Explanation:
Let the first wave is :
[tex]y_1=A\sin\omega t[/tex]
And another wave is :
[tex]y_2=A\sin (\omega t+\phi)[/tex]
[tex]\phi[/tex] is phase difference between waves
Let y is the resultant of these two waves. So,
[tex]y =y_1+y_2[/tex]
The waves reflected from the upper and lower surfaces of the medium, it means that the resultant to be zero. So,
[tex]\cos(\dfrac{\phi}{2})=0\\\\\cos(\dfrac{\phi}{2})=\cos(\dfrac{\pi}{2})\\\\\phi=\pi[/tex]
So, the phase difference between the two waves is [tex]\pi[/tex].
A millionairess was told in 1992 that she had exactly 15 years to live. However, if she immediately takes off, travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:
Answer:
The expected year is 2017.
Explanation:
Total years that the millionaire to live = 15 years
Travel away from the earth at = 0.8 c
This is a time dilation problem so if she travels at 0.8 c then her time will pass at slower. Below is the following calculation:
[tex]T = \frac{T_o}{ \sqrt{1-\frac{V^2}{c^2}}} \\T = \frac{15}{ \sqrt{1-\frac{0.8^2}{c^2}}} \\T = 25 years[/tex]
Thus the doctors are expecting to celebrate in the year, 1992 + 25 = 2017
2. In the ice cream making process, after the pasteurization of the base mixture, the syrup should be cooled to 4 °C to avoid the proliferation of pathogenic microorganisms. A new thermometer was attached to the tank; however, it marked a temperature in another unit: Rankine. What should be the value indicated on the thermometer for the process to be carried out under the same conditions?
3. During the class in the laboratory, the manometer coupled to the analysis equipment indicates a vacuum of 638 mmHg. What should be the absolute pressure in kPa and psi, knowing that the local barometric pressure is 101.3 kPa?
Answer:
2. 500 R
3. 16.3 kPa, 2.36 psi
Explanation:
2. Convert Celsius to Fahrenheit.
1.8 (4°C) + 32 = 39.2°F
Convert Fahrenheit to Rankine
39.2°F + 459.67 = 498.87 R
Rounding to one significant figure, the temperature is 500 R.
3. Absolute pressure = gauge pressure + atmospheric pressure
P = Pg + Pa
First, convert mmHg to kPa (remember that a vacuum is negative gauge pressure).
-638 mmHg × (101.3 kPa / 760 mmHg) = -85.0 kPa
So the absolute pressure is:
P = -85.0 kPa + 101.3 kPa
P = 16.3 kPa
Converting to psi:
P = 16.3 kPa × (14.7 psi / 101.3 kPa)
P = 2.36 psi
The moment of inertia for a rod that rotates about the axis perpendicular to the rod and passing through one end is: . If the axis of rotation passes through the center of the rod, then the moment of inertia is: . Give a physical explanation for this difference in terms of the way the mass of the rod is distributed with respect to the axis in the two cases.
Answer:
Explanation:
he moment of inertia for a rod that rotates about the axis perpendicular to the rod and passing through one end is: m L²/ 3 where m is mass and L is length of rod
If the axis of rotation passes through the center of the rod, then the moment of inertia is: m L² / 12
So for the former case , moment of inertia is higher that that in the later case .
In the former case , the axis is at one extreme end . Hence range of distance of any point on the rod from axis is from zero to L .
In the second case , as axis passes through middle point , this range of distance of any point on the rod from axis is from zero to L / 2 .
Since range of distance from axis is less , moment of inertia too will be less because
Moment of inertia = Σ m r² where r is distance of mass m from axis .
what is rotation? a. the orbit of a satelite around a central body b. the motion of two objects around each other c the spinning of an object on its axis d the motion of an object around a central body
Answer:
C. The spinning of an object on its axis.
Explanation:
The definition of rotation answers this question.
A good way to remember is that when you rotate, you turn, and when you revolve, you move around.
A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v= 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?
Answer:
The wait time is [tex]t_w = 3.4723 \ s[/tex]
Explanation:
From the question we are told that
The distance of the hot air balloon above the ground is [tex]z = 50 \ m[/tex]
The distance of the hot air balloon from the target is [tex]k = 100 \ m[/tex]
The speed of the wind is [tex]v = 15 \ m/s[/tex]
Generally the time it will take the balloon to hit the ground is
[tex]t = \sqrt{ \frac{2 * z }{g} }[/tex]
where g is acceleration due to gravity with value [tex]g = 9.8 m/s^2[/tex]
substituting values
[tex]t = \sqrt{ \frac{2 * 50 }{9.8} }[/tex]
[tex]t = 3.194 \ s[/tex]
Now at the velocity the distance it will travel before it hit the ground is mathematically represented as
[tex]d = v * t[/tex]
substituting values
[tex]d = 15 * 3.194[/tex]
[tex]d = 47.916 \ m[/tex]
Now in order for the balloon to hit the target on the ground it will need to travel b distance on air before the balloonist drops it and this b distance can be evaluated as
[tex]b = k - d[/tex]
substituting values
[tex]b =100 -47.916[/tex]
[tex]b = 52.084 \ m[/tex]
Hence the time which the balloonist need to wait before dropping the balloon is mathematically evaluated as
[tex]t_w = \frac{b}{v}[/tex]
substituting values
[tex]t_w = \frac{52.084}{15}[/tex]
[tex]t_w = 3.4723 \ s[/tex]
A spring is attached to the ceiling and pulled 11 cm down from equilibrium and released. The amplitude decreases by 19% each second. The spring oscillates 10 times each second. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.
Answer:
D(t)= 11x2.09t cos( 20pi,t)
Explanation:
Pls attached file
if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutual force between them
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;
[tex]F = \frac{k|q_1||q_2|}{r^2}[/tex]
where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²
[tex]F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N[/tex]
Therefore, the mutual force between the two point charges is 319.64 N
An electromagnetic wave is traveling through vacuum in the positive x direction. Its electric field vector is given by E⃗ =E0sin(kx−ωt)j^,where j^ is the unit vector in the y direction.
What is the Poynting vector S⃗ (x,t), that is, the power per unit area associated with the electromagnetic wave described in the problem introduction?
Given that,
The electric field is given by,
[tex]\vec{E}=E_{0}\sin(kx-\omega t)\hat{j}[/tex]
Suppose, B is the amplitude of magnetic field vector.
We need to find the complete expression for the magnetic field vector of the wave
Using formula of magnetic field
Direction of [tex](\vec{E}\times\vec{B})[/tex] vector is the direction of propagation of the wave .
Direction of magnetic field = [tex]\hat{j}[/tex]
[tex]B=B_{0}\sin(kx-\omega t)\hat{k}[/tex]
We need to calculate the poynting vector
Using formula of poynting
[tex]\vec{S}=\dfrac{E\times B}{\mu_{0}}[/tex]
Put the value into the formula
[tex]\vec{S}=\dfrac{E_{0}\sin(kx-\omega t)\hat{j}\timesB_{0}\sin(kx-\omega t)\hat{k}}{\mu_{0}}[/tex]
[tex]\vec{S}=\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}[/tex]
Hence, The poynting vector is [tex]\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}[/tex]
A 22.0 cm diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. The loop is rotated so that its plane is parallel to the field direction in 0.20 s. What is the average induced emf in the loop?
Answer:
The average induced emf in the loop is 0.3 V
Explanation:
Given:
d = 22 cm
Magnetic field is B = 1.5 T
Change in time Δt = 0.20 sec.
Radius of loop = r = d/2 = 11 x [tex]10^{-2}[/tex] m
According to the faraday's law, Induced emf is given by e = - (ΔФ / Δt)
where Ф = magnetic flux
Ф = BAcos0 (in this occasion, Ф=0)
where area A = pi * r²
note that we have to neglect negative sign due to its lenz law...so
B * pi * r
e = ----------------
Δt
1.5 ( 3.1416) 11 x [tex]10^{-2}[/tex] )²
e = ------------------------------------
0.20
e = 0.3 V
Therefore, the average induced emf in the loop is 0.3 V
Two positive charges are located at x = 0, y = 0.3m and x = 0, y = -.3m respectively. Third point charge q3 = 4.0 μC is located at x = 0.4 m, y = 0.
A) Make a careful sketch of decent size that illustrates all force vectors with directions and magnitudes.
B) What is the resulting vector of the total force on charge q1 exerted by the other two charges using vector algebra?
Answer:
0.46N
Explanation:
See attached file
An ideal spring of negligible mass is 11.00cm long when nothing is attached to it. When you hang a 3.05-kg weight from it, you measure its length to be 12.40cm .
If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.
Express your answer numerically. If there is more than one answer, enter each answer, separated by a comma.
=
Answer
0.2067m or 0.2067m
Explanation;
Let lenght of spring= Lo= 11cm=0.110m
It is hang from a mass of
3.05-kg having a length of L1= 12.40cm= 0.124m
Force required to stretch the spring= Fkx
But weight of mass mg= kx then K= Mg/x
K= 3.05-kg× 9.8)/(0.124m-.110m)
K=2135N
But potential Energy U= 0.5Kx
X=√ 2U/k
√(2*10)/2135
X=0.0967m
The required new length= L2= L0 ±x
=
.110m ± 0.0967m
X= 0.2067m or 0.2067m hence the total lenghth
A solid disk of radius 1.4 cm and mass 430 g is attached by a wire to one of its circular faces. It is twisted through an angle of 10 o and released. If the spring has a torsion constant of 370 N-m/rad, what is the frequency of the motion
Answer:
f= 4,186 10² Hz
Explanation:
El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por
w = √ k/I
donde ka es constante de torsion de hilo e I es el momento de inercia del disco
El momento de inercia de indican que giran un eje que pasa por enronqueces
I= ½ M R2
reduzcamos las cantidades al sistema SI
R= 1,4 cm = 0,014 m
M= 430 g = 0,430 kg
substituimos
w= √ (2 k/M R2)
calculemos
w = RA ( 2 370 / (0,430 0,014 2)
w = 2,963 103 rad/s
la velocidad angular esta relacionada con la frecuencia por
w =2pi f
f= w/2π
f= 2,963 10³/ (2π)
f= 4,186 10² Hz
"Neon signs need 12,000 V to operate. If a transformer operates off a 240 V source and has 1000 turns in its primary coil, how may turns must the secondary coil have
Answer:
50000 turns
Explanation:
Vp / Vs = Np / Ns
240 / 12000 = 1000 / Ns
Ns = 50000 turns
Which is true about a concave mirror? Incident rays that are parallel to the central axis are dispersed but will be perceived as originating from a point on the near side of the mirror.
Answer:
'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
Explanation:
The question is incomplete, find the complete question in the comment section.
Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. The focus point is close to the curved mirror than the centre of curvature.
During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. All incident light striking the surface all converges at a point on the central axis known as the focus.
Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
A customs inspector was suspecting that some of the 12 plastic spheres, which were shipped out of the country, had something in them. Each sphere weighted the same and had hard walls everywhere. Inspector thought that it was possible to hide something inside each sphere. He was correct, and was able to use a simple experiment in determining which sphere had diamonds inside. How did he do it?
Answer:
use a hammer to hit it
Explanation:
if u hit it u will be able to hear the shattered noise
The AC voltage source supplies an rms voltage of 146 V at frequency f. The circuit has R = 110 Ω, XL = 210 Ω, and XC = 110 Ω. At the instant the voltage across the generator is at its maximum value, what is the magnitude of the current in the circuit?
Answer:
1.03A
Explanation:
For computing the magnitude of the current in the circuit we need to do the following calculations
LCR circuit impedance
[tex]Z = \sqrt{R^2 + (X_L - X_c)^2} \\\\ = \sqrt{110^2 + (210 - 110)^2}[/tex]
= 148.7Ω
Now the phase angle is
[tex]\phi = tan^{-1} (\frac{X_L - X_C}{R}) \\\\ = tan^{-1} (\frac{210 - 110}{110})\\\\ = 42.3^{\circ}[/tex]
Now the rms current flowing in the circuit is
[tex]I_{rms} = \frac{V_{rms}}{Z} \\\\ = \frac{146}{148.7}[/tex]
= 0.98 A
The current flowing in the circuit is
[tex]I = I_{rms}\sqrt{2} \\\\ = (0.98) (1.414)[/tex]
= 1.39 A
And, finally, the current across the generator is
[tex]I'= I cos \phi[/tex]
[tex]= (1.39) cos 42.3^{\circ}[/tex]
= 1.03A
Hence, the magnitude of the circuit current is 1.03A
In Young's double slit experiment if the maximum intensity of light is Imax, then the intensity at path difference λ/2 will be
A .Imax
B .2Imax
C .4Imax
D. None
Answer:
Explanation:
When the path difference is half the wave length or λ /2 , destructive interference takes place which results in reduced or zero intensity in case equal intensity waves interfere as in Young's double slit experiment
Hence dark fringe is formed at that place where intensity is zero .
Hence the right option is D
A variable force of 6x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 18 ft. (Round your answer to two decimal places.) ft-lb
Answer:
931.00ft-lb
Explanation:
Pls see attached file
The work done in moving the object from x = 1 ft to x = 18 ft is 935 ft-lb.
What is work?
Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.
Given that force = 6x - 2 pounds.
So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]
= [ 3x² - 2x]¹⁸₁
= 3(18² - 1² ) - 2(18-1) ft-lb
= 935 ft-lb.
Hence, the work done is 935 ft-lb.
Learn more about work here:
https://brainly.com/question/18094932
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If the magnetic field steadily decreases from BBB to zero during a time interval ttt, what is the magnitude III of the induced current
Answer:
Using ohms law
The current is found from Ohm's Law.
I = V /R = E /R = Bxy /Rt.
A container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37, what is the index of refraction of the second fluid
Answer:
The refractive index of fluid 2 is 1.78
Explanation:
Refractive index , n = real depth/apparent depth
For the first fluid, n = 1.37 and apparent depth = 9.00 cm.
The real depth of the container is thus
real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm
To find the refractive index of fluid index of fluid 2, we use the relation
Refractive index , n = real depth/apparent depth.
Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.
So, n = 12.33 cm/6.86 cm = 1.78
So the refractive index of fluid 2 is 1.78
Since the same container is used, real depth of fluid 1 is equal to the real depth of fluid 2. The index of refraction of the second fluid is 1.8
Given that a container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37,
Then,
Index of refraction = [tex]\frac{Real depth}{Apparent depth}[/tex]
Real depth = Index of refraction x apparent depth
Since the same container is used, we can make an assumption that;
real depth of fluid 1 = real depth of fluid 2
That is,
1.37 x 9 = n x 6.86
Where n = Index of refraction for the second fluid.
make n the subject of formula
n = 12.33 / 6.86
n = 1.79
Therefore, the index of refraction of the second fluid is 1.8 approximately.
Learn more about refraction here: https://brainly.com/question/10729741
A spring balance is attached with string to the piece of aluminum in the preceding problem. What reading will the balance register when the metal is submerged
A cylindrical rod 120 mm long and having a diameter of 15 mm is to be deformed using a tensile load of 35,000 N. It must not experience either plastic deformation or a diameter reduction of more than 1.2x10-2 mm. Of the materials listed below, which are possible candidates? Justify your choice(s).
Material Modulus of Elasticity (GPa) Yield Strength (MPa) Poisson's ratio
Titanium alloy 70 250 0.33
Steel alloy 105 850 0.36
Magnesium alloy 205 550 0.27
Aluminium alloy 45 170 0.20
Answer:
The only suitable candidate is steel.
Explanation:
We are given;
Initial length;L_o = 120 mm = 0.12 m
Diameter;d_o = 15 mm = 0.015 m
Radius;r = 0.015/2 m = 0.0075 m
Applied force;F = 35000 N
Allowable Diameter reduction;Δd = 1.2 x 10^(-2) mm = 0.012 mm
Formula for Poisson’s ratio is;
v = ε_x/ε_z
Where;
ε_x is lateral strain = = Δd/d_o
ε_z is axial strain = σ/E
Thus;
v = (Δd/d_o)/(σ/E)
Making Δd the subject, we have;
Δd = (vσd_o)/E
Where σ is force per unit area
σ = F/A = 35000/πr² = 35000/(π0.0075)² = 198 × 10^(6) Pa
Now, let's find Δd for each of the metals given;
For Titanium alloy, E = 70 GPa = 70 × 10^(9) Pa and Poisson’s ratio (v) = 0.33
Thus;
Δd_ti = (0.33 × 198 × 10^(6) × 0.015)/70 × 10^(9) = 0.000014m = 0.014 mm
Similarly, for steel;
Δd_st = (0.36 × 198 × 10^(6) × 0.015)/(105 × 10^(9)) = 0.0000102 m = 0.0102 mm
For magnesium;
Δd_mn = (0.27 × 198 × 10^(6) × 0.015)/(205 × 10^(9)) = 0.0000039 m = 0.0039 mm
For aluminum;
Δd_al = (0.2 × 198 × 10^(6) × 0.015)/(45 × 10^(9)) = 0.0000132 m = 0.0132 mm
Since they must not experience a diameter reduction of more than 0.012 mm, then the only alloy that has a reduction diameter less than 0.012 is steel which has 0.0102.
Thus,the only possible candidate is stee.
A bug is sitting on the edge of a rotating disk. At what angular velocity will the bug slide off the disk if its radius is 0.241 m, the coefficient of static friction between the bug and disk is 0.321, and the coefficient of kinetic friction is 0.102
Answer:
ω = 3.61 rad/sec
Explanation:
Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.
μmg = mv^2/r = mω^2r
Thus;
μg = ω^2r
ω^2 = μg/r
ω = √(μg/r)
ω = √(0.321 * 9.8)/0.241
ω = √(13.05)
= 3.61 rad/sec
Wiley Coyote has missed the elusive road runner once again. This time, he leaves the edge of the cliff at 52.4 m/s with a purely horizontal initial trajectory. If the canyon is 141 m deep, how far from the base of the cliff does the coyote land
Answer:
280.86 mExplanation:
Range is defined as the distance covered in the horizontal direction. In projectile, range is expressed as x = vt where;
x is the range
v is the velocity of the runner
t is the time taken
Before we can get the range though, we need to find the time taken t using the relationship S = ut + 1/2gt²
if u = 0
S = 1/2gt²
2S = gt²
t² = 2S/g
t = √2S/g
t = √2(141)/9.8
t = √282/9.8
t = 5.36secs
The range x = 52.4*5.36
x =280.86 m
Hence, the coyort lies approximately 280.86 m from the base of the cliff
A coil of 160 turns and area 0.20 m2 is placed with its axis parallel to a magnetic field of initial magnitude 0.40 T. The magnetic field changes uniformly from 0.40 T in the +x direction to 0.40 T in the -x direction in 2.0 s. If the resistance of the coil is 16 Ω, at what rate is power generated in the coil?
Answer:
The rate at which power is generated in the coil is 10.24 Watts
Explanation:
Given;
number of turns of the coil, N = 160
area of the coil, A = 0.2 m²
magnitude of the magnetic field, B = 0.4 T
time for field change = 2 s
resistance of the coil, R = 16 Ω
The induced emf in the coil is calculated as;
emf = dΦ/dt
where;
Φ is magnetic flux = BA
emf = N (BA/dt)
emf = 160 (0.4T x 0.2 m²)/dt
emf = 12.8 V/s
The rate power is generated in the coil is calculated as;
P = V²/ R
P = (12.8²) / 16
P = 10.24 Watts
Therefore, the rate at which power is generated in the coil is 10.24 Watts
show that energy dissipated due to motion of a conductor in the magnetic field is due to mechanical energy.
Answer:
P = BILv = Iε, it shows that the mechanical energy due to the force on the conductor equals the electrical energy dissipated due to the motion of the conductor in the magnetic field and so, energy is conserved.
Explanation:
The force, F on a conductor of length, L in a magnetic field of magnetic field strength, B and current, I flowing through it is given by
F = BIL.
Now, if the conductor has a velocity, v, the energy dissipated by this force is P
P = Fv = BIL × v = BILv.
Now, we know that the induced e.m.f due to the motion of the conductor is given by ε = BLv.
From P above, P = BILv = I(BLv)
substituting ε = BLv into P, we have
P = Iε
Thus, P is the electrical energy dissipated due to the motion of the conductor.
Now since P = BILv = Iε, it shows that the mechanical energy due to the force on the conductor equals the electrical energy dissipated due to the motion of the conductor in the magnetic field and so, energy is conserved.
A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then the speed of efflux of non viscous water through the opening will be
Answer:
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
Explanation:
Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ([tex]P_{atm}[/tex]), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:
[tex]P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}[/tex]
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Water total pressures inside the tank and at ground level, measured in pascals.
[tex]\rho[/tex] - Water density, measured in kilograms per cubic meter.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Water speeds inside the tank and at the ground level, measured in meters per second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Heights of the tank and ground level, measured in meters.
Given that [tex]P_{1} = P_{2} = P_{atm}[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]z_{1} = 6.9\,m[/tex] and [tex]z_{2} = 4.9\,m[/tex], the expression is reduced to this:
[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)[/tex]
And final speed is now calculated after clearing it:
[tex]v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}[/tex]
[tex]v_{2} \approx 6.263\,\frac{m}{s}[/tex]
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
To work on your car at night, you use an extension cord to connect your work light to a power outlet near the door. How would the illumination provided by the light be affected by the length of the extension cord
Answer:
The longer the cord, the lower the illumination
Explanation:
The illumination provided by the light bulb will be reduced as the length of the extension cord increases. This is because the resistance provided by the wire increases with its length.
Long wires have more electrical resistance than shorter ones.
Let us consider this formula:
Resistance =[tex]\frac{\rho L}{A}[/tex]
From this formula, we can see that as the length increases, the resistance to current flow offered by the wire increases also provided the resistivity and cross-sectional area of the wire remain constant. As a result of this, the illumination will drop.
1.5 kg of air within a piston-cylinder assembly executes a Carnot power cycle with maximum and minimum temperatures of 800 K and 300 K, respectively. The heat transfer from the air during the isothermal compression is 80 kJ. At the end of the isothermal compression, the volume is 0.2 m3. Determine the volume at the beginning of the isothermal compression, in m3. Assume the ideal gas model for air and neglect kinetic and potential energy effects.
Answer:
Explanation:
Carton cycle consists of four thermodynamic processes . The first is isothermal expansion at higher temperature , then adiabatic expansion which lowers the temperature of gas . The third process is isothermal compression at lower temperature and the last process is adiabatic compression which increases the temperature of the gas to its original temperature .
So the given process of isothermal compression must have been done at the temperature of 300K , keeping the temperature constant .
Work done on gas at isothermal compression is equal to heat transfer .
work done on gas = 80 x 10³ J
work done on gas = n RT ln v₁ / v₂
n is number of moles v₁ and v₂ are initial and final volume
molecular weight of gas = 28.97 g
1.5 kg = 1500 / 28.97 moles
= 51.77 moles
work done on gas = n RT ln v₁ / v₂
Putting the values in the equation above
80 x 10³ = 51.78 x 8.31 x 300 x ln v₁ / .2
ln v₁ / .2 = .62
v₁ / .2 = 1.8589
v₁ = 0.37 m³
Two 2.0 g plastic buttons each with + 65 nC of charge are placed on a frictionless surface 2.0 cm (measured between centers) on either side of a 5.0 g button charged to +250 nC. All three are released simultaneously.
a. How many interactions are there that have a potential energy?
b. What is the final speed of the left 2.0 g plastic button?
c. What is the final speed of the right 2.0 g plastic button?
d. What is the final speed of the 5.0 g plastic button?
Answer:
A) 3 interactions
B) 2.78 m/s
C) 2.78 m/s
D) 0 m/s
Explanation:
Given Data :
plastic buttons ( 2 ) = 2.0 g each
charge on each plastic buttons = +65 nC
A) The interactions that have potential energy can be expressed as
V net = v1 + v2 + v3
V net = 0 + [tex]\frac{kq1q2}{r12} + \frac{kq1q3}{r13} + \frac{kq2q3}{r23}[/tex]
hence the total interactions is (3)
B) The final speed of the left 2.0 g plastic button
U = [tex]\frac{9*10^9* 10^-18}{0.02} [ 65*250 + 65*250 + (65*65)/2 ][/tex]
U = 0.0155 J
also U = [tex]2( 1/2 mv^2 )[/tex]
U = mv^2 = 0.0155 J
therefore ; v = [tex]\sqrt{0.0155/(2*10^-3)}[/tex] = 2.78 m/s
C) final speed of the right 2.0 g plastic button = Final speed of the right 2.0 g plastic button
v = 2.78 m/s
D) The final speed of the 5.0 g plastic button
= 0 m/s