Answer:
[tex]P= 60.5 \frac{psia ft^3}{s} *\frac{1 Btu}{5.404 psia ft^3} *\frac{1 hp}{0.7068 Btu/s}= 15.839 hp[/tex]
Explanation:
Notation
For this case we have the following pressures:
[tex] p_1 = 15 psia[/tex] initial pressure
[tex]p_2 = 70 psia[/tex] final pressure
[tex] V^{*} = 1.1 ft^3/s[/tex] represent the volumetric flow
[tex] rho[/tex] represent the density
[tex]m^{*}[/tex] represent the mass flow
Solution to the problem
From the definition of mass flow we have the following formula:
[tex] m^{*} = \rho V^{*}[/tex]
For this case we can calculate the total change is the sytem like this:
[tex] \Delta E= \frac{p_2 -p_1}{\rho}[/tex]
Since we just have a change of pressure and we assume that all the other energies are constant.
The power is defined as:
[tex] P = m^* \Delta E[/tex]
And replacing the formula for the change of energy we got:
[tex]P = m V^* \frac{p_2 -p_1}{\rho} = V^* (p_2 -p_1)[/tex]
And replacing we have this:
[tex] P= (70-15) psia * 1.1 \frac{ft^3}{s} =60.5 \frac{psia ft^3}{s}[/tex]
And we can convert this into horsepower like this:
[tex]P= 60.5 \frac{psia ft^3}{s} *\frac{1 Btu}{5.404 psia ft^3} *\frac{1 hp}{0.7068 Btu/s}= 15.839 hp[/tex]
A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air resistance, so the brick is in free fall. (a) How tall, in meters, is the building? (b) What is the magnitude of the brick’s velocity just before it reaches the ground? (c) Sketch ay-t, vy-t, and y-t graphs for the motion of the brick.
Answer: Height of building = 17.69m, velocity of brick = 18.6m/s
Explanation: From the question, the body has a zero initial speed, thus initial velocity (u) all through the motion is zero.
By ignoring air resistance makes it a free fall motion thus making it to accelerate constantly with a value of [tex]a = 9.8m/s^{2}[/tex].
Time taken to fall = 1.90s
a)
thus the height of the building is calculated using the formulae below
[tex]H = ut + \frac{1}{2} gt^{2}[/tex]
but u = 0 , hence
[tex]H = \frac{1}{2} gt^{2}[/tex]
[tex]H = \frac{1}{2} *9.8* 1.9^{2} \\\\H = 17.69m[/tex]
b)
to get the value of velocity (v) as the brick hits the ground, we use the formulae below
[tex]v^{2} = u^{2} + 2aH[/tex]
but u= 0, hence
[tex]v^{2} = 2gH\\[/tex]
[tex]v^{2} = 2 * 9.8 * 17.69\\v = \sqrt{2 *9.8* 17.69} \\v = 18.62m/s[/tex]
find the attachment in this answer for the accleration- time graph, velocity- time graph and distance time graph
a. The height of building, in meters, is equal to 17.69 meters.
b. The magnitude of the brick’s velocity just before it reaches the ground is equal to 18.72 m/s.
Given the following data:
Time = 1.90 secondsInitial velocity = 0 m/sWe know that acceleration due to gravity (a) for an object in free fall is equal to 9.8 meter per seconds square.
a. To determine the height of building, in meters, we would use the second equation of motion:
Mathematically, the second equation of motion is given by the formula;
[tex]S = ut + \frac{1}{2} at^2[/tex]
Where:
S is the distance covered.u is the initial velocity.a is the acceleration.t is the time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]S = 0(1.90) + \frac{1}{2} \times 9.8 \times 1.90^2\\\\S = 0 + 4.9 \times 3.61[/tex]
Distance, S = 17.69 meters.
b. To determine the magnitude of the brick’s velocity just before it reaches the ground, we would use the third equation of motion;
[tex]V^2 = U^2 + 2aS[/tex]
Where:
V is the final speed.U is the initial speed.a is the acceleration.S is the distance covered.
Substituting the given parameters into the formula, we have;
[tex]V^2 = 0^2 + 2 \times 9.8 \times 17.69\\\\V^2 = 350.26\\\\V = \sqrt{350.26}[/tex]
V = 18.72 m/s
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A protester carries his sign of protest, starting from the origin of an xyz coordinate system, with the xy plane horizontal. He moves 70 m in the negative direction of the x axis, then 19 m along a perpendicular path to his left, and then 21 m up a water tower. In unit-vector notation, what is the displacement of the sign from start to end
Answer:
[tex]-70\hat{i} m+19\hat{j}m+21\hat{k} m[/tex]
Explanation:
We are given that
Displacement along x- axis =-[tex]70\hat{i}[/tex] m
Displacement along y-axis=[tex]19\hat{j}[/tex]m
Displacement along z-axis=[tex]21\hat{k} m[/tex]
Where [tex]\hat{i},\hat{j},\hat{k}[/tex] are unit vector along x, y and z-axis
We have to find the displacement from start to end in unit vector notation.
Total displacement from start to end=Displacement along x-axis+displacement along y-axis+displacement along z- axis
Total displacement from start to end=[tex]-70\hat{i} m+19\hat{j}m+21\hat{k} m[/tex]
Hence, the displacement of the sign from start to end=[tex]-70\hat{i} m+19\hat{j}m+21\hat{k} m[/tex]
An Atwood machine is constructed using a disk of mass 2 kg and radius 24.8 cm. What is the acceleration of the system? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s 2 .
Answer:
a = (m₂-m₁) / (m₂ + m₁ + ½ m)
a = (m₂-m₁) / (m₂ + m₁ + 1)
Explanation:
An Atwood machine consists of two masses of different m1 and m2 value that pass through a pulley, in this case with mass. Let's use Newton's second law for this problem.
Assume that m₂> m₁, so the direction of descent of m₂ is positive, this implies that the direction of ascent of m₁ is positive
Equation of the side of m₁
T₁ - W₁ = m₁ a
Equation of the side of m₂
W₂ - T₁ = m₂ a
Mass pulley equation m; by convention the counterclockwise rotation is positive
τ = I α
T₁ R - T₂ R = I (-α)
The moment of inertia of a disk is
I = ½ m R²
Angular and linear acceleration are related
a = α R
α = a / R
The rotation is clockwise, so it is negative
We replace
(T₁ –T₂) R = ½ m R² (-a / R)
T₁ -T₂ = - ½ m a
Let's write our three equations together
T₁ - m₁ g = m₁ a
m₂ g - T₂ = m₂ a
T₁ –T₂ = -½ m a
Let's multiply the last equation by (-1) and add
m₂ g - m₁ g = m₂ a + m₁ a + ½ m a
a = (m₂-m₁) / (m₂ + m₁ + ½ m)
calculate
a = (m₂ - m₁)/ (m₁ +m₂ + 1)
Based on the calculations, the acceleration of this system is equal to 0.58 [tex]m/s^2[/tex].
Given the following data:
Mass of disk = 2 kg.Radius of disk = 24.8 cm.Acceleration of gravity = 9.8 [tex]m/s^2[/tex]How to calculate the acceleration of the system.First of all, we would determine the moment of inertia of this disk by using this formula:
[tex]I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 2 \times (0.248)^2\\\\I=0.0615\;Kgm^2[/tex]
Next, we would use a free body diagram to determine the tensional forces acting on the disk by applying Newton's Second Law of Motion as follows:
For the first force:
[tex]F_1g-F_{T1}=m_1a\\\\m_1g-F_{T1}=m_1a\\\\1.61(9.8)-F_{T1}=1.61a\\\\15.8-F_{T1}=1.61a\\\\F_{T1}=15.8-1.61a[/tex]
For the second force:
[tex]F_{T2}-F_2g=m_2a\\\\F_{T2}-m_2g=m_2a\\\\F_{T2}-1.38(9.8)=1.38a\\\\F_{T2}-13.5=1.38a\\\\F_{T2}=13.5+1.38a[/tex]
For the torque, we have:
[tex]\sum T =I\alpha \\\\F_{T1}r-F_{T2}r=I\alpha\\\\(F_{T1}-F_{T2})r=I\alpha\\\\(15.8-1.61a-[13.5+1.38a])0.248=0.0615 \times \frac{a}{0.248} \\\\(15.8-1.61a-13.5-1.38a)0.248=0.0615 \times \frac{a}{0.248} \\\\(2.3-2.99a)0.248=0.0615 \times \frac{a}{0.248}\\\\0.5704-0.7415a=\frac{0.0615a}{0.248}\\\\0.1415-0.1839a=0.0615a\\\\0.1839a+0.0615a=0.1415\\\\0.2454a=0.1415\\\\a=\frac{0.1415}{0.2454}[/tex]
Acceleration, a = 0.58 [tex]m/s^2[/tex]
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A point charge is placed at the center of a spherical Gaussian surface. The electric flux is changed: A. if the sphere is replaced by a cube of the same volume Explain B. if the sphere is replaced by a cube of one-tenth the volume C. if the point charge is moved off center, but still inside the original sphere D. if the point charge is moved to just outside the sphere E. if a second point charge is placed just outside the sphere
Answer:
Explanation:
Point charge is Placed at the center of a spherical Gaussian surface and according to Gauss law electric flux through Gaussian surface is given by [tex]\frac{1}{\epsilon _0}[/tex] times the charge enclosed
where [tex]\epsilon _0=[/tex]permittivity of free space
So unless we take charge outside of the surface flux will not change for the Gaussian surface
So Statement A,B,C and E are false
Option D is correct because we take the charge outside of the Gaussian surface
The electric flux depends on the charge enclosed by the Gaussian surface and the location of the charges. If the volume of the surface changes, but the charge remains inside, the flux remains the same. If the charge is moved off-center or outside the surface, or if a second charge is added outside the surface, the flux will change.
Explanation:The electric flux depends on the charge enclosed by the Gaussian surface. So, if the sphere is replaced by a cube of the same volume with the charge at the center, the electric flux will not change. However, if the cube is one-tenth the volume of the original sphere, the electric flux will decrease by a factor of 10, since the charge is still at the center.
If the point charge is moved off center, but still inside the original sphere, the electric flux will not change, as long as the charge remains enclosed by the Gaussian surface.
If the point charge is moved to just outside the sphere, the electric flux will increase, as some electric field lines are now passing through the Gaussian surface.
If a second point charge is placed just outside the sphere, the electric flux through the Gaussian surface will increase even more, as the electric field lines from both charges are now passing through the surface.
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A tuner first tunes the A string very precisely by matching it to a 440 Hz tuning fork. She then strikes the A and E strings simultaneously and listens for beats between the harmonics. What beat frequency between higher harmonics indicates that the E string is properly tuned
Answer:
The beat frequency is 2 Hz.
Explanation:
Given that,
Frequency of A = 440 Hz
Frequency of E = 659 Hz
Suppose, piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note A should have a frequency of 440 Hz and the note E should be at 659 Hz.
We need to calculate the third harmonic of A
Using formula of harmonic
[tex]f_{3}=n\times f[/tex]
Put the value into the formula
[tex]f_{3} =3\times440[/tex]
[tex]f_{3}=1320\ Hz[/tex]
We need to calculate the second harmonic of E
Using formula of harmonic
[tex]f_{2}=n\times f[/tex]
Put the value into the formula
[tex]f_{2} =2\times659[/tex]
[tex]f_{2}=1318\ Hz[/tex]
We need to calculate the beat frequency
Using formula of beat frequency
[tex]f_{b}=f_{3}-f_{2}[/tex]
Put the value into the formula
[tex]f_{b}=1320-1318[/tex]
[tex]f_{b}=2\ Hz[/tex]
Hence, The beat frequency is 2 Hz.
The beat frequency indicating proper tuning of the E string relative to the A string tuned to 440 Hz should be zero. This corresponds to a situation where the harmonics of both strings match exactly, with the second harmonic of the E string (659.26 Hz) aligning with the third harmonic of the A string (660 Hz), reducing the beat frequency to zero.
Explanation:To determine the beat frequency that indicates proper tuning for the E string of a guitar, we must consider the fundamental frequencies of the A and E strings. The A string is tuned to 440 Hz. The E string above it should be tuned to a fundamental frequency of 329.63 Hz.
When the A and E strings are played together, harmonics of these frequencies are heard. The second harmonic of the E string would be 2 * 329.63 Hz = 659.26 Hz. This is very close to the 660 Hz, which is the third harmonic of the A string (3 * 440 Hz = 1320 Hz). For the strings to resonate without beats, the harmonics should match as closely as possible. Any difference in these harmonic frequencies would cause beats corresponding to the frequency difference.
Therefore, if the second harmonic of the E string (659.26 Hz) matches the third harmonic of the A string (660 Hz) closely, there will be no beats. In this ideal case, the beat frequency would be 660 Hz - 659.26 Hz = 0.74 Hz. However, for the E string to be perfectly in tune, we would want the beat frequency to be zero, indicating no difference between the expected harmonic frequency and the actual harmonic frequency.
A hydraulic turbine is used to generate power by using the water in a dam. The elevation difference between the free surfaces upstream and downstream of the dam is 131 m. The water is supplied to the turbine at a rate of 201 kg/s. If the shaft power output from the turbine is 234 kW, the efficiency of the turbine is_________.
Answer:
0.906
Explanation:
Let g = 9.81 m/s2. We can calculate the rate of change in potential energy when m = 201kg of water is falling down a distance of h = 131m per second
[tex]\dot{E_p} = \dot{m}gh = 201*9.81*131 = 258307 J/s (W) = 258.307 kW[/tex]
So the efficiency of the water turbine is the ratio of output power over input power:
[tex]\frac{234}{258.307} = 0.906[/tex]
This question involves the concepts of efficiency and potential energy.
The efficiency of the turbine is "90.6%".
The efficiency of the turbine can be given by the following formula:
[tex]Efficiency = \frac{Turbine\ shaft\ power}{Potential\ Energy\ power\ of\ Water}\\\\[/tex]
where,
Turbine shaft power = 234 KW
Potential energy power of water = mgh/t = (201 kg/s)(9.81 m/s²)(131 m)
Potential energy power of water = 258307.11 W = 258.31 KW
Therefore,
[tex]Efficiency=\frac{234\ KW}{258.31\ KW}[/tex]
Efficiency = 0.906 = 90.6%
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a projectile is fired in the earths gravitational field with a horizontal velocity in v = 9.00m/s. how far does it go in the horizontal direction in .550 s?
Answer:
Distance traveled by projectile in horizontal direction will be 4.95 m
Explanation:
We have given horizontal velocity of the projectile v = 9 m/sec
We have to find the distance traveled by projectile in 0.550 sec in horizontal direction.
We know that distance is equal to multiplication of speed and time
So distance traveled in horizontal direction will be equal to [tex]d=v\times t=9\times 0.550=4.95m[/tex]
So distance traveled by projectile in horizontal direction will be 4.95 m
Answer:
4.95 m
Explanation:
Horizontal velocity of projectile, u = 9 m/s
time, t = 0.550 s
The horizontal distance, d = horizontal velocity x time
d = 9 x 0.55
d = 4.95 m
Thus, the horizontal distance traveled is 4.95 m.
A gas has a pressure of 48atm in a 15.5L container. It was found that at 25∘C the gas occupied a volume of 25L and had a pressure of 22atm. What was the initial temperature in degrees Celsius?
Answer:
130.165636364°C
Explanation:
P = Pressure
V = Volume
n = Number of moles
R = Gas constant = 0.082 L atm/mol K
From ideal gas law we have
[tex]PV=nRT\\\Rightarrow n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{22\times 25}{0.082\times (25+273.15)}\\\Rightarrow n=22.496451696\ moles[/tex]
[tex]PV=nRT\\\Rightarrow T=\dfrac{PV}{nR}\\\Rightarrow T=\dfrac{48\times 15.5}{22.496451696\times 0.082}\\\Rightarrow T=403.315636364\ K[/tex]
The initial temperature is [tex]403.315636364-273.15=130.165636364\ ^{\circ}C[/tex]
Answer: 130 degrees Celsius
Explanation:
P1V1 / T1 = P2V2 / T2
Let the subscript 1 represent the initial 15.5L of gas and the subscript 2 represent the gas at the final volume of 25L. Rewrite the temperature in Kelvin by adding 273.
P1=48atm, V1=15.5L, P2=22atm, V2=25L, T2=298K, and T1 is unknown.
Substitute the known values into the combined gas law equation.
P1V1 / T1=P2V2 / T2 → (48atm)(15.5L) / T1 = (22atm)(25L) / 298K
Solve the equation for T1, simplify, and round to the nearest degree.
T1 = P1V1T2 / P2V2
T1 = (48atm)(15.5L)(298K)(22atm)(25L)
T1 = 403K
To get the temperature in Celsius, subtract 273.
T1 = 130∘C
A proton is initially at rest. After some time, a uniform electric field is turned on and the proton accelerates. The magnitude of the electric field is 1.36 105 N/C.
(a) What is the speed of the proton after it has traveled 3.00 cm?
(b) What is the speed of the proton after it has traveled 30.0 cm?
Answer:
a) 8.83*10⁵ m/s b) 2.80*10⁶ m/s
Explanation:
a) Assuming no other forces acting on the proton, the acceleration on it is produced by the electric field.
By definition, the force due to the electric field is as follows:
F = q*E = e*E (1)
where e is the elementary charge, the charge carried by only one proton, and is e = 1.6*10⁻¹⁹ C.
According to Newton's 2nd law, this force is at the same time, the product of the mass of the proton, times the acceleration a:
F = mp*a (2)
From (1) and(2), being left sides equal, right sides must be equal too:
[tex]F = e*E = mp*a[/tex]
Solving for a:
[tex]a = \frac{e*E}{mp} =\frac{1.6e-19C*1.36e5N/C}{1.67e-27kg} =1.3e13 m/s2[/tex]
⇒ a = 1.3*10¹³ m/s²
As we have the value of a (which is constant due to the field is uniform), the displacement x, and we know that the initial velocity is 0, in order to get the value of the speed, we can use the following kinematic equation:
[tex]vf^{2} -vo^{2} = 2*a*x[/tex]
Replacing by v₀ = 0, a= 1.3*10¹³ m/s² and x = 0.03 m, we can find vf as follows:
[tex]vf =\sqrt{2*(1.3e13 m/s2)*0.03m} = 8.83e5 m/s[/tex]
⇒ vf = 8.83*10⁵ m/s
b) We can just repeat the equation from above, replacing x=0.03 m by x=0.3 m, as follows:
[tex]vf =\sqrt{2*(1.3e13 m/s2)*0.3m} = 2.80e6 m/s[/tex]
⇒ vf = 2.80*10⁶ m/s
To find the speed of the proton after traveling a certain distance in a uniform electric field, we can use the equations of motion and the equation for force. By plugging in the known values and solving for acceleration, we can then use the equation for speed to find the desired velocity.
To find the speed of the proton, we can use the equation:
F = ma
where F is the force, m is the mass, and a is the acceleration.
The force on the proton is given by:
F = qE
where q is the charge of the proton and E is the electric field strength.
Plugging in the values, we get:
qE = ma
Solving for acceleration, we have:
a = qE/m
For the first part of the question, with a distance of 3.00 cm, we can use the equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity (which is 0), a is the acceleration, and s is the distance.
Plugging in the known values, we have:
v^2 = 0 + 2 * (qE/m) * s
Simplifying:
v = sqrt(2 * (qE/m) * s)
For the second part of the question, with a distance of 30.0 cm, we can use the same equation:
v^2 = u^2 + 2as
Plugging in the known values:
v^2 = 0 + 2 * (qE/m) * s
Simplifying:
v = sqrt(2 * (qE/m) * s)
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A cannon ball is shot straight upward with a velocity of 72.50 m/s. How high is the cannon ball above the ground 3.30 seconds after it is fired? (Neglect air resistance.)
Answer:
Explanation:
Given
Cannon is fired with a velocity of [tex]u=72.50\ m/s[/tex]
Using Equation of motion
[tex]y=ut+\frac{1}{2}at^2[/tex]
where
[tex]y=displacement[/tex]
[tex]u=initial\ velocity[/tex]
[tex]a=acceleration[/tex]
[tex]t=time[/tex]
after time [tex]t=3.3 s[/tex]
[tex]y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2[/tex]
[tex]y=239.25-53.36[/tex]
[tex]y=185.89\ m[/tex]
So after 3.3 s cannon ball is at a height of 185.89 m
Distinguish between a meteor, a meteoroid, and a meteorite.
Answer:
The distinction can be understood by their individual definitions given below.
Explanation:
A meteoroid is a small rocky/metallic body that can be found in outer space (space beyond the Earth's atmosphere). Their sizes are much smaller than asteroids (often called planetoids) and even more smaller than that of any planets or their moons. They generally originate from comets, asteroids (fragments of them) and even from planets or moons when there occurs heavy collisions.
A meteor is basically what we know to be "shooting stars". When a meteoroid, asteroid, etc. passes through the Earth's atmosphere, they heat up and begin to glow because of the frictional force experienced due to gas molecules in the atmosphere. But the important thing is that they do not reach the surface of the Earth as they completely burn out long before coming close. If some object does manage to reach the Earth's surface, we then call it a meteorite.
(These definitions are not restricted to the Earth but applies to all panets and moons.)
(Also check the gif provided here: https://en.wikipedia.org/wiki/Meteoroid)
What are the three longest wavelengths for standing sound waves in a 120-cm-long tube that is (a) open at both ends and (b) open at one end, closed at the other?
Answer
given,
length of tube = 120 cm
a) Open at both ends.
distance between two successive nodes or anti nodes = λ/2
for first possibility = λ/2 = 120
λ =2 x 120 = 240 cm
for second Possibility, distance between two node or antinode acting symmetrically
λ = 120 cm
for three nodes in the tube, distance between them is equal to
[tex]\dfrac{3}{2}\lambda = 120[/tex]
[tex]\lambda = 80 cm[/tex]
b) Open at one end
First possibility, The distance between one node or anti node
[tex]\dfrac{\lambda}{4} = 120[/tex]
λ = 480 cm
Second Possibility, distance between two node or anti node is equal to
[tex]\dfrac{3}{4}\lambda = 120[/tex]
λ = 160 cm
Third possibility, distance between three nodes and three anti nodes is equal to
[tex]\dfrac{5}{4}\lambda = 120[/tex]
λ = 100 cm
A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a height H, directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. (b) Find the value of H in terms of v0 and g such that at the instant when the balls collide, the first ball is at the highest point of its motion.
Answer:
(a) [tex]t=\frac{H}{v_0}[/tex]
(b) [tex]H=\frac{v_0^2}{g}[/tex]
Explanation:
Let the two balls collide at a height x from the ground. Therefore, ball 2 travels a distance of (H-x) before colliding with ball 1.
Using the following Newton's law of motion,
[tex]S=ut+\frac{1}{2}at^2[/tex]
where,
[tex]S[/tex] = displacement
[tex]u[/tex] = initial velocity
[tex]a[/tex] = acceleration
[tex]t[/tex] = time
we can write the equations of motion of the two balls(ball 1 and ball 2 respectively):
[tex]x=v_0t-\frac{1}{2}gt^2[/tex] ......(1) ([tex]a=-g[/tex], ball is moving against gravity)
[tex]H-x=\frac{1}{2} gt^2[/tex] .......(2) (initial velocity is zero; [tex]a=+g[/tex])
Substituting [tex]x[/tex] from equation (1) in (2),
[tex]H-v_0t+\frac{1}{2}gt^2=\frac{1}{2}gt^2[/tex]
or, [tex]t=\frac{H}{v_0}[/tex] ......(a)
(b) Now, it is said that the collision will occur when ball 1 is at it's highest point. That is, it's final velocity must be zero.
This time we shall have to use another equation of motion given by,
[tex]v^2=u^2+2aS[/tex]
where, [tex]v[/tex] = final velocity
therefore, we get for ball 1,
[tex]0=v_0^2-2gx[/tex] ([tex]u=v_0,v=0,a=-g[/tex])
or, [tex]x=\frac{v_0^2}{2g}[/tex]
Putting the value of [tex]x[/tex] in equation (2) and rearranging, we get,
[tex]\frac{g}{2v_0^2}H^2-H+\frac{v_0^2}{2g}=0[/tex]
which is a quadratic equation, whose solution is given by,
[tex]H=\frac{+1\pm\sqrt{(-1)^2-(4\times\frac{g}{2v_0^2} \times\frac{v_0^2}{2g}) } }{2\times\frac{g}{2v_0^2} }[/tex]
[tex]=\frac{v_0^2}{g}[/tex]
(a) The time at which the balls collide is H/[tex]v_{0}[/tex]
(b) The height H is equal to [tex]\frac{v_{0} ^{2} }{g}[/tex]
Let the balls collide at a height x above the ground.
Then the distance traveled by the ball thrown above is x.
And the distance traveled by the ball dropped from height H is (H-x).
(i) Both the balls will take the same time to travel respective distances in order to collide.
[tex]H-x=\frac{1}{2}gt^{2}[/tex]
[tex]x = v_{0}t - \frac{1}{2}gt^{2}[/tex]
We get:
[tex]x=v_{0}t-(H-x)[/tex]
[tex]t=\frac{H}{v_{0}}[/tex] , is the time after which the balls collide.
(ii) Let the ball thrown up attains its maximum height x at the time of thecollision
[tex]v^{2} = u^{2}-2gx[/tex] here v is the final velocity which is 0 when the ball attains maximum height
[tex]0=v_{0} ^{2}-2gx[/tex]
[tex]x=\frac{v_{0} ^{2} }{2g}[/tex] is the maximum height attained.
Now, the ball thrown downward travels distance (H-x) just before collision:
[tex]H-x=\frac{1}{2}gt^{2}[/tex]
[tex]H-\frac{v_{0} ^{2} }{2g}=\frac{1}{2}g\frac{H^{2} }{v_{0} ^{2} }[/tex]
Solving the quadratic equation we get:
[tex]H=\frac{v_{0} ^{2} }{g}[/tex]
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An airplane in a holding pattern flies at constant altitude along a circular path of radius 3.50 km. If the airplane rounds half the circle in 1.50 3 102 s, determine the magnitude of its (a) displacement and (b) average velocity during that time. (c) What is the airplane’s average speed during the same time interval?
Explanation:
A.
Displacement,S; Θ = S/r
= 180/360 * 2π * 3500
= 10995.6 m
Θ = 10995.6/3500
= 3.142
B.
Angular speed,w = Θ/t
= 3.142/1.503102
= 2.09 rad/s
Velocity = w * r
= 2.09 × 3500
= 7315 m/s.
C.
The same as B.
The airplane's Motion in a Circle displacement is 7.0km, its average velocity is 0.0467 km/s, and its average speed is 0.0737 km/s.
The airplane is flying half a circle, which is essentially semi-circular motion. Under this condition, we can calculate the values asked in the question as follows:
(a) Displacement: Displacement is a vector quantity and represents the shortest distance between the initial and the final points of an object's path. But since the airplane is rounding half the circle, the displacement is essentially the diameter of the circular path. The diameter will be two times the radius, therefore 2*3.50km = 7.00 km.
(b) Average velocity: Average velocity is the total displacement divided by the total time. So, the average velocity would be 7.00km / (1.50*10^2) s = 0.0467 km/s.
(c) Average speed: Average speed is defined as the total distance traveled by the object divided by the total time taken. Here, the airplane travels half the circumference of the circle in the given time. The formula for the circumference of a circle is 2*pi*r, so half the circle's circumference will be pi*3.50 km. Then, Average speed = (pi*3.50 km) / (1.50*10^2) s = 0.0737 km/s.
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Which of the following liquids will have the lowest freezing point/ A) pure h20 B)aqueous KF(0.50 m) C)aqueous sucrose (0.60 m) D)aqueous glucose (0.60 m) E) aqueous Fel3 (0.24 m)
The aqueous solution with the highest concentration of particles will have the lowest freezing point.
Explanation:The freezing point of a solution is determined by the concentration of solute particles in the solution. The greater the concentration of particles, the lower the freezing point. In this case, the aqueous solution with the lowest freezing point will have the highest concentration of particles.
Out of the options given, the aqueous solution with the highest concentration of particles is 0.60 m aqueous sucrose. Therefore, aqueous sucrose (0.60 m) will have the lowest freezing point.
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A square steel bar has a length of 9.8 ftft and a 2.6 inin by 2.6 inin cross section and is subjected to axial tension. The final length is 9.80554 ftnt . The final side length is 2.59952 in in . What is Poisson's ratio for the material? Express your answer to three significant figures.
To solve this problem we will apply the concept related to the Poisson ratio for which the longitudinal strains are related, versus the transversal strains. First we need to calculate the longitudinal strain as follows
[tex]\epsilon_x = \frac{l_f-l_i}{l_i}[/tex]
[tex]\epsilon_x = \frac{(9.80554)-(9.8)}{9.8}[/tex]
[tex]\epsilon_x = 0.0005653[/tex]
Second we will calculate the lateral strain as follows
[tex]\epsilon_y = \frac{a_f-a_i}{a_i}[/tex]
[tex]\epsilon_y = \frac{2.59952-2.6}{2.6}[/tex]
[tex]\epsilon_y = -0.0001846153[/tex]
The Poisson's ratio is the relation between the two previous strain, then,
[tex]\upsilon = -\frac{\epsilon_y}{\epsilon_x}[/tex]
[tex]\upsilon = -\frac{(-0.0001846153)}{0.0005653}[/tex]
[tex]\upsilon = 0.3265[/tex]
Therefore the Poisson's ratio for the material is 0.3265
At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what distance from the line will the field strength to be 2000 N/C
To find the distance from the line where the electric field strength is 2000 N/C, we can use the formula for the electric field of an infinitely long charged wire. The distance from the line where the electric field strength is 2000N/C is d = lambda / 2 x 10^6 m.
Explanation:To find the distance from the line where the electric field strength is 2000 N/C, we can use the formula for the electric field of an infinitely long charged wire:
E = k * (lambda / d)
Where E is the electric field strength, k is the Coulomb's constant, lambda is the charge density of the wire, and d is the distance from the wire. In this case, since the line of charge is infinite, the charge density is simply the charge per unit length.
To solve for the distance d, we can rearrange the formula: d = k * (lambda / E)
Plugging in the given values, the distance d is:
d = (9.0 x 10^9 Nm^2/C^2) * (lambda / 2000 N/C)
So, the distance from the line where the electric field strength is 2000 N/C is d = lambda / 2 x 10^6 m.
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Which of the following does not involve work? 1. A child is pushed on a swing. 2. A golf ball is struck. 3. A weight lifter does military presses (lifting weights over his head.) 4. A professor picks up a piece of chalk from the floor. 5. A runner stretches by pushing against a wall.
From the work theorem, this is defined as the amount of force applied on an object displaced on a longitudinal unit. Mathematically this is
[tex]W = \vec{F} \times \vec{d}[/tex]
Here,
F = Force vector
d = Displacement vector
Of all the options presented, only in the last one there is no change in distance, so the work done there is zero.
The correct option is 5.
Work in Physics is when a force causes displacement. Of the choices, the example of a runner stretching by pushing against a wall does not involve work, as there is no displacement.
Explanation:In the context of Physics, 'work' is defined as a force causing displacement on an object. Essentially, work is done when a force acts upon an object to cause or prevent motion. Among the options provided, 5. A runner stretches by pushing against a wall does not involve 'work'. This is because, despite the runner exerting a force against the wall, there is no displacement of the wall in response to this force.
Work (W) is calculated by multiplying the force (F) that is applied to an object and the distance (d) that the object is moved, i.e., W = F * d. In this case, since the displacement (d) is zero, the work done is also zero.
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You are on your balcony and notice some bad squirrels digging in your garden directly below. You start tossing your pistachios at them to get them out of your yard. If the height of your guard rail is 43 inches above the deck, which is at a standard height of 10 feet above the ground, and you toss the pistachios straight down with a speed of 7.4 m/s, then how long does it take for the pistachio to reach the ground?
Answer:
0.43 s
Explanation:
We have the following parameters:
Initial velocity, u = 7.4 m/s
Acceleration of gravity, g = 9.8 [tex]m/s^2[/tex]
Distance, s = 43 in + 10 ft = 1.092 m + 3.048 m = 4.14 m
Time, t = ?
Using the equation of motion [tex]s=ut +\frac{1}{2}gt^2[/tex], we have
[tex]4.14 = 7.4t + 0.5\times9.8t^2[/tex]
[tex]4.9t^2 + 7.4t - 4.14 =0[/tex]
Using the quadratic formula [tex]\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex] where a = 4.9, b = 7.4 and c = - 4.14, and solving for the positive value of t only, we have
[tex]t = 0.43[/tex] s
A 200-m-wide river has a uniform flow speed of 0.99 m/s through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of 4.4 m/s with respect to the water. There is a clearing on the north bank 35 m upstream from a point directly opposite the clearing on the south bank.
a. At what angle, measured relative to the direction of flow of the river, must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank?
b. How long will the boat take to cross the river and land in the clearing?
Answer:
a. 1.174 rad[/tex] or 67.3 degree
b. t = 49.28 s
Explanation:
Let [tex]v_v[/tex] be the vertical component of the boat velocity with respect the the river, pointing North. Let [tex]v_h[/tex] be the horizontal component of the boat velocity with respect to the river, pointing West, aka upstream. Since the total velocity of the boat is 4.4m/s
[tex]v_v^2 + v_h^2 = 4.4^2 = 19.36[/tex]
The time it takes for the boat to cross 200m-wide river at [tex]v_v[/tex] rate is
[tex]t = 200 / v_v[/tex] or [tex]v_v = 200 / t[/tex]
This is also the time it takes for the boat to travel 35m upstream, horizontally, at the rate of [tex] v_h - 0.99[/tex] m/s
[tex]t = \frac{35}{v_h - 0.99}[/tex]
[tex]v_h - 0.99 = 35/t[/tex]
[tex]v_h = 35/t + 0.99[/tex]
We can substitute [tex]v_v,v_h[/tex] into the total velocity equation to solve for t
[tex]\frac{200^2}{t^2} + (\frac{35}{t} + 0.99)^2 = 19.36[/tex]
[tex]\frac{40000}{t^2} + \frac{35^2}{t^2} + 2*0.99*\frac{35}{t} + 0.99^2 = 19.36[/tex]
From here we can multiply both sides by [tex]t^2[/tex]
[tex]40000 + 1225 + 69.3t + 0.9801t^2 = 19.36t^2[/tex]
[tex]18.38 t^2 - 69.3t - 41225 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{69.3\pm \sqrt{(-69.3)^2 - 4*(18.3799)*(-41225)}}{2*(18.38)}[/tex]
[tex]t= \frac{69.3\pm1742.31}{36.7598}[/tex]
t = 49.28 or t = -45.51
Since t can only be positive we will pick t = 49.28
[tex]v_h = 35 / t + 0.99 = 35 / 49.38 + 0.99 = 1.7 m/s[/tex]
The angle, relative to the flow of river direction is
[tex]cos(\alpha) = \frac{v_h}{v} = \frac{1.7}{4.4} = 0.3864[/tex]
[tex]\alpha = cos^{-1}(0.3864) = 1.174 rad[/tex] or 67.3 degree
A train leaves the train station at noon and travels at a constant speed of vt = 50 mi/hr on a straight track. 2 hr later, a car leaves the same train station and travels in the same direction at vc = 70 mi/hr on a road next to the train track. How far from the station is the place where the car catches up with the train? x =
Final answer:
The distance from the train station where the car catches up with the train is 250 miles.
Explanation:
To find the distance from the train station where the car catches up with the train, we need to determine the time it takes for the car to catch up.
Let's assume the car catches up with the train after t hours. Since the train leaves 2 hours before the car, the train has already been traveling for t + 2 hours.
Distance traveled by the train = speed of the train × time = 50t + 100 miles
Distance traveled by the car = speed of the car × time = 70t miles
Since the car catches up with the train, the distances traveled by both must be equal. Equating the distances, we get:
50t + 100 = 70t
20t = 100
t = 5
Now, substitute the value of t back into one of the distance formulas to find the distance from the train station:
Distance from the train station = speed of the train × time = 50 × 5 = 250 miles
Final answer:
The car catches up with the train in 5 hours, and they are 350 miles from the station when the car catches up.
Explanation:
The problem you presented is related to finding when two objects, moving at different constant speeds, meet. To solve this, we use the concept of relative speed.
The train travels for 2 hours before the car starts, so by that time, the train would have covered a distance of (50 mi/hr × 2 hr) = 100 miles. When the car starts, it has to cover not only the distance to the train but also the additional distance the train covers as the car approaches.
Let's consider the time it takes for the car to catch up with the train to be 't' hours. In that time, the car would travel 70t miles, and the train would travel an additional 50t miles.
Since the car needs to cover the initial 100 miles plus whatever distance the train covers in 't' hours, we can set up the equation:
70t = 100 + 50t
Subtracting 50t from both sides gives us:
20t = 100
Dividing both sides by 20:
t = 5
So, the car catches up with the train in 5 hours. To find out how far from the station they are when the car catches up with the train:
x = 70 mi/hr × 5 hr = 350 miles
A man wandering in the desert walks 2.7 miles in the direction S 35°W. He then turns 90° and walks 3.5 miles in the direction N 55° W. At that time, how far is he from his starting point, and what is his bearing from his starting point?
To solve this problem we will make a diagram in the Cartesian plane that will allow us to find and understand more accurately the displacement and the angle of rotation.
According to Pythagoras, the distance traveled would be equivalent to
[tex]d = \sqrt{(2.7)^2+(3.5)^2}[/tex]
[tex]d = 4.4 miles[/tex]
The individual had a displacement of 4.4 thousand from the starting point.
Now the angle [tex]\theta[/tex] plus the previously given angle will allow us to find the direction of travel.
[tex]tan\theta = \frac{\text{Opposite side}}{\text{Adjacent side}}[/tex]
[tex]tan\theta = \frac{3.5}{2.7}[/tex]
[tex]\theta = tan^{-1} (\frac{3.5}{2.7})[/tex]
[tex]\theta = 52.35\°[/tex]
[tex]\angle =[/tex] [tex]\theta + 35 = 52.35+35 = 87.35\°[/tex]
Therefore the net direction of the man is S 87.35° W
Suppose that at room temperature, a certain aluminum bar is 1.0000 m long. The bar gets longer when its temperature is raised. The length l of the bar obeys the following relation: l=1.0000+2.4×10−5T, where T is the number of degrees Celsius above room temperature. What is the change of the bar's length if the temperature is raised to 16.1 ∘C above room temperature?
Answer:
The change of the bar's length is [tex] 3.9\times10^{-4} m[/tex]
Explanation:
The bar length is a function of temperature T above room temperature:
[tex] L(T)=1.0000+2.4\times10^{-5} T[/tex]
So, if we evaluate at T= 16.1 C above room temperature
[tex]L(16.1)=1.0000+2.4\times10^{-5} (16.1)[/tex]
[tex]L=1.00039 m [/tex]
Now we can find the change of the bar length with the difference of L and Lo (the length at room temperature)
[tex]L- L_0=1.00039-1.0000 = 3.9\times10^{-4} m[/tex]
A single-turn rectangular wire loop measures 6.00 cm wide by 10.0 cm long. The loop carries a current of 3.00 A. The loop is in a uniform magnetic field with B = 4.00 times 10^-3 T. Taking torques about an axis, parallel to either side of the rectangular loop, that maximizes the torque, what is the magnitude of the torque exerted by the field on the loop if the direction of the magnetic field is described as the following? a) parallel to the short sides of the loop N middot m b) parallel to the long sides of the loop N middot m c) perpendicular to the plane of the loop N middot m
Answer:
Explanation:
Single turn implies that N=1
The rectangle is 6cm (0.06m) wide and 10 cm (0.1m) long.
Current in loop is 3A
Magnetic Field B= 4×10^-3T
To take the maximum torque then, sinθ=1
Then
τ=NiABsinθ
Where N is number of turns, and in this case N is 1
i is current in coil and it this case it is 3Amps.
A is area of of coil
And in this case it is a rectangle
Area of rectangle is Lenght × breadth
A= 0.06×0.1= 0.006m^2
And B is magnetic field given as B=4×10^-3T
And θ is angle between the field and the normal to the coil.
a. Torque parallel to the short side of the loop,
The length of the short side is 6cm=0.06m.
This is actually the maximum possible torque, when the field is in the plane of the loop.
τ=NiABsinθ
τ=1×3×0.006×4×10^-3
τ=0.072 ×10^-3
τ=7.2 ×10^-5Nm
b. Torque parallel to the long side of the loop,
The length of the short side is 10cm=0.1m.
This is actually the maximum possible torque, when the field is in the plane of the loop.
τ=NiABsinθ
τ=3×0.006×4×10^-3
τ=0.072 ×10^-3
τ=7.2 ×10^-5Nm.
c. Torque perpendicular to the plane. When the field is perpendicular to the loop the torque is zero.
If the torque is perpendicular to the plane, we said theta is the angle between the normal and the magnetic field
Then if the torque is perpendicular to the normal, then the angle between the torque and the normal is 0, the sinθ = 0
τ=NiABsinθ
Since sinθ =0
Then,
τ=0Nm
A nonconducting sphere has radius R = 2.36 cm and uniformly distributed charge q = +2.50 fC. Take the electric potential at the sphere's center to be V0 = 0. What is V at radial distance r = 1.45 cm?
Explanation:
It is known that inside a sphere with uniform volume charge density the field will be radial and has a magnitude E that can be expressed as follows.
E = [tex]\frac{q}{4 \p \epsilon_{o}R^{3}}r[/tex]
V = [tex]-\int_{0}^{r}E. dr[/tex]
= [tex]-\int_{0}^{r}\frac{q}{4 \p \epsilon_{o}R^{3}}r dr[/tex]
= [tex](\frac{q}{4 \p \epsilon_{o}R^{3}})(\frac{r^{2}}{2})[/tex]
= [tex]\frac{-qr^{2}}{8 \pi \epsilon_{o}R^{3}}[/tex]
At r = 1.45 cm = [tex]1.45 \times 10^{-2}[/tex] (as 1 m = 100 cm)
V = [tex]\frac{2.50 \times 10^{-15} \times 1.45 \times 10^{-2}}{8 \times 3.1416 \times 8.85 \times 10^{-12} \times 2.36 \times 10^{-2}}[/tex]
= [tex]6.905 \times 10^{-6}[/tex] mV
Thus, we can conclude that value of V at radial distance r = 1.45 cm is [tex]6.905 \times 10^{-6}[/tex] mV.
To practice Problem-Solving Strategy 2.1 Motion with constant acceleration You are driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. Your reaction time before stepping on the brakes is 0.50 s , and the maximum deceleration of your car is 10 m/s2 . How much distance is between you and the deer when you come to a stop
a. How much distance is between you and the deer when you come to a stop?
b. What is the maximum speed you could have and still not hit the deer?
Answer:
a) [tex]\Delta s=5\ m[/tex] is the distance between deer and the vehicle
b) [tex]u'=22.36\ m.s^{-1}[/tex] is the maximum speed the driver can be at and still not hit the deer.
Explanation:
Given:
initial speed of driving, [tex]u=20\ m.s^{-1}[/tex]distance of deer from the vehicle, [tex]x=35\ m[/tex]reaction time taken to step onto the brakes, [tex]t'=0.5\ s[/tex]maximum deceleration of the car, [tex]a_m=-10\ m.s^{-2}[/tex]a)
Now the distance travelled after application of the brakes till the vehicle stops:
[tex]v^2=u^2+2a_m.s[/tex]
(assuming that the brakes are applied with maximum acceleration)
where:
[tex]s=[/tex] displacement of the vehicle after braking till it stops
[tex]v=[/tex] final velocity of the vehicle = 0 (stops)
putting the values:
[tex]0^2=20^2-2\times 10\times s[/tex]
[tex]s=20\ m[/tex]
Now before the application of the brakes 0.5 second is taken to react and the vehicle travels during this time as well.
So, distance covered before applying the brakes:
[tex]s'=u.t'[/tex]
[tex]s'=20\times 0.5[/tex]
[tex]s'=10\ m[/tex]
The distance between the deer and the vehicle:
[tex]\Delta s=x-(s+s')[/tex]
[tex]\Delta s=35-(20+10)[/tex]
[tex]\Delta s=5\ m[/tex]
b)
The maximum speed the driver can have with the vehicle and still not hit the deer is given as:
[tex]v^2=u'^2+2. a_m.(x-s')[/tex]
because s' is the distance covered before braking during the reaction time.
[tex]0^2=u'^2-2\times 10\times (35-10)[/tex]
[tex]u'=22.36\ m.s^{-1}[/tex] is the maximum speed the driver can be at and still not hit the deer.
Using the equations of motion under constant acceleration, a) the distance between the driver and the deer when the car comes to a stop is 5 m and b) the maximum speed the driver could have and still not hit the deer is approximately 23.45 m/s.
Explanation:The subject of the question, Problem-Solving Strategy 2.1 Motion with constant acceleration, involves using the equations of motion under constant acceleration. Let's break down the problem into two parts:
How much distance is between you and the deer when you come to a stop? In this scenario, you first drive at 20 m/s for 0.50 s before stepping on the brakes. So, the distance travelled during this time is v*t = 20 m/s * 0.50 s = 10 m. Then, you decelerate at 10 m/s². As you finally come to stop, the additional distance travelled can be found by the formula (v² - u²) / 2a, which gives (0 - (20²)) / 2*(-10) = 20 m. So, the total distance covered is 10 m + 20 m = 30 m. Therefore, you come to a stop 5 m away from the deer because the deer was initially 35 m away.What is the maximum speed you could have and still not hit the deer? For this, you need to calculate the stopping distance for the car in relation to the deer at 35 m and find the initial speed where the stopping distance equals the distance to the deer. If the car travels distance D in the driver's reaction time, then it travels (35 - D) while braking. The braking distance = (v² - u²) / 2a => v² = 2aD, where D = (35 - v*0.50). Solving for v in the quadratic equation gives the maximum speed, approx 23.45 m/s to not hit the deer.Learn more about Motion with Constant Acceleration here:https://brainly.com/question/25307462
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Automobile traveling at 65 mph constant on the road described below. Find rate at which radar must rotate when theta = 15 deg. Ans: theta_dot = 0.219 rad/s.
Answer:
The rate at which radar must rotate is 0.335 rad/s.
Explanation:
Given that,
Velocity = 65 m/h = 29.0576 m/s
Angle = 15°
Suppose, the radius given by
[tex]r=(100\cos2\theta)\ m[/tex]
We need to calculate the rate at which radar must rotate
Using formula of linear velocity
[tex]v=r\omega[/tex]
[tex]\omega=\dfrac{v}{r}[/tex]
Where, v = velocity
r = radius
Put the value into the formula
[tex]\omega=\dfrac{29.0576}{100\cos30}[/tex]
[tex]\omega=0.335\ rad/s[/tex]
Hence, The rate at which radar must rotate is 0.335 rad/s.
A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each section of the string moves with simple harmonic motion at a frequency of 8 Hz. Find the power propagated along the string.
Answer:
Power of the string wave will be equal to 5.464 watt
Explanation:
We have given mass per unit length is 0.050 kg/m
Tension in the string T = 60 N
Amplitude of the wave A = 5 cm = 0.05 m
Frequency f = 8 Hz
So angular frequency [tex]\omega =2\pi f=2\times 3.14\times 8=50.24rad/sec[/tex]
Velocity of the string wave is equal to [tex]v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec[/tex]
Power of wave propagation is equal to [tex]P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt[/tex]
So power of the wave will be equal to 5.464 watt
1 mole of air undergoes a Carnot cycle. The hot reservoir is at 800 oC and the cold reservoir is at 25 oC. The pressure ranges between 0.2 bar and 60 bar. Determine the net work produced, and the efficiency of the cycle.
Answer:
The net work produced is 30.37 KJ
The efficiency of cycle is 72.3%
Explanation:
For the net work produced, we have the formula:
Work = (Th - Tc)(Sh - Sc)(M)
Where,
Th = higher temperature = 800° C + 273 = 1073 k
Tc = lower temperature = 25° C + 273 = 298 k
Sh = specific entropy at higher temperature
Sc = specific entropy at lower temperature
M = molar mass of air
Using, ideal gas table to find entropy. The table is attached.
therefore,
Work = (1073 k - 298 k)(3.0485235 KJ/kg.k - 1.69528 KJ/kg.k)(0.02896 kg)
Work = (775 k)(1.3532435 KJ/kg.k)(0.02896 kg)
Work = 30.37 KJ
Now, for the efficiency (n), we have a formula:
n = 1 - Tc/Th
n = 1 - (298 k)/(1073 k)
n = 0.723 = 72.3 %
A vector that is 7.1 units long and a vector that is 5.0 units long are added. Their sum is a vector 7.6 units long. (a) Show graphically at least one way that the vectors can be added. (Do this on paper. Your instructor may ask you to turn in this work.) (b) Using your sketch in Part (a), determine the angle between the original two vectors. °
Answer:
[tex]\theta=119.8138^{\circ}[/tex]
Explanation:
Given:
magnitude of first vector, [tex]A=7.1\ units[/tex]magnitude of second vector, [tex]B=5.0\ units[/tex]resultant of magnitude of the two vectors, [tex]R=7.6\ units[/tex]a)
From the vector addition rule we know:
[tex]R^2={A^2+B^2+2A.B\cos\theta}[/tex]
where
[tex]\theta=[/tex] angle between the two vectors with their tail at common point.
b)
Putting respective values:
[tex]7.6^2=7.1^2+5^2+7.1\times 5\times \cos \theta[/tex]
[tex]\theta=119.8138^{\circ}[/tex]
To add the two vectors, draw them graphically and connect the tip of one vector to the tail of the other. The angle between the original two vectors can be determined using the adjacent and hypotenuse sides of a right triangle.
Explanation:Part (a)To add the two vectors, draw the first vector as a line segment with a length of 7.1 units and draw the second vector as a line segment with a length of 5.0 units. Place the tail of the second vector at the tip of the first vector. The sum of the two vectors is the line segment connecting the tail of the first vector to the tip of the second vector. This sum vector should have a length of 7.6 units, as mentioned in the question.
Part (b)Using the sketch from Part (a), draw a horizontal line from the tail of the first vector to the tip of the second vector. Next, draw a vertical line from the tip of the first vector to the tip of the second vector. Connect the tail of the first vector to the tip of the second vector to form a right triangle. The angle between the original two vectors is the angle opposite to the horizontal line. To determine this angle, we can use the trigonometric inverse function. Let's call the angle 'theta'. We have the adjacent side as 5.0 units and the hypotenuse as 7.6 units. Using the arccosine function (inverse cosine), we can calculate theta using the equation: theta = arccos(5.0/7.6).
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