
'N' is the speed in rpm
Applying the given values we get
[tex]\omega =\frac{1000}{60}\times 2\pi=104.72rad/sec[/tex]
hence the power equals
[tex]Power=200\times 104.72=20944Watts[/tex]
Now since we know that 1 Horse power equals 746 Watts hence 20944 Watts equals
[tex]Power_{H.P}=\frac{20944}{746}=28.075H.P[/tex]
What is a shearing stress? Is there a force resulting from two solids in contact to which is it similar?
Answer:
Shearing stresses are the stresses generated in any material when a force acts in such a way that it tends to tear off the material.
Generally the above definition is valid at an armature level, in more technical terms shearing stresses are the component of the stresses that act parallel to any plane in a material that is under stress. Shearing stresses are present in a body even if normal forces act on it along the centroidal axis.
Mathematically in a plane AB the shearing stresses are given by
[tex]\tau =\frac{Fcos(\theta )}{A}[/tex]
Yes the shearing force which generates the shearing stresses is similar to frictional force that acts between the 2 surfaces in contact with each other.
A piston-cylinder apparatus has a piston of mass 2kg and diameterof
10cm resting on a body of water 30 cm high atmospheric pressureis
101 kpa, and the temperature of water 50 degrees Celsius. What is
the mass of water in the container.
Answer:
M =2.33 kg
Explanation:
given data:
mass of piston - 2kg
diameter of piston is 10 cm
height of water 30 cm
atmospheric pressure 101 kPa
water temperature = 50°C
Density of water at 50 degree celcius is 988kg/m^3
volume of cylinder is [tex]V = A \times h[/tex]
[tex]= \pi r^2 \times h[/tex]
[tex]= \pi 0.05^2\times 0.3[/tex]
mass of available in the given container is
[tex]M = V\times d[/tex]
[tex] = volume \times density[/tex]
[tex]= \pi 0.05^2\times 0.3 \times 988[/tex]
M =2.33 kg
The flow curve for a certain metal has parameters: strain-hardening
exponent is 0.22 and strength coefficient is54,000
lb/in2 . Determine:
a) the true stress at a true strain = 0.45
b) the true strain at a true stress = 40,000
lb/in2.
Answer:(a)[tex]45,300.24 lb/in/^2[/tex]
(b)0.255
Explanation:
Given
Strain hardening exponent(n)=0.22
Strength coefficient(k)[tex]=54000 lb/in/^2[/tex]
and we know
[tex]\sigma =k\left ( \epsilon \right )^n[/tex]
where
[tex]sigma =true\ stress[/tex]
[tex]\epsilon =true\ strain[/tex]
(a)True strain=0.45
[tex]\sigma =54000\times 0.45^{0.22}[/tex]
[tex]\sigma =45,300.24 lb/in^2[/tex]
(b)true stress[tex]=40,000 lb/in^2[/tex]
[tex]40000=54000\times \epsilon ^{0.22}[/tex]
[tex]\epsilon ^{0.22}=0.7407[/tex]
[tex]\epsilon =0.7407^{4.5454}=0.255[/tex]
Yield and tensile strengths and modulus of elasticity . with increasing temperature. (increase/decrease/independent)
Answer:
Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.
Explanation:
The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.
Mathematically we can write,
[tex]E(t)=E_o[1-a\frac{T}{T_m}][/tex]
where,
E(t) is the modulus of elasticity at any temperature 'T'
[tex]E_o[/tex] is the modulus of elasticity at absolute zero.
[tex]T_{m}[/tex] is the mean melting point of the material
Hence we can see that with increasing temperature modulus of elasticity decreases.
In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.
A large water jet with a discharge of 2m^3 /s rises 90m above the ground. The exit nozzle diameter to achieve this must be. (a) 0.246m (b) 0.318m (c) 1.3m (d) 0.052m (e) None of the above
Answer:
The correct answer is option 'a': 0.046 meters.
Explanation:
We know that the exit velocity of a jet of water is given by Torricelli's law as
[tex]v=\sqrt{2gh}[/tex]
where
'v' is velocity of head
'g' is acceleration due to gravity
'h' is the head under which the jet falls
Now since the jet rises to a head of 90 meters above ground thus from conservation of energy principle it must have fallen through a head of 90 meters.
Applying the values in above equation we get the exit velocity as
[tex]v=\sqrt{2\times 9.81\times 90}=42.02m/s[/tex]
now we know the relation between discharge and velocity as dictated by contuinity equation is
[tex]Q=V\times Area[/tex]
Applying values in the above equation and solving for area we get
[tex]Area=\frac{Q}{v}=\frac{2}{42.02}=0.0476m^{2}[/tex]
The circular area is related to diameter as
[tex]Area=\frac{\pi D^{2}}{4}\\\\\therefore D=\sqrt{\frac{4\cdot A}{\pi }}=\sqrt{\frac{4\times 0.0476}{\pi }}=0.246m[/tex]
Thus the diameter of the nozzle is 0.246 meters
What is mass flow of 200 lbm/min in kg/sec?
Answer:
Mass flow in kg /sec will be 1.511 kg/sec
Explanation:
We have given mass flow = 200 lbm/min
We have to convert lbm/min into kg /sec
We know that 1 lbm = 0.4535 kg
So for converting lbm to kg we have multiply with 0.4535
So 200 lbm = 200×0.4535 =90.7 kg
Now we know that 1 minute = 60 sec
So 200 lbm/min [tex]=\frac{200\times 0.4535kg}{60sec}=1.511kg/sec[/tex]
So mass flow in kg /sec will be 1.511 kg/sec
If the dry-bulb temperature is 95°F and the wet-bulb temperature is also 78°F, what is the relative humid- ity? What is the dew point? What is the humidity ratio? What is the enthalpy?
Answer:
Relative humidity 48%.
Dew point 74°F
humidity ratio 118 g of moisture/pound of dry air
enthalpy 41,8 BTU per pound of dry air
Explanation:
You can get this information from a Psychrometric chart for water, like the one attached.
You enter the chart with dry-bulb and wet-bulb temperatures (red point in the attachment) and following the relative humidity curves you get approximately 48%.
To get the dew point you need to follow the horizontal lines to the left scale (marked with blue): 74°F
for the humidity ratio you need to follow the horizontal lines but to the rigth scale (marked with green): 118 g of moisture/pound of dry air
For enthalpy follow the diagonal lines to the far left scale (marked with yellow): 41,8 BTU per pound of dry air
A tow truck is using a cable to pull a car up a 15 degree hill. If the car weighs 4000 lbs and the cable has a diameter of .75 inches, find the stress in the cable when the truck comes to a stop while on the hill. Ignore friction between the car and the pavement.
Answer:40.603 MPa
Explanation:
Given
Car weighs (m)4000 lbs [tex]\approx 1814.37 kg[/tex]
diameter of cable(d)[tex]=0.75 in.\approx 19.05 mm[/tex]
Hill angle [tex]\theta =15^{\circ}[/tex]
Now
tension in cable will bear the weight of car acting parallel to rope which is [tex]mgsin\theta [/tex]
Thus
[tex]T=mgsin\theta [/tex]
[tex]T=1814.37\times 9.81\times sin(15)=11,574.45 N[/tex]
T=11.57 kN
thus stress([tex]\sigma [/tex])=[tex]\frac{T}{A}[/tex]
where A=cross section of wire[tex]=285.059 mm^2[/tex]
[tex]\sigma =\frac{11574.45}{285.059}=40.603 MPa[/tex]
a baseball is thrown downward from a 50 ft tower with an
initialspeed of 18 ft/s.
what is the speed when the ball hits the ground and the time
oftrave?
Answer:
Final velocity will be 36.11 m/sec
Time required by ball to heat the ground = 1.297 sec
Explanation:
We have given height = 50 feet
Initial velocity u = 18 ft/sec
Acceleration due to gravity [tex]g=32.8ft/sec^2[/tex]
We have to find final velocity v
From third equation of motion
[tex]v^2=u^2+2gh[/tex]
So [tex]v^2=18^2+2\times 32.8\times 50=3543.82[/tex]
v = 59.53m/sec
From first equation of motion we know that
v= u+gt
So [tex]59.53=18+32.8\times t[/tex]
t = 1.297 sec
A brittle intermetallics specimen is tested with a bending test. The specimen's width 0.45 in and thickness 0.20 in. The length of the specimen between supports 2.5 in. Determine the transverse rupture strength if failure occurs at a load 1200 lb.
Answer:
250 kpsi
Explanation:
Given:
Width of the specimen, w = 0.45 in
Thickness of the specimen, t = 0.20 in
length of the specimen between supports, L = 2.5 in
Failure load, F = 1200 lb
Now,
The transverse rupture strength [tex]\sigma_t=\frac{1.5FL}{wt^2}[/tex]
on substituting the respective values, we get
[tex]\sigma_t=\frac{1.5\times1200\times2.5}{0.45\times0.2^2}[/tex]
or
[tex]\sigma_t=250,000\ psi\ =\ 250 kpsi[/tex]
The information on a can of pop indicates that the can contains 360 mL. The mass of a full can of pop is 0.369 kg, while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at Express your results in SI units.
Answer:
Specific weight of the pop, [tex]w_{s} = 8619.45 N/m^{3}[/tex]
Density of the pop, [tex]\rho_{p} = 8790.76 kg/m^{3}[/tex]
[tex]g_{s} = 8.79076[/tex]
[tex]w_{w} = 9782.36 N/m^{3}[/tex]
Given:
Volume of pop, V = 360 mL = 0.36 L = [tex]0.36\times 10^{-3} m^{3}[/tex]
Mass of a can of pop , m = 0.369 kg
Weight of an empty can, W = 0.153 N
Solution:
Now, weight of a full can pop, W
W' = mg = [tex]0.369\times 9.8 = 3.616 N[/tex]
Now weight of the pop in can is given by:
w = W' - W = 3.616 - 0.513 = 3.103 N
Now,
The specific weight of the pop, [tex]w_{s} = \frac{weight of pop}{volume of pop}[/tex]
[tex]w_{s} = \frac{3.103}{0.36\times 10^{- 3}} = 8619.45 N/m^{3}[/tex]
Now, density of the pop:
[tex]\rho_{p} = \frac{w_{s}}{g}[/tex]
[tex]\rho_{p} = \frac{86149.45}{9.8} = 8790.76 kg/m^{3}[/tex]
Now,
Specific gravity, [tex]g_{s} = \frac{\rho_{p}}{density of water, \rho_{w}}[/tex]
where
[tex]g_{s} = \frac{8790.76}{1000} = 8.79076[/tex]
Now, for water at [tex]20^{\circ}c[/tex]:
Specific density of water = [tex]998.2 kg/m^{3}[/tex]
Specific gravity of water = [tex]0.998 kg/m^{3}[/tex]
Specific weight of water at [tex]20^{\circ}c[/tex]:
[tex]w_{w} = \rho_{20^{\circ}}\times g = 998.2\times 9.8 = 9782.36 N/m^{3}[/tex]
If a plus sight of 12.03 ft is taken on BM A, elevation 312.547 ft, and a minus sight of 5.43 ft is read on point X, calculate the HI and the elevation of point X.
Answer:
Therefore, height of instrument is 324.577 ft
Therefore, elevation of point x is 330 m
Explanation:
Given that
Plus sight on BM = 12.03 ft
Minus sight is = 5.43 ft
Elevation = 312.547 ft
Height of instrument is H.I
H.I = elevation on bench mark + plus sight
= 312.547 + 12.03 = 324.577 ft
Therefore, height of instrument is 324.577 ft
Elevation at point x is = H.I - minus sight
= 324.577 - (- 5.43)
= 330.00 m
Therefore, elevation of point x is 330 m
a piston moves a 25kg hammerhead vertically down 1m from rest to a
velocity of 50m/s in a stamping machine.
what is the total change in energy of the hammerhead?
Answer:
Total change in energy = 31 KJ.
Explanation:
Mass m=25 kg
Height h = 1 m
Initial velocity = 0
Final velocity = 50 m/s
Energy at initial condition
[tex]E_1=mgh+\dfrac{1}{2}mv^2[/tex]
[tex]E_1=25\times 10\times 1+0[/tex]
[tex]E_1=250\ J[/tex]
Energy at final condition
[tex]E_2=0+\dfrac{1}{2}\times 25\times 50^2[/tex]
[tex]E_2=31250\ J[/tex]
So the change in energy = 31250 -250 J
The total change in energy = 31000 J
Answer:
Change in energy will be 31.25 KJ
Explanation:
We have given mass of the piston = 25 kg
Initial velocity u = 0 m/sec
Final velocity v = 50 m/sec
Kinetic energy is given by [tex]KE=\frac{1}{2}mv^2[/tex]
We have to find the change in energy
So change in energy = final KE - initial KE
[tex]\Delta E=\frac{1}{2}m(v^2-u^2)[/tex]
[tex]\Delta E=\frac{1}{2}25\times (50^2-0^2)=31250j=31.25KJ[/tex]
So change in energy will be 31.25 KJ
A spacecraft component occupies a volume of 8ft^3 and weighs 25 lb at a location where the acceleration of gravity is 31.0 ft/s^2. Determine its weight, in pounds, and its average density, in lbm/ft^3, on the moon, where g=5.57 ft/s^2.
Answer:
The weight is 4.492 lb
The density is [tex]0.5614 lbm/ft^{3}[/tex]
Solution:
As per the question:
Volume of spacecraft component, [tex]V_{s} = 8ft^{3}[/tex]
Mass of the component of spacecraft, [tex]m_{s} = 25 lb[/tex]
Acceleration of gravity at a point on Earth, [tex]g_{E} = 31.0 ft/s^{2}[/tex]
Acceleration of gravity on Moon, [tex]g_{M} = 5.57 ft/s^{2}[/tex]
Now,
The weight of the component, [tex]w_{c} = m_{c}\times \frac{g_{M}}{g_{E}}[/tex]
[tex]w_{c} = 25\times \frac{5.57}{31.0}[/tex]
[tex]w_{c} = 4.492 lb[/tex]
Now,
Average density, [tex]\rho_{avg}[/tex]
[tex]\rho_{avg} = \farc{w_{s}}{V_{s}}[/tex]
[tex]\rho_{avg} = \farc{4.492}{8} = 0.5614 lbm/ft^{3}[/tex]
Answer:
density=3.125pounds/ft^3
weight=4.35lbf
Explanation:
Density is a property of matter that indicates how much mass a body has in a given volume.
It is given by the following equation.
ρ=m/v (1)
where
ρ=density
m=mass
v=volume
The weight on the other hand is the force which the earth (or the moon) attracts to a body with mass, this force is given by the following equation
W=mg (2)
W=weight
m=mass
g=gravity
to solve this problem we have to calculate the mass of the component using the ecuation number 2
W=mg
m=w/g
w=25lbf=804.35pound .ft/s^2
g=32.2ft/s^2
m=804.35/32.2=25 pounds
density
ρ=m/v (1)
ρ=25pounds/8ft^3
ρ=3.125pounds/ft^3 =density
weight in the moon
W=mg
W=(25pounds)(5.57ft/s^2.)=139.25pound .ft/s^2=4.35lbf
What is the origin of the "horsepower"? Why would anyone wish to express power in the unit of horsepower? How many watts are in one horsepower?
Answer:
Horsepower unit was first time used by James Watt in 1782. The story refers that James Watt uses worked with pony to charge coal from the mines. According to that story, he have the need of a unit to measure the force from one of this animals. He founds that they can move 22.000 lbs per minute, so he (arbitrarily) increase this measure in 50% been the unit Horsepower in 33.000 lb/feet per minute.
Explanation:
This measure unit can measure "work" or "force". In the SI correspond to move up 75 Kg, to 1 meter high, in one second.
1 HP = (330 lb) x (100 feet)/1min = 33000 lb x feet/min
Why would anyone wish to express power in HP?
Is a practical unit, because reduce the amount of digits in a specific value. Also, his use is very spread specially in mechanical applications.
1 HP= 746 W (0,746 kW).
As shown, a load of mass 10 kg is situated on a piston of diameter D1 = 140 mm. The piston rides on a reservoir of oil of depth h1 = 42 mm and specific gravity S = 0.8. The reservoir is connected to a round tube of diameter D2 = 5 mm and oil rises in the tube to height h2. Find h2. Assume the oil in the tube is open to atmosphere and neglect the mass of the piston.
Use Pascal's law and the hydrostatic pressure formula to solve for the height of the oil column, considering the specific gravity of the oil and the mass applied on the piston.
Explanation:The question requires the application of principles from fluid mechanics to calculate the height h2 to which an oil rises in a tube, using the specific gravity of the oil and the mass of a load on a piston. Assuming the system is in equilibrium and by using Pascal's law, the pressure applied by the mass onto the piston is transmitted equally throughout the fluid. Therefore, the pressure exerted by the mass on the larger piston is balanced by the hydrostatic pressure of the oil column of height h2 in the smaller tube.
To solve for h2, we can use the formula for hydrostatic pressure P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column. Given the specific gravity S of the oil, we can find its density (ρ) by multiplying S by the density of water (1000 kg/m³). Then, by setting the hydrostatic pressure of the oil column equal to the pressure exerted by the mass and solving for h2, we find the height of the oil in the tube.
Define drag and lift forces.
Explanation:
Drag is the force which is generated parallel and also in opposition to direction of the travel for the object which is moving through the fluid.
Lift is the force which is generated perpendicular to direction of the travel for the object moving through the fluid.
Both of the two forces which are the lift and the drag force act through center of the pressure of object.
What is the range of a 32-bit unsigned integer?
Answer:
0 to 4294967295
Explanation:
Unsigned integers have only positive numbers and zero. Their range goes from zero to (2^n)-1.
In the case of a 32-bit unsigned integer this would be.
(2^32) - 1 = 4294967296 - 1 = 4294967295
So the range goes from 0 to 4294967295 or form zero to about 4.3 billion.
The minus one term is because the zero takes one of the values.
Higher molecular weight results in worst mechanical properties. a)-True b)- false?
Answer:
b)False
Explanation:
Higher molecular weight will increase the mechanical properties because if molecular weight is more then it will require more energy break the molecule .It means that mechanical properties is also improve.
Higher molecular weight will also improve the resistance to corrosion and reduce the chances of material from oxidation.
Higher molecular weight will also increase the viscosity of material and reduce the fluidity.
Various factors to be considered in deciding the factor of safety?
Answer with Explanation:
There are various factors that needed to be taken into account while deciding the factor of safety some of which are summarized below as:
1) Importance of the structure: When we design any structure different structures have different importance in our society. Take an example of hospital, in case a natural disaster struck's a place the hospital should be the designed to withstand the disaster as it's role in the crisis management following a disaster is well understood. Thus while designing it we need it to have a higher factor of safety against failure when compared to a local building.
2) Errors involved in estimation of strength of materials: when we design any component of any machine or a structure we need to have an exact idea of the behavior of the material and know the value of the strength of the material. But many materials that we use in structure such as concrete in buildings have a very complex behavior and we cannot estimate the strength of the concrete absolutely, thus we tend to decrease the strength of the concrete more if errors involved in the estimation of strength are more to give much safety to the structure.
3) Variability of the loads that may act on the structure: If the loads that act on the structure are highly variable such as earthquake loads amd dynamic loads then we tend to increase the factor of safety while estimating the loads on the structure while designing it.
4) Economic consideration: If our project has abundant funds then we can choose a higher factor of safety while designing the project.
An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C, what will be the percentage change in its volume? If the vat has a diameter of 2 m, how much will the water level rise due to this temperature increase?
Answer:
percentage change in volume is 2.60%
water level rise is 4.138 mm
Explanation:
given data
volume of water V = 500 L
temperature T1 = 20°C
temperature T2 = 80°C
vat diameter = 2 m
to find out
percentage change in volume and how much water level rise
solution
we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure
and we know that is also in term of change in density also
so
E = [tex]-\frac{dp}{dV/V}[/tex] ................1
And [tex]-\frac{dV}{V} = \frac{d\rho}{\rho}[/tex] ............2
here ρ is density
and we know ρ for 20°C = 998 kg/m³
and ρ for 80°C = 972 kg/m³
so from equation 2 put all value
[tex]-\frac{dV}{V} = \frac{d\rho}{\rho}[/tex]
[tex]-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}[/tex]
dV = 0.0130 m³
so now % change in volume will be
dV % = [tex]-\frac{dV}{V}[/tex] × 100
dV % = [tex]-\frac{0.0130}{500*10^{-3} }[/tex] × 100
dV % = 2.60 %
so percentage change in volume is 2.60%
and
initial volume v1 = [tex]\frac{\pi }{4} *d^2*l(i)[/tex] ................3
final volume v2 = [tex]\frac{\pi }{4} *d^2*l(f)[/tex] ................4
now from equation 3 and 4 , subtract v1 by v2
v2 - v1 = [tex]\frac{\pi }{4} *d^2*(l(f)-l(i))[/tex]
dV = [tex]\frac{\pi }{4} *d^2*dl[/tex]
put here all value
0.0130 = [tex]\frac{\pi }{4} *2^2*dl[/tex]
dl = 0.004138 m
so water level rise is 4.138 mm