Answer:
a) [tex]\Delta S_{in} = 2.791\,\frac{J}{K}[/tex], b) [tex]\Delta S_{out} = 6.897\,\frac{J}{K}[/tex], c) [tex]S_{gen} = 4.106\,\frac{J}{K}[/tex], d) Due to irreversibilities due to temperature differences.
Explanation:
a) The change in entropy of the hot reservoir is:
[tex]\Delta S_{in} = \frac{2400\,J}{860\,K}[/tex]
[tex]\Delta S_{in} = 2.791\,\frac{J}{K}[/tex]
b) The change in entropy of the cold reservoir is:
[tex]\Delta S_{out} = \frac{2400\,J}{348\,K}[/tex]
[tex]\Delta S_{out} = 6.897\,\frac{J}{K}[/tex]
c) The total change in entropy of the Universe is modelled after the Second Law of Thermodynamics. Let assume that process is steady:
[tex]\Delta S_{in} - \Delta S_{out} + S_{gen} = 0[/tex]
[tex]S_{gen} = \Delta S_{out} - \Delta S_{in}[/tex]
[tex]S_{gen} = 6.897\,\frac{J}{K} - 2.791\,\frac{J}{K}[/tex]
[tex]S_{gen} = 4.106\,\frac{J}{K}[/tex]
d) Since irreversibilities create entropy as process goes by. The main source of irreversibilities is the existence of temperature differences.
A soccer player is benched for being late to the game. In a fit of anger, she drops her ball from the top of the Physics building. It falls 4.9 meters after 1.0 second has elapsed. How much farther does it fall in the next 2.0 seconds
Answer:
The distance is [tex]S = 39.2 \ m[/tex]
Explanation:
From the question we are told that
The distance covered after t = 1 s is [tex]d = 4.9 \ m[/tex]
According to the equation of motion
[tex]v^2 = u^2 + 2ad[/tex]
Now u = 0 m/s since before the drop the ball was at rest
[tex]v^2 = 2ad[/tex]
here [tex]a =g = 9.8 \ m/s^2[/tex]
So
[tex]v = 9.8 m/s[/tex]
Also from equation of motion we have that
[tex]S = ut + \frac{1}{2} at^2[/tex]
Now at t = 2 s , as given from the question
Then u = v = 9.8 m/s
And
[tex]S = 9.8 * 2 + \frac{1}{2} * (9.8) * (2^2)[/tex]
[tex]S = 9.8 * 2 + \frac{1}{2} * (9.8) * (2^2)[/tex]
[tex]S = 39.2 \ m[/tex]
Which statement BEST explains the relationship between voltage, current, and power?
A. If voltage increases and everything else remains constant, then power will increase.
B. If voltage increases and everything else remains constant, then power will decrease.
C. If current decreases and everything else remains constant, then power will increase.
D. Voltage and power are inversely related.
A string is stretched between fixed supports separated by 72.0 cm. It is observed to have resonant frequencies of 370 and 555 Hz, and no other resonant frequencies between these two.(a) What is the lowest resonant frequency for this string?(b) What is the wave speed for this string?
Answer:
(a) f = 185 Hz
(b) v = 266.4 m/s
Explanation:
(a) The lowest frequency can be calculated by using the following formula for the calculation of the modes (resonant frequencies) in a string:
[tex]f_n=\frac{nv}{2L}[/tex]
[tex]f_n=nf[/tex]
n: order of the mode
v: velocity of the waves in the string
L: length of the string = 72.0cm = 0.72m
fn: frequency of the n-th mode
With the information about two consecutive modes you can find the lowest resonant frequency. First you find the resonant mode n:
[tex]f_n=nf\\\\f_{n-1}=(n-1)f\\\\\frac{f_n}{f_{n-1}}=\frac{n}{n-1}[/tex]
you solve the previous equation for n:
[tex](n-1)f_n=nf_{n-1}\\\\555n-555=370n\\\\n=3[/tex]
With this information you can calculate the lowest resonant frequency:
[tex]f_n=nf\\\\f=\frac{f_n}{n}=\frac{555}{3}=185Hz[/tex]
b) You have information about two consecutive modes fn, fn-1. Then, you can calculate the velocity of the waves:
[tex]f_{n}-f_{n-1}=n\frac{v}{2L}-(n-1)\frac{v}{2L}\\\\f_n-f_{n-1}=\frac{v}{2L}\\\\v=2L(f_n-f_{n-1})[/tex]
fn = 555 Hz
fn-1: 370 Hz
[tex]v=2(0.72m)(555-370)Hz=266.4\frac{m}{s}[/tex]´
hence, the velocityof the waves in the string is 266.4 m/s
Carbon is added to iron to make steel. Steel is stronger than either carbon or iron by itself.
What does this example show?
Answer:
This example shows that alloys are stronger than either of it's parent materials by themselves.
Explanation:
Since carbon is added to iron to make steel, it means steel is an alloy of iron and carbon.
This is because an alloy is a mixture of two or more elements, where at least one element is a metal.
Now, steel is stronger than either carbon or or iron by itself because Steel contains atoms of other elements including carbon and iron. These atoms have different sizes to iron carbon atoms, so they distort the layers of atoms in the pure iron and carbon. This means that a greater force is required for the layers to slide over each other in steel, so steel is harder than pure iron.
Q) Suppose, you are in a sporting event. You notice that everyone stands up when it’s his turn,
creating a wave that moves through the crowd and they sit back down again after a while. This wave
move around the stadium without moving the people around it. Considering this situation, justify
your answer about nature of wave.
Answer:
The nature of the wave formed is a transverse progressive wave.
Explanation:
A wave is a disturbance that travels through a material medium without permanent displacement of the particles of the medium. The two major types are: transverse and longitudinal.
A transverse wave is one in which the direction of vibration of the particles of the medium is perpendicular to the direction of propagation of the wave. Examples are: water wave, light wave etc. While a longitudinal wave is one in which the direction of vibrations of the particles of the medium is parallel with the direction of propagation of the wave, creating a region of rarefaction and compression. Examples are; sound wave, wave in a rope, wave in a slinky etc.
The cited wave formed in the given question is a transverse wave because each person stands and sits after some time to create a moving (progressive) wave without them moving from their positions.
Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.
Car A: 500 kg, 10 m/s,
Car B: 2000 kg, 5 m/s,
Car C: 500 kg, 20 m/s,
Car D: 1000 kg, 20 m/s,
Car E: 4000 kg, 5 m/s, and
Car F: 1000 kg, 10 m/s.
(a) Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.
(b) Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.
Answer:
a)Car E = Car D > (Car F = Car B = Car C) > Car A
b)Car E = Car D > (Car F = Car B = Car C) > Car A
Explanation:
Car A: mass = 500 kg; speed = 10 m/s
Car B: mass = 2000 kg;speed = 5 m/s
Car C:mass = 500 kg; speed = 20 m/s
Car D: mass = 1000 kg; speed = 20 m/s
Car E:mass = 4000 kg; speed = 5 m/s
Car F: mass = 1000 kg; speed = 10 m/s
Part a) Now we know that momentum of each car is product of mass and velocity , so we will have
CarA:
[tex]P_1 = m \times v\\P_1 = (500)(10)\\P_1 = 5 \times 10^3 kg m/s[/tex]
Car B:
[tex]P_2 = m v\\P_2 = (2000)(5)\\P_2 = 10^4 kg m/s[/tex]
Car C:
[tex]P_3 = m v\\P_3 = (500)(20)\\P_3 = 10^4 kg m/s[/tex]
Car D:
[tex]P_4 = m v\\P_4 = (1000)(20)\\P_4 = 2\times 10^4 kg m/s[/tex]
Car E:
[tex]P_5 = m v\\P_5 = (4000)(5)\\P_5 = 2\times 10^4 kg m/s[/tex]
Car F:
[tex]P_6 = m v\\P_6 = (1000)(10)\\P_6 = 10^4 kg m/s[/tex]
So the momentum is given as ,
Car E = Car D > (Car F = Car B = Car C) > Car A
Part b)Impulse is given as change in momentum so here we can say that final momentum of all the cars will be zero as they all stops and hence the impulse is same as initial momentum of the car
so the order of impulse from largest to least is given as
Car E = Car D > (Car F = Car B = Car C) > Car A
During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a velocity of 2.60 m/s, grabs and holds onto him so that they move off together with a velocity of 1.30 m/s. If the mass of the tackler is 122 kg, determine the mass of the receiver. Assume momentum is conserved.
Answer:
122kgExplanation:
Using the law of conservation of momentum which states that 'the sum of momentum of bodies before collision is equal to their sum after collision. The bodies will move together with a common velocity after collision.
Momentum = Mass * Velocity
Before collision;
Momentum of receiver m1u1= 0 kgm/s (since the receiver is standing still)
Momentum of the tackler
m2u2 = 2.60*122 = 317.2 kgm/s
where m2 and u2 are the mass and velocity of the tacker respectively.
Sum of momentum before collision = 0+317.2 = 317.2 kgm/s
After collision
Momentum of the bodies = (m1+m2)v
v = their common velocity
m1 = mass of the receiver
Momentum of the bodies = (122+m1)(1.30)
Momentum of the bodies = 158.6+1.30m1
According to the law above;
317.2 = 158.6+1.30m1
317.2-158.6 = 1.30m1
158.6 = 1.30m1
m1 = 158.6/1.30
m1 = 122kg
The mas of the receiver is 122kg
A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled has an acceleration of 2.10 m/s2. Part A By how much does the spring stretch if it pulls on the sled horizontally
Answer:
Stretch in the spring = 0.1643 (Approx)
Explanation:
Given:
Mass of the sled (m) = 9 kg
Acceleration of the sled (a) = 2.10 m/s ²
Spring constant (k) = 115 N/m
Computation:
Tension force in the spring (T) = ma
Tension force in the spring (T) = 9 × 2.10
Tension force in the spring (T) = 18.9 N
Tension force in the spring = Spring constant (k) × Stretch in the spring
18.9 N = 115 N × Stretch in the spring
Stretch in the spring = 18.9 / 115
Stretch in the spring = 0.1643 (Approx)
A merry-go-round is shaped like a uniform disk and has moment of inertia of 50,000 kg m 2 . It is rotating so that it has an angular momentum of 10,000 (kg m 2 radians/s) and its outer edge has a speed of 2 m/s. What is its radius, in m
Answer:
r = 20 m
Explanation:
The formula for the angular momentum of a rotating body is given as:
L = mvr
where,
L = Angular Momentum = 10000 kgm²/s
m = mass
v = speed = 2 m/s
r = radius of merry-go-round
Therefore,
10000 kg.m²/s = mr(2 m/s)
m r = (10000 kg.m²/s)/(2 m/s)
m r = 5000 kg.m ------------- equation 1
Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:
I = (1/2) m r²
where,
I = moment of inertia = 50000 kg.m²
Therefore,
50000 kg.m² = (1/2)(m r)(r)
using equation 1, we get:
50000 kg.m² = (1/2)(5000 kg.m)(r)
(50000 kg.m²)/(2500 kg.m) = r
r = 20 m
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?
Sort the following quantities as known or unknown. Take the horizontal direction to be along the x axis.
mQ: the mass of the quarterback
mB: the mass of the football
(vQx)i: the horizontal velocity of quarterback before throwing the ball
(vBx)i: the horizontal velocity of football before being thrown
(vQx)f: the horizontal velocity of quarterback after throwing the ball
(vBx)f: the horizontal velocity of football after being thrown
Answer:
vBxf = 0.08625m/s
Explanation:
This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.
[tex]p_f=p_i[/tex]
pf: final momentum
pi: initial momentum
The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.
Then, you have:
[tex]m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}[/tex] (1)
mQ: the mass of the quarterback = 80kg
mB: the mass of the football = 0.43kg
(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s
(vBx)i: the horizontal velocity of football before being thrown = 0m/s
(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?
(vBx)f: the horizontal velocity of football after being thrown = 15 m/s
You replace the values of the variables in the equation (1), and you solve for (vBx)f:
[tex]0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}[/tex]
Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.
Hence, the speed of the quarterback after he throws the ball is 0.08625m/s
A glass sphere carrying a uniformly distributed charge of +Q is surrounded by an initially neutralspherical plastic shell. (Assume the charge +Q is uniformly distributed across thesurface of the glass sphere.)
Required:
a. Qualitatively, indicate the polarization of the plastic.
1. The plastic will polarize so as to have positive charge +Qon its inner surface and negativecharge −Q on its outer surface.
2. Dipoles in the plastic will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center.
4. Dipoles in the plastic will polarize and orient themselves radially, with their positiveendspointing toward the center.
b. Qualitatively, indicate the polarization of the inner glass sphere. Explain briefly.A net charge −Q from the dipoles will be uniformly distributed through the volume of the sphere.
1. There will be no polarization inside the glass sphere since the net electric field there iszero.
2. Dipoles in the glass will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the glass will polarize and orient themselves radially, with their positive endspointing toward the center.
c. Is the electric field at location Poutside the plastic shell larger, smaller, or the same as itwould be if the plastic weren't there? Explain briefly.
1. Larger, because a net positive charge is created from the polarization of the shell.
2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.
3. Smaller, because the negative charges displaced during polarization are closer to Pthanthe positive charges.
4. Smaller, because the plastic shell shields location Pfrom the charge +Q, such that the netfield at Pis zero.
5. The same, because no net charge is created from the polarization of the field.
Answer:
(A) 3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center
(B) 2. There will be no polarization inside the glass sphere since the net electric field there is zero.
Explanation: charges are only distributed on the surface of the charged hollow conductor. The core must have zero charge.
(C) 2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.
A balloon with a radius of 16 cm has an electric charge of 4.25 10 –9 C.
Determine the electric field strength at a distance of 40.0 cm from the balloon’s centre.
Answer:
239 N/C
Explanation:
Electric field strength at distance R from a charge Q is given by the expression
E = k Q / R² where Q is charge , R is distance of charge from the point . k is a constant .
R = 40 cm , Q = 4.25 x 10⁻⁹
Putting the given values
E = 9 x 10⁹ x 4.25 x 10⁻⁹ / ( 40 x 10⁻²)²
= 239 N/C .
Un levantador de pesas puede generar 3000 N de fuerza ¿Cuál es el peso máximo que puede levantar con una palanca que tiene un brazo de la fuerza de 2 m y un brazo de resistencia de 50 cm?
Responder: 12000N
Explicación: Usando la fórmula para encontrar la eficiencia de una máquina. Eficiencia = ventaja mecánica / relación de velocidad × 100%
Dado MA = Carga / Esfuerzo
Relación de velocidad = distancia recorrida por esfuerzo (brazo de fuerza) / distancia recorrida por carga (brazo de resistencia)
MA = Carga / 3000
VR = 2 / 0.5 VR = 4
Asumiendo que la eficiencia es 100% 100% = (Carga / 3000) / 4 × 100%
1 = (Carga / 3000) / 4
4 = Carga / 3000
Carga = 4 × 3000
Carga = 12000N
Esto significa que el peso máximo que se puede levantar es 12000N
first law of equilibrium
Answer:
for an object to be in equilibrium, it must be experiencing no acceleration. Both the net force and the net torque must be zero.
Hope I helped
Answer:
An object in static equilibrium has zero net force acting upon it.
The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes. This can be written as
F = F1+ F2+ F3+ F4+. . . = 0
An arrow is shot from a height of 1.55 m toward a cliff of height H. It is shot with a velocity of 26 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.99 s later.
(a) Draw a sketch of the given example. Include the x-y coordinate system.
(b) What is the height of the cliff?
(c) What is the maximum height reached by the arrow along its trajectory?
(d) What is the arrow's impact speed just before hitting the cliff?
Answer:
Explanation:
vertical component of the velocity of arrow
= 26 sin 60 = 22.516 m
height reached by it after 3.99 s
h = ut - 1/2 g t²
= 22.516 x 3.99 - .5 x 9.8 x 3.99²
= 89.83 - 78
11.83 m
Total height of cliff = 1.55 + 11.83
= 13.38 m
c ) maximum height covered s
v² = u² - 2gs
0 = u² - 2gs
s = u² / 2g
= 22.516² / 2 x 9.8
= 25.86
maximum height reached
= 25.86 + 1.55
= 27.41 m
d )
vertical speed after 3.99 s
v = u - gt
= 22.516 - 9.8 x 3.99
= -16.586
Horizontal component will remain unchanged
Horizontal component = 26 cos 60
= 13 m /s
Resultant of two velocities
= √ 13²+ 16.568²
= 21 m /s
A meter stick hurtles through space at a speed of 0.95c relative to you, with its length perpendicular to the direction of motion. You measure its length to be equal to:_______
a. 0 m.
b. 0.05 m.
c. 0.95 m.
d. 1.00 m.
e. 1.05 m.
Answer:
d. 1.00 m
Explanation:
In 1905, Einstein proposed special theory of relativity of light.
This theory had a number of consequences or results. One of them is called "Length Contraction".
According to this consequence, whenever an object travels at a speed comparable to the speed of light, its length decreases.
But this decrease in length is only seen in the dimension, which is parallel to the direction of motion of the body. All other dimensions of the object remains same.
In the given situation, the length of meter stick is not parallel to the direction of motion, but it is perpendicular. Hence, the length of meter stick will be same as the length of meter stick at rest. Hence, the correct option will be:
d. 1.00 m
3. A ray of light incident on one face of an equilateral glass prism is refracted in such a way that it emerges from the opposite surface at an angle of 900 to the normal. Calculate the i. angle of incidence. ii. minimum deviation of the ray of light passing through the prism [n_glass=1.52]
Answer:
i) angle of incidence;i = 29.43°
ii) δm = 38.92°
Explanation:
Prism is equilateral so angle of prism (A) = 60°
Refractive index of glass; n_glass = 1.52
A) Let's assume the incident angle = i and Critical angle = θc
We know that, sin θc = 1/n
Thus;
sin θc = 1/n_glass
θc = sin^(-1) (1/n_glass)
θc = sin^(-1) (1/1.52)
θc = 41.14°
Now, the angle of prism will be the sum of external angle that is critical angle and reflected angle.
Thus;
A = r + θc
r = A - θc
So;
r = 60° - 41. 14°
r = 18.86°
From, Snell's law. If we apply it to this question, we will have;
(sin i)/(sin r) = n_glass
Where;
i is angle of incidence and r is angle of reflection.
Let's make i the subject;
i = sin^(-1) (n_glass × sin r)
i = sin^(-1) (1.52 × sin 18.86)
i = sin^(-1) 0.4914
i = 29.43°
B) The formula to calculate minimum deviation would be from;
μ = [sin ((A + δm)/2)]/(sin A/2)
Where;
μ is Refractive index
δm is minimum angle of deviation
A is angle of prism
Now Refractive index is given by a formula; μ = (sin i)/(sin r)
So; μ = (sin 29.43)/(sin 18.86)
μ = 1.52
Thus;
1.52 = [sin ((60 + δm)/2)]/(sin 60/2)
1.52 * sin 30 = sin ((60 + δm)/2)
0.76 = sin ((60 + δm)/2)
sin^(-1) 0.76 = ((60 + δm)/2)
49.46 × 2 = (60 + δm)
98.92 - 60 = δm
δm = 38.92°
Water, in a 100-mm-diameter jet with speed of 30 m/s to the right, is deflected by a cone that moves to the left at 14 m/s. Determine (a) the thickness of the jet sheet at a radius of 230 mm. and (b) the external horizontal force needed to m
Answer:
Explanation:
The velocity at the inlet and exit of the control volume are same [tex]V_i=V_e=V[/tex]
Calculate the inlet and exit velocity of water jet
[tex]V=V_j+V_e\\\\V=30+14\\\\V=44m/s[/tex]
The conservation of mass equation of steady flow
[tex]\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0[/tex]
[tex]A_i\ \texttt {is the inlet area of the jet}[/tex]
[tex]A_e\ \texttt {is the exit area of the jet}[/tex]
since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal
The expression for thickness of the jet
[tex]A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}[/tex]
R is the radius
t is the thickness of the jet
D_j is the diameter of the inlet jet
[tex]t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm[/tex]
(b)
[tex]R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)][/tex]
[tex]1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j[/tex]
[tex]R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N[/tex]
The negative sign indicate that the direction of the force will be in opposite direction of our assumption
Therefore, the horizontal force is -7603N
A 2 kg object is subjected to three forces that give it an acceleration −→a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj. If two of the three forces, are −→F1 = (30.0N)ˆi + (16.0N)ˆj and −→F2 = −(12.0N)ˆi + (8.00N)ˆj, find the third force.
Answer:
[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]
Explanation:
You have three forces F1, F2 an F3 that produce the following acceleration:
a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj
you know that force F1 and F2 are:
F1 = (30.0N)ˆi + (16.0N)ˆj
F2 = −(12.0N)ˆi + (8.00N)ˆj
and the force F3 is unknown:
F3 = F3x ˆi + F3y ˆj
The second Newton law is given by the following equation:
[tex]\vec{F}=m\vec{a}[/tex]
F: the total force = F1 +F2 + F3
m: mass of the object = 2 kg
By the properties of vectors you have:
[tex]\vec{F_1}+\vec{F_2}+\vec{F_3}=m\vec{a}\\\\(30.0-12.0+F_{3x})N\hat{i}+(16.0+8.00+F_{3y})N\hat{j}=(2.0kg)[(-8.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}]\\\\(18.0+F_{3x})N\hat{i}+(24.0+F_{3y})\hat{j}=(-16.00N)\hat{i}+(12.0N)\hat{j}[/tex]
Both x and y component must be equal in the previous equality, then you have:
[tex]18.0N+F_{3x}=-16.00N\\\\F_{3x}=-34.00N\\\\24.0N+F_{3y}=12.0N\\\\F_{3y}=-12.00N[/tex]
Hence, the vector F3 is:
[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]
A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. 6.34). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.80 m/s.
(a) What is the tension in the cord in the original situation when the block has speed v = 0.70 m/s? (b) What is the tension in the cord in the final situation when the block has speed v = 2.80 m/s? (c) How much work was done by the person who pulled on the cord?
Answer:
a) 0.147 N
b) 9.408 N
c) 9.261 N
Explanation:
The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force [tex]\frac{mv^{2} }{r}[/tex]. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.
9. How do air masses move?
Answer:
Air masses move with the global pattern of winds. In most of the United States, air masses generally move from west to east. They may move along with the jet stream in more complex and changing patterns.
Convert from scientific notation to standard form
9.512 x 10-8
The temperature at the surface of the Sun is approximately 5,300 K, and the temperature at the surface of the Earth is approximately 293 K. What entropy change of the Universe occurs when 6.00 103 J of energy is transferred by radiation from the Sun to the Earth?
Answer:
The entropy change of the Universe that occurs is 19.346 J/K
Explanation:
Given;
temperature of the sun, [tex]T_s[/tex] = 5,300 K
temperature of the Earth, [tex]T_E[/tex] = 293 K
radiation energy transferred by the sun to the earth, E = 6000 J
The sun loses Q of heat and therefore decreases its entropy by the amount
[tex]\delta S_{sun} = \frac{-Q}{T_s}[/tex]
The earth gains Q of heat and therefore increases its entropy by the amount
[tex]\delta S_{Earth} = \frac{-Q}{T_E}[/tex]
The total entropy change is:
[tex]\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\ = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K[/tex]
Therefore, the entropy change of the Universe that occurs is 19.346 J/K
A parallel-plate capacitor in air has a plate separation of 1.30 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 255 V and dis-connected from the source. The capacitor is then immersed in distilled water. Determine a) the charge on the plates before and after immersion.b) the capacitance and potential difference after immersion.c) the change in energy of the capacitor.
Answer:
Explanation:
capacitance of air capacitor
C = ε₀ A / d
ε₀ is permittivity of medium , A is plate area , d is distance between plate .
C = 8.85 x 10⁻¹² x 25 x 10⁻⁴ / 1.3 x 10⁻²
= 170.19 x 10⁻¹⁴ F .
charge on the capacitor when it is charged to potential of 255 V
= CV , C is capacitance and V is potential
= 170.19 x 10⁻¹⁴ x 255
= 4.34 x 10⁻¹⁰ C .
After it is disconnected from the source , and it is immersed in water , charge on it remains the same .
So its charge when immersed in water will be constant at 4.34 x 10⁻¹⁰ C.
b )
When it is immersed in water its capacity increases k times where k is dielectric constant of water which is 80 .
capacitance of capacitor in water = 80 x 170.19 x 10⁻¹⁴ F
= 13615.2 x 10⁻¹⁴ F .
= 1.36 x 10⁻¹⁰ F
potential difference = charge / capacitance
= 4.34 x 10⁻¹⁰ / 1.36 x 10⁻¹⁰
= 3.2 V
c )
Energy of capacitor = 1/2 C V²
Initial energy = 1/2 x 170.19 x 255² x 10⁻¹⁴
= 55.33 x 10⁻⁹ J
Final energy = 1/2 x 1.36 x 10⁻¹⁰ x 3.2²
= .7 x 10⁻⁹ J .
decrease of energy = 54.63 x 10⁻⁹ J .
A population _____ follows a period of
Answer:
a population increase
Explanation:
During the 20th century, the world population increased from 1.65 billion to 6 billion. In 1970, the world's population was half that of today. In less than 15 years, 47% of the population will live in areas already under heavy water stress. In Africa, between 75 and 250 million people will face growing shortages in 2020 due to climate change. The scarcity of some arid and semi-arid regions will have a decisive impact on migration.
Photoelectric effect:
A. What is the maximum kinetic energy of electrons ejected from barium (W0=2.48eV) when illuminated by white light, lambda=410-750nm?
B. The work functions for sodium, cesium, copper, and iron are 2.3, 2.1, 4.7, and 4.5eV, respectively. Which of these metals will not emit electrons when visible light shines on it?
Answer:
A. K = 0.546 eV
B. cooper and iron will not emit electrons
Explanation:
A. This is a problem about photoelectric effect. Then you have the following equation:
[tex]K=h\nu-\Phi=h\frac{c}{\lambda} -\Phi[/tex] (1)
K: kinetic energy of the ejected electron
Ф: Work function of the metal = 2.48eV
h: Planck constant = 4.136*10^{-15} eV.s
λ: wavelength of light = 410nm - 750nm
c: speed of light = 3*10^8 m/s
As you can see in the equation (1), higher the wavelength, lower the kinetic energy. Then, the maximum kinetic energy is obtained with the lower wavelength (410nm). Thus, you replace the values of all variables :
[tex]K=(4.136*10^{-15}eV)\frac{3*10^8m/s}{410*10^{-9}m}-2.48eV\\\\K=0.546eV[/tex]
B. First you calculate the energy of the photon with wavelengths of 410nm and 750nm
[tex]E_1=(4.136*10^{-15}eV)\frac{3*10^{8}m/s}{410*10^{-9}m}=3.02eV\\\\E_2=(4.13610^{-15}eV)\frac{3*10^{8}m/s}{750*10^{-9}m}=1.6544eV[/tex]
You compare the energies E1 and E2 with the work functions of the metals and you can conclude:
sodium = 2.3eV < E1
cesium = 2.1 eV < E1
cooper = 4.7eV > E1 (this metal will not emit electrons)
iron = 4.5eV > E1 (this metal will not emit electrons)
Question 7 of 10
The coefficient of kinetic friction between a couch and the floor is 0.4. If the
couch has a mass of 35 kg and you push it with a force of 200 N. what is the
net force on the couch as it slides?
O A. 337 N
B. 143 N
O C. 343 N
O D. 63 N
Answer:
D
Explanation:
Now the net force is the applied force minus the frictional force; this is expressed mathematically as:
Fnet= Fappplied - Ffrictional
Now the frictional force is given as ;
Coefficient of friction × normal reaction
Normal reaction is the weight of the human acting in opposite direction.
Normal reaction of the human is ;
35 × 9.8 = 343N { note that weight = m× g and g= 9.8m/S2, a known standard }
Hence the Frictional force =343×0.4 =137.20N
Hence Fnet = 200-137.20 = 62.8N
Fnet = 63N to the nearest whole
The net force on the couch as it slides is 63N.
What is frictional force?
When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.
The friction force is given by
f = coefficient of friction x Normal force
Given, the coefficient of kinetic friction between a couch and the floor is 0.4. If the couch has a mass of 35 kg and you push it with a force of 200 N.
Normal reaction is the weight of the human acting in opposite direction.
Normal reaction N =35 × 9.8 = 343N
Frictional force f =0.4 x 343
f =137.20N
The net force will be
Fnet= Fappplied - Ffrictional
Fnet = 200-137.20 = 62.8N
Fnet = 63N
Thus, the net force on the couch as it slides is 63N.
Learn more about friction force.
https://brainly.com/question/1714663
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Why do some nucleus emit electrons?
Answer:
In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus. In beta-plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.
Explanation:
Hope this helps!
A flat coil of wire is used with an LC-tuned circuit as a receiving antenna. The coil has a radius of 0.30 m and consists of 420 turns. The transmitted radio wave has a frequency of 1.3 MHz. The magnetic field of the wave is parallel to the normal of the coil and has a maximum value of 1.7 x 10-13 T. Using Faraday's Law of electromagnetic induction and the fact that the magnetic field changes from zero to its maximum value in one-quarter of a wave period, find the magnitude of the average emf induced in the antenna in this time.
Answer:
The average emf induce is [tex]V = 2.625 * 10^{-5} \ V[/tex]
Explanation:
From the question we are told that
The radius of the coil is [tex]r = 0.30 \ m[/tex]
The number of turns is [tex]N = 420 \ turns[/tex]
The frequency of the transition radio wave is [tex]f = 1.3\ MHz = 1.3 *10^{6} Hz[/tex]
The magnetic field is [tex]B_,{max} = 1.7 * 10^{-13} \ T[/tex]
The time taken for the magnetic field to go from zero to maximum is [tex]\Delta T = \frac{T}{4}[/tex]
The period of the transmitted radio wave is [tex]T = \frac{1}{f}[/tex]
So
[tex]\Delta T = \frac{T}{4} = \frac{1}{4 f}[/tex]
The potential difference can be mathematically represented as
[tex]V = NA (\frac{\Delta B}{\Delta T} )[/tex]
[tex]V = NA ([B_{max} - B_{min} ] * 4f)[/tex]
Where [tex]B_{min} = 0T[/tex]
substituting values
[tex]V = 420 * (\pi *(0.30)^2) * (1.7 *10^{-13} * 4 * 1.3 *10^{6})[/tex]
[tex]V = 2.625 * 10^{-5} \ V[/tex]
A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atmatm pressure is 1.00028.
Required:
How many bright-dark-bright fringe shifts are observed as the cell fills with air?
Answer:
55.3
Explanation:
The computation of the number of bright-dark-bright fringe shifts observed is shown below:
[tex]\triangle m = \frac{2d}{\lambda} (n - 1)[/tex]
where
d = [tex]3.95 \times 10^{-2}m[/tex]
[tex]\lambda = 400 \times 10^{-9}m[/tex]
n = 1.00028
Now placing these values to the above formula
So, the number of bright-dark-bright fringe shifts observed is
[tex]= \frac{2 \times3.95 \times 10^{-2}m}{400 \times 10^{-9}m} (1.00028 - 1)[/tex]
= 55.3
We simply applied the above formula so that the number of bright dark bright fringe shifts could come
