Answer:
scenario A and B
Explanation:
The Doppler effect is the change in frequency of a wave by the relative movement of the source and the observer. It is described by the expression
f ’= (v + v₀) / (v - [tex]v_{s}[/tex]) f₀
Where f₀ is the emitted frequency, v the speed of the wave, v₀ and [tex]v_{s}[/tex] the speed of the observer and the source, respectively, the signs are for when they are approaching and in the case of being away the signs are changed.
Consequently, from the above for the Doppler effect to exist there must be a relative movement of the source and the observer.
Let's examine the scenarios
A) True. You agree with the equation shown
B) True. Only the signs should be changed and it is described by the equation shown
C) False. If there is no relative movement there is no Doppler effect
A light ray enters a glass enclosed fish tank. From air it enters the glass at 20.° with respect to the surface, then emerges into the water. The index for glass is 1.50 and for water 1.33.
(a) What is the angle of refraction in the glass?
(b) What is the angle of refraction in the water?
(c) Is there any incident angle in air for which the ray will not enter the water due to total internal reflection?
Answer:
a) [tex]\angle r_{ag}=38.79^{\circ}[/tex]
b) [tex]\angle r_{gw}=44.95^{\circ}[/tex]
c) not possible
Explanation:
Given:
angle of incidence on the air-glass interface, [tex]\angle i_{ag}=90-20=70^{\circ}[/tex]
refractive index of glass with respect to air, [tex]n_g=1.5[/tex]
refractive index of water with respect to air, [tex]n_a=1.33[/tex]
a)
For angle of refraction in glass we use Snell's law:
[tex]n_g=\frac{sin\ i_{ag}}{sin\ r{ag}}[/tex]
[tex]1.5=\frac{sin\ 70}{sin\ r_{ag}}[/tex]
[tex]\angle r_{ag}=38.79^{\circ}[/tex]
b)
Now we have angle of incident for glass-water interface, [tex]\angle i_{gw}=\angle r_{ag}=38.79^{\circ}[/tex]
And the refractive index of water with respect to glass:
[tex]n_{gw}=\frac{n_w}{n_g}[/tex]
[tex]n_{gw}=\frac{1.33}{1.5}[/tex]
[tex]n_{gw}=0.8867 [/tex]
Therefore, angle of refraction in the water:
[tex]n_{gw}=\frac{sin\ i_{gw}}{sin\ r_{gw}}[/tex]
[tex]0.8867 =\frac{sin\ 38.79}{sin\ r_{gw}}[/tex]
[tex]\angle r_{gw}=44.95^{\circ}[/tex]
c)
For total internal reflection through water the light must enter the glass-water interface at an angle greater than the critical angle.
So,
[tex]n_{gw}=\frac{sin\ i_{(gw)_c}}{sin\ 90}[/tex]
[tex]0.8867 =\frac{sin\ i_{(gw)_c}}{1}[/tex]
[tex]i_{(gw)_c}=62.46^{\circ}[/tex]
Now this angle will become angle of refraction for the air-glass interface.
Hence,
[tex]n_g=\frac{sin\ i_{ag}}{sin\ i_{(gw)_c}}[/tex]
[tex]1.5=\frac{sin\ i_{ag}}{0.8867 }[/tex]
[tex]sin\ i_{ag}=1.33005[/tex]
Since we do not have any angle for sine for which the value exceeds 1, therefore it is not possible that the light will reflect from the water.
A piccolo and a flute can be approximated as cylindrical tubes with both ends open. The lowest fundamental frequency produced by one kind of piccolo is 516.1 Hz, and that produced by one kind of flute is 257.0 Hz. What is the ratio of the piccolo's length to the flute's length?
Answer:
0.49806
Explanation:
v = Velocity of wave
L = Length of tube
p denotes piccolo
f denotes flute
The fundamental frequency with both ends open is given by
[tex]f=\dfrac{v}{2L}\\\Rightarrow L=\dfrac{v}{2f}[/tex]
It can be seen that
[tex]L\propto \dfrac{1}{f}[/tex]
So,
[tex]\dfrac{L_p}{L_f}=\dfrac{f_f}{f_p}\\\Rightarrow \dfrac{L_p}{L_f}=\dfrac{257}{516} \\\Rightarrow \dfrac{L_p}{L_f}=0.49806[/tex]
The ratio of the piccolo's length to the flute's length is 0.49806
A monochromatic light passes through a narrow slit and forms a diffraction pattern on a screen behind the slit. As the wavelength of the light decreases, the diffraction pattern
a. spreads out with all the fringes getting wider.
b. becomes dimmer.
c. spreads out with all the fringes getting alternately wider and then narrower.
d. shrinks with all the fringes getting narrower.
e. remains unchanged.
We need to apply the definition of Young's double slit experiment. For fringe width of bright and dark fringe we have that
[tex]\beta = \frac{D\lambda}{d}[/tex]
Or expressed in terms of the wavelength we have that
[tex]\lambda = \frac{\beta d}{D}[/tex]
Where,
[tex]\lambda[/tex]= Wavelength
[tex]\beta[/tex]= Fringe width
d = Slit separation
D = Distance between slit and screen
From the ratios given in the equation, we have that as the wavelength decreases, the pattern determined for the diffraction pattern shrinks, which therefore causes all fringes to get narrower.
Final answer:
When the wavelength of monochromatic light decreases, the single-slit diffraction pattern shrinks with fringes getting narrower (option d). A decrease in slit width results in a wider diffraction pattern. More lines per centimeter on a diffraction grating cause bands to spread farther from the central maximum.
Explanation:
When a monochromatic light passes through a narrow slit, it forms a diffraction pattern due to the phenomenon of diffraction. The pattern consists of a series of bright and dark bands. The bright areas are known as maxima, while the dark areas are known as minima.
If the wavelength of the light decreases, the diffraction pattern shrinks, with all the fringes getting narrower (answer choice d). This is because the angle of diffraction is directly related to the wavelength, and as the wavelength becomes shorter, the angle at which light is bent decreases, causing the fringes to become narrower and the overall pattern to shrink.
Similarly, if the width of the slit producing a single-slit diffraction pattern is reduced, the pattern produced changes as well, with the bands spreading out and becoming wider. This is due to the inverse relationship between slit width and the angle of diffraction. A smaller slit width results in a larger spread of the diffraction pattern.
Lastly, if pure-wavelength light falls on a diffraction grating that has more lines per centimeter, the interference pattern will have bands that spread farther from the central maximum because a higher number of slits per unit area increases the diffractive effects.
A flat plate is oriented parallel to a 45 m/s airflow at 20°C and atmospheric pressure. The plate is L = 1 m in the flow direction and 0.5 m wide. On one side of the plate, the boundary layer is tripped at the leading edge, and on the other side there is no tripping device. Find the total drag force on the plate.
Answer:
4.192 N
Explanation:
Step 1: Identify the given parameters
Velocity of airflow = 45m/s
air temperature = 20⁰
plate length and width = 1m and 0.5m respectively.
Step 2: calculate drag force due to shear stress, [tex]F_{s}[/tex]
[tex]F_{s} = C_{f} \frac{1}{2} (\rho{U_{o}}WL)[/tex]
Note: The density and kinematic viscosity of air at 20⁰ at 1 atm, is 1.2 kg/m³ and 1.5 X 10⁻⁵ N.s/m²
⇒The Reynolds number ([tex]R_{eL}[/tex]) based on the length of the plate is
[tex]R_{eL} =\frac{VXL}{U}[/tex]
[tex]R_{eL} =\frac{45X1}{1.5 X 10^{-5}}[/tex]
[tex]R_{eL}[/tex] = 3 X10⁶ (flow is turbulent, Re ≥ 500,000)
⇒The average shear stress coefficient ([tex]C_{f}[/tex]) on the "tripped" side of the plate is
[tex]C_{f} = \frac{0.074}{(R_{e})^\frac{1}{5}}[/tex]
[tex]C_{f} = \frac{0.074}{(3 X10^6)^\frac{1}{5}}[/tex]
[tex]C_{f}[/tex] = 0.0038
⇒The average shear stress coefficient ([tex]C_{f}[/tex]) on the "untripped" side of the plate is
[tex]C_{f} = \frac{0.523}{in^2(0.06XR_{e})} -\frac{1520}{R_{e}}[/tex]
[tex]C_{f} = \frac{0.523}{in^2(0.06 X 3X10^6)} -\frac{1520}{3X10^6}[/tex]
[tex]C_{f}[/tex] = 0.0031
The total drag force = [tex]\frac{1}{2}(1.2 X 45^2 X 1 X 0.5 (0.0038 +0.0031)[/tex]
The total drag force is 4.192 N
For small amplitudes of oscillation the motion of a pendulum is simple harmonic. Consider a pendulum with a period of 0.550 s Find the ground-level energy. Express your result in joules Find the ground-level energy. Express your result in election volts
To solve this problem we will use the concepts related to the expression of energy for harmonic oscillator. From our given values we have that the period is equivalent to
[tex]T = 0.55s[/tex]
Therefore the frequency will be the inverse of the period and would be given as
[tex]f= \frac{1}{T}[/tex]
[tex]f = \frac{1}{0.55}[/tex]
[tex]f = 1.82s^{-1}[/tex]
The ground state energy of the pendulum is,
[tex]E = \frac{1}{2} hv[/tex]
[tex]E = \frac{1}{2}(6.626*10^{-34}J\cdot s)(1.82s^{-1})[/tex]
[tex]E = 6.03*10^{-34}J[/tex]
The ground state energy in eV,
[tex]E = 6.03 * 10^{-34}J(\frac{1eV}{1.6*10^{-19}J})[/tex]
[tex]E = 3.8*10^{-15}eV[/tex]
The energy difference between adjacent energy levels,
[tex]\Delta E = hv[/tex]
[tex]\Delta E = (6.626*10^{-34}J\cdot s)(1.82s)[/tex]
[tex]\Delta E = 12.1*10^{-34}J[/tex]
How much heat energy is required to vaporize a 1.0-g ice cube at 0°C? The heat of fusion of ice is 80 cal/g.
The heat of vaporization of water is 540 cal/g, and cwater = 1.00 cal/g⋅°C.
a. 620 cal
b. 720 cal
c. 820 cal
d. 1 kcal
The heat required to convert 1.0-g of ice cube at 0°C to vapor is 720 cal.
We know that when heat is supplied to a substance, change of state occurs as the object moves from a particular state of matter to another. The heat required to convert 1.0-g of ice cube at 0°C to vapor is obtained from;
H = mLfus + mcdT + mLvap
H = (1 g * 80 cal/g) + (1 g * 1.00 cal/g⋅°C * (100 - 0)) + (1 g * 540 cal/g)
H = 80 cal + 100 cal + 540 cal
H = 720 cal
The heat required to convert 1.0-g of ice cube at 0°C to vapor is 720 cal.
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The total amount of heat required to vaporize a 1-g ice cube at 0°C considering the stages of heat of fusion, heating up water, and heat of vaporization is 720 cal.
Explanation:To calculate the total amount of heat energy required to vaporize a 1.0-g ice cube at 0°C, we need to consider the heat added to turn the ice into water (heat of fusion), heat to raise the temperature of that water from 0°C to 100°C, and finally heat to turn the water into steam (heat of vaporization).
First, we use the heat of fusion of ice, which is 80 cal/g, thus the heat needed to turn 1g of ice into water at 0°C is 1g * 80 cal/g = 80 cal.
Secondly, to raise the water temperature from 0 to 100°C, we use cwater = 1.00 cal/g⋅°C, thus the required heat is 1g * 100°C * 1.00 cal/g⋅°C = 100 cal.
Finally, we use the heat of vaporization of water which is 540 cal/g, for 1g of water, the heat needed is 1g * 540 cal/g = 540 cal.
Adding all those up give us a total heat of 80 cal + 100 cal + 540 cal = 720 cal.
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A loop-the-loop has a circular arc, with a marble that can run along a track and traverse the entire inside of the loop. When the marble is precisely at the top of the inside loop, it has its minimum speed. ~v What do we know about the direction of the net force on the marble at this point, the top of the loop? a. The net force is instantaneously in the direction of the marble’s velocity b. The net force is instantaneously in the opposite direction to the marble’s velocity ~v. c. The net force is vertically downward. d. The net force is vertically upward
Answer:
c. The net force is vertically downward.
Explanation:
At the top of the loop, the only external force that keeps the ball moving around the loop, is the centripetal force.
Now, this centripetal force, is not a " new" force, it's just the vector sum of the two external forces (neglecting friction) , that act simultaneously on the marble, making it to change its speed, in magnitude and direction: the gravity force (which it is always downward), and the normal force(which is always perpendicular to the contact surface, preventing that the marble comes trough the surface), in this case between the marble and the track, which, at the top of the loop, points down too.
So, the net force, exactly at the top of the loop, is vertically downward.
Fluid flows over a smooth cylinder. The diameter of the cylinder is D and the length normal to the flow direction is L. The drag coefficient is defined as: The drag coefficient is essentially constant with a value of 1.1 in the range of Reynolds numbers of 103 to 105. In this range, at a velocity of 2 m/s the drag force is 3 N. When the velocity is doubled to 4 m/s the drag force is:
Answer:
Explanation:
Given
Coefficient of drag [tex]C_d=1.1[/tex]
Reynolds number [tex]Re.no.=103 to 105[/tex]
velocity [tex]v=2 m/s [/tex]
[tex]F_d=3 N[/tex]
if velocity if 2v i.e. [tex]4 m/s [/tex]
[tex]F_d=\frac{1}{2}C_d\rho A v^2----1[/tex]
keeping other factors as constant
[tex]F'_d=\frac{1}{2}C_d\rho A (2v)^2----2[/tex]
dividing 1 and 2
[tex]\frac{F_d}{F'_d}=\frac{v^2}{2v^2}[/tex]
[tex]F'_d=4F_d[/tex]
[tex]F'_d=4\times 3=12 N[/tex]
The specific heat capacity of aluminum is about twice that of iron. Consider two blocks of equal mass, one made of aluminum and the other one made of iron, initially in thermal equilibrium.
Heat is added to each block at the same constant rate until it reaches a temperature of 500 K. Which of the following statements is true?
a. The iron takes less time than the aluminum to reach the final temperature.
b. The aluminum takes less time than the iron to reach the final temperature.
c. The two blocks take the same amount of time to reach the final temperature.
Answer:a
Explanation:
Given
Specific heat capacity of aluminium is twice that of iron.
[tex]c_{Al}=2c_{iron}[/tex]
Also mass of two blocks is equal
Rate of heat added is also same
[tex]\dot{Q_{Al}}=\dot{Q_{iron}}[/tex]
[tex]\frac{Q_{Al}}{t_{Al}}=\frac{Q_{iron}}{t_{iron}}[/tex]
[tex]\frac{mc_{Al}\Delta T}{t_{Al}}=\frac{mc_{iron}\Delta T}{t_{iron}}[/tex]
[tex]\frac{2c_{iron}}{t_{Al}}=\frac{c_{iron}}{t_{iron}}[/tex]
[tex]t_{Al}=2t_{iron}[/tex]
Thus Time taken by aluminium block will be more
A woman of mass 44 kg jumps off the bow of
a 49 kg canoe that is intially at rest.
If her velocity is 2.5 m/s to the right, what
is the velocity of the canoe after she jumps?
Answer in units of m/sˆı.
Answer:
2.2 m/s to the left
Explanation:
Momentum is conserved, so:
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
0 = (44 kg) (2.5 m/s) + (49 kg) v
v = -2.2 m/s
The canoe will move in the opposite direction of the woman with a velocity of approximately 2.24 m/s.
The velocity of the canoe after the woman jumps off can be determined by using the principle of conservation of momentum, which states that the total momentum before an event must equal the total momentum after the event when no external forces act on the system. In this scenario, the system consists of the woman and the canoe.
Before the woman jumps, the system is at rest, so its initial momentum is zero. When the woman jumps off the canoe to the right with a velocity of 2.5 m/s, by conservation of momentum, the canoe must move in the opposite direction to maintain the total momentum at zero.
The momentum possessed by the woman is given by the product of her mass and velocity (mw×vw).
Similarly, the momentum of the canoe is the product of its mass and its velocity in the opposite direction (mc×vc).
Momenta are equal in magnitude and opposite in direction:
mw ×vw = mc ×vc
Therefore, vc = (mw ×vw) / mc
Substituting the given values:
vc = (44 kg ×2.5 m/s) / 49 kg
= 2.24 m/s
The canoe will move to the left (opposite to the woman's direction) with a velocity of approximately 2.24 m/s.
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnification of 400x with an objective lens that has a focal length of 0.40 cm. The distance between the eyepiece and objective lenses is 12 cm. 1) Find the focal length of the eyepiece lens assuming a near point of 25 cm (the closest an object can be and still be seen in focus). Do not neglect any values in your calculation. (Express your answer to two significant figures.)
Answer:
[tex]f_{e}[/tex] = 1.9 cm
Explanation:
The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective
M = M₀ [tex]m_{e}[/tex]
Where M₀ is the magnification of the objective and [tex]m_{e}[/tex] is the magnification of the eyepiece.
The eyepiece is focused to the near vision point (d = 25 cm)
[tex]m_{e}[/tex] = 25 / [tex]f_{e}[/tex]
The objective is focused on the distances of the tube (L)
M₀ = -L / f₀
Substituting
M = - L/f₀ 25/[tex]f_{e}[/tex]
1) Let's look for the focal length of the eyepiece (faith)
[tex]f_{e}[/tex] = - L 25 / f₀ M
M = 400X = -400
[tex]f_{e}[/tex] = - 12 25 /0.40 (-400)
[tex]f_{e}[/tex] = 1.875 cm
Let's approximate two significant figures
[tex]f_{e}[/tex] = 1.9 cm
Final answer:
The focal length of the eyepiece lens in the microscope is approximately -0.01 mm.
Explanation:
To find the focal length of the eyepiece lens, we can use the formula for the magnification of a compound microscope, which is given by:
M = -fobjective/feyepiece
Where M is the overall magnification, fobjective is the focal length of the objective lens, and feyepiece is the focal length of the eyepiece lens.
Given that the overall magnification is 400x and the focal length of the objective lens is 0.40 cm, we can rearrange the formula to solve for the focal length of the eyepiece lens:
feyepiece = -fobjective/M = -0.40 cm/400 = -0.001 cm = -0.01 mm.
Therefore, the focal length of the eyepiece lens is approximately -0.01 mm.
A glass tea kettle containing 500 g of water is on the stove. The portion of the tea kettle that is in contact with the heating element has an area of 0.090 m2 and is 1.5 mm thick.
At a certain moment, the temperature of the water is 75°C, and it is rising at the rate of 3°C per minute.
What is the temperature of the outside surface of the bottom of the tea kettle?
Neglect the heat capacity of the kettle, and assume that the inner surface of the kettle is at the same temperature as the water inside.
Answer:
77.08 C
Explanation:
[tex]m[/tex] = mass of the water = 500 g = 0.5 kg
[tex]c[/tex] = specific heat of water = 4186 J/(kg °C)
[tex]\Delta T[/tex] = Rate of change of temperature = 3 °C /min = (3/60 ) °C /s = 0.05 °C /s
[tex]k[/tex] = thermal conductivity of glass = 0.84
[tex]A[/tex] = Area of the element = 0.090 m²
[tex]t[/tex] = thickness of the element = 1.5 mm = 0.0015 m
[tex]T_{i}[/tex] = Temperature inside = 75 °C
[tex]T_{o}[/tex] = Temperature outside = ?
Using conservation of energy
Heat gained by water = Heat transferred through glass
[tex]m c \Delta T = \frac{kA(T_{o} - T_{i})}{t} \\(0.5) (4186 (0.05) = \frac{(0.84)(0.090)(T_{o} - 75)}{0.0015} \\104.65 = (50.4)(T_{o} - 75)\\T_{o} = 77.08 C[/tex]
Water is pumped from a lake to a storage tank 15 m above at a rate of 70 L/s while consuming 15.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of the pump-motor unit and (b) the pressure difference between the inlet and the exit of the pump.
Answer:
(a) 66.9%
(b) 147.14 kPa
Explanation:
Given:
Elevation of the water tank z = 15 m
Water volume flow rate V = 70 L/s = 0.07 m³/s
Input electric power consumption by the pump, Welec, in = 15.4 kW
Assuming there are no frictional losses in the pipes and changes in kinetic energy, the efficiency of the pump-motor will be;
η = ΔEmech ÷ Welec,
Where;
η is the overall efficiency
ΔEmech is the workdone to move the water pumped from the lake to a storage tank 15m above
Welec is the Input electric power consumption by the pump
Solving for ΔEmech ;
ΔEmech = mgh
mass, m = density × volume
Density of water = 1000 kg/m³
m = 1000 kg/m³ × 0.07 m³/s
m = 70 kg/s
∴ ΔEmech = 70 × 9.8 × 15
= 10.3 kW
Substituting the values of ΔEmech and Welec to calculate the overall efficiency
η = (10.3 kW ÷ 15.4 kW) × 100 %
= 0.6688 × 100 %
= 66.88 %
= 66.9 %
The overall efficiency of the pump-motor unit is = 66.9 %
(b) The pressure difference between the inlet and the exit of the pump is calculated to be;
Pressure = ΔEmech ÷ V
= 10.3 ÷ 0.07
= 147.14 kPa
This question involves the concept of potential energy, pressure difference, and electrical work.
(a) Efficiency of pump-motor unit is "66.9 %".
(b) The pressure difference between the inlet and the exit of the pump is "147.15 KPa".
(a) Pump EfficiencyThe efficiency of the pump-motor unit can be given by the following formula:
[tex]\eta = \frac{W_{P.E}}{W_{elect}}[/tex]
where,
[tex]\eta[/tex] = efficiency = ?P.E = power due to potential energy = [tex]\frac{mgh}{t}=\frac{\rho Vgh}{t}[/tex][tex]\rho[/tex] = density of water = 1000 kg/m³g = 9.81 m/s²h = height = 15 m[tex]\frac{V}{t}[/tex] = volume flow rate = 70 L/s = 0.07 m³/s[tex]W_{elect}[/tex] = electrical power input = 15.4 KW = 15400 WTherefore,
[tex]\eta = \frac{\rho Vgh}{t\ W_{elect}}\\\\\eta=\frac{(1000\ kg/m^3)(0.07\ m^3/s)(9.81\ m/s^2)(15\ m)}{15400\ W}\\\\\eta =0.669 = 66.9\ \%[/tex]
(b) PRESSURE DIFFERENCEThe pressure difference between inlet and outlet of the pump can be found using the following equation:
[tex]\Delta P = \rho gh\\\\\Delta P = (1000\ kg/m^3)(9.81\ m/s^2)(15\ m)[/tex]
[tex]\Delta P = 147150\ Pa = 147.15\ KPa[/tex]
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A wire 95.0 mm long is bent in a right angle such that the wire starts at the origin and goes in a straight line to x = 40.0 mm , y = 0, and then in another straight line from x = 40.0 mm , y = 0 to x = 40.0 mm , y = 55.0 mm . The wire is in an external uniform 0.300-T magnetic field in the +z direction, and the current through the wire is 5.10 A , directed from the origin into the wire.
A/Determine the magnitude of the magnetic force exerted by the external field on the wire.
B/Determine the direction of the magnetic force exerted by the external field on the wire measured clockwise from the positive x axis.
C/The wire is removed and replaced by one that runs directly from the origin to the locationx = 20.0mm , y = 35.0mm . If the current through this wire is also 4.90A , what is the magnitude of the magnetic force exerted by the external field on the wire?
D/The wire is removed and replaced by one that runs directly from the origin to the locationx = 20.0mm , y = 35.0mm . If the current through this wire is also 4.90A , what is the magnitude of the direction force exerted by the external field on the wire measured clockwise from the positive x axis.?
Answer:
0.1040512455 N
[tex]36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]
0.05925 N
[tex]29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]
Explanation:
I = Current
B = Magnetic field
Separation between end points is
[tex]l=\sqrt{40^2+55^2}\\\Rightarrow l=68.00735\ mm[/tex]
Effective force is given by
[tex]F=IlB\\\Rightarrow F=5.1\times 68.00735\times 10^{-3}\times 0.3\\\Rightarrow F=0.1040512455\ N[/tex]
The force is 0.1040512455 N
[tex]tan\theta=\dfrac{55}{40}\\\Rightarrow \theta=tan^{-1}\dfrac{55}{40}\\\Rightarrow \theta=53.97^{\circ}[/tex]
The angle the force makes is given by
[tex]\alpha=\theta-90\\\Rightarrow \alpha=53.97-90\\\Rightarrow \alpha=-36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]
The direction is [tex]36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]
[tex]F=IlB\\\Rightarrow F=4.9\times \sqrt{20^2+35^2}\times 10^{-3}\times 0.3\\\Rightarrow F=0.05925\ N[/tex]
The force is 0.05925 N
[tex]tan\theta=\dfrac{35}{20}\\\Rightarrow \theta=tan^{-1}\dfrac{35}{20}\\\Rightarrow \theta=60.26^{\circ}[/tex]
[tex]\alpha=\theta-90\\\Rightarrow \alpha=60.26-90\\\Rightarrow \alpha=-29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]
The direction is [tex]29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]
A. Magnitude of the force: [tex]\( 0.104 \, \text{N} \)[/tex]
B. Direction of the force: [tex]\( 323.8^\circ \)[/tex]
C. Magnitude of the force: [tex]\( 0.0593 \, \text{N} \)[/tex]
D. Direction of the force: [tex]\( 330^\circ \)[/tex]
Part A: Determine the magnitude of the magnetic force exerted by the external field on the wire
Given: Length of wire segments: [tex]\( L_1 = 40.0 \, \text{mm} = 0.040 \, \text{m} \) and \( L_2 = 55.0 \, \text{mm} = 0.055 \, \text{m} \)[/tex]
- Magnetic field, [tex]\( B = 0.300 \, \text{T} \)[/tex]
- Current, [tex]\( I = 5.10 \, \text{A} \)[/tex]
The force on a current-carrying wire in a magnetic field is given by [tex]\( \mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \).[/tex]
For the first segment along the x-axis:
- [tex]\( \mathbf{L}_1 = 0.040 \, \text{m} \, \hat{i} \)[/tex]
[tex]\( \mathbf{B} = 0.300 \, \text{T} \, \hat{k} \)[/tex]
[tex]\[ \mathbf{F}_1 = I (\mathbf{L}_1 \times \mathbf{B}) = 5.10 \, \text{A} \times (0.040 \, \hat{i} \times 0.300 \, \hat{k}) \][/tex]
[tex]\[ \mathbf{F}_1 = 5.10 \times 0.040 \times 0.300 \, \hat{j} \][/tex]
[tex]\[ \mathbf{F}_1 = 0.0612 \, \text{N} \, \hat{j} \][/tex]
For the second segment along the y-axis:
[tex]- \( \mathbf{L}_2 = 0.055 \, \text{m} \, \hat{j} \)[/tex]
[tex]- \( \mathbf{B} = 0.300 \, \text{T} \, \hat{k} \)[/tex]
[tex]\[ \mathbf{F}_2 = I (\mathbf{L}_2 \times \mathbf{B}) = 5.10 \, \text{A} \times (0.055 \, \hat{j} \times 0.300 \, \hat{k}) \][/tex]
[tex]\[ \mathbf{F}_2 = 5.10 \times 0.055 \times (-0.300) \, \hat{i} \][/tex]
[tex]\[ \mathbf{F}_2 = -0.08415 \, \text{N} \, \hat{i} \][/tex]
The net force [tex]\(\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2\)[/tex]:
[tex]\[ \mathbf{F} = -0.08415 \, \hat{i} + 0.0612 \, \hat{j} \][/tex]
The magnitude of the force is:
[tex]\[ |\mathbf{F}| = \sqrt{(-0.08415)^2 + (0.0612)^2} \][/tex]
[tex]\[ |\mathbf{F}| = \sqrt{0.00708 + 0.00375} \][/tex]
[tex]\[ |\mathbf{F}| = \sqrt{0.01083} \][/tex]
[tex]\[ |\mathbf{F}| \approx 0.104 \, \text{N} \][/tex]
Part B: The direction of the force can be found using the angle [tex]\( \theta \)[/tex] with respect to the positive x-axis:
[tex]\[ \theta = \tan^{-1} \left( \frac{F_y}{F_x} \right) \][/tex]
[tex]\[ \theta = \tan^{-1} \left( \frac{0.0612}{-0.08415} \right) \][/tex]
[tex]\[ \theta = \tan^{-1} \left( -0.727 \right) \][/tex]
[tex]\[ \theta \approx -36.2^\circ \][/tex]
Since this angle is measured counterclockwise from the positive x-axis, the direction clockwise is:
[tex]\[ 360^\circ - 36.2^\circ = 323.8^\circ \][/tex]
Part C: Determine the magnitude of the magnetic force exerted by the external field on the wire
For the wire directly from the origin to [tex]\( x = 20.0 \, \text{mm}, y = 35.0 \, \text{mm} \):[/tex]
- Length [tex]\( L = \sqrt{(20.0 \, \text{mm})^2 + (35.0 \, \text{mm})^2} = \sqrt{(0.020 \, \text{m})^2 + (0.035 \, \text{m})^2} \)[/tex]
- Length [tex]\( L = \sqrt{0.0004 + 0.001225} = \sqrt{0.001625} \)[/tex]- Length [tex]\( L \approx 0.0403 \, \text{m} \)[/tex]
[tex]\[ \mathbf{L} = 0.0403 \, \text{m} \][/tex]
Given current [tex]\( I = 4.90 \, \text{A} \)[/tex]:
[tex]\[ \mathbf{F} = I L B \sin(\theta) \][/tex]
[tex]\[ \theta = \angle \text{between } \mathbf{L} \text{ and } \mathbf{B} = 90^\circ \][/tex]
[tex]\[ \mathbf{F} = 4.90 \times 0.0403 \times 0.300 \][/tex]
[tex]\[ \mathbf{F} \approx 0.0593 \, \text{N} \][/tex]
Part D: - Current: [tex]\( 4.90 \, \text{A} \)[/tex]
- Magnetic field: [tex]\( 0.300 \, \text{T} \, \hat{k} \)[/tex]
- Displacement vector: [tex]\( \mathbf{L} = 0.020 \, \hat{i} + 0.035 \, \hat{j} \)[/tex]
- Force vector: [tex]\( \mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \)[/tex]
[tex]\[ \mathbf{L} \times \mathbf{B} = (0.020 \, \hat{i} + 0.035 \, \hat{j}) \times 0.300 \, \hat{k} \][/tex]
[tex]\[ \mathbf{F} = 4.90 (0.020 \, \hat{i} + 0.035 \, \hat{j}) \times 0.300 \, \hat{k} \][/tex]
[tex]\[ \mathbf{F} = 4.90 \times (0.020 \times 0.300 \, \hat{j} - 0.035 \times 0.300 \, \hat{i}) \][/tex]
[tex]\[ \mathbf{F} = 4.90 \times (0.006 \, \hat{j} - 0.0105 \, \hat{i}) \][/tex]
[tex]\[ \mathbf{F} = 0.0294 \, \hat{j} - 0.05145 \, \hat{i} \][/tex]
The magnitude of the force:
[tex]\[ |\mathbf{F}| = \sqrt{(0.0294)^2 + (-0.05145)^2} \approx 0.0593 \, \text{N} \][/tex]
The angle with respect to the x-axis:
[tex]\[ \theta = \tan^{-1} \left( \frac{0.0294}{-0.05145} \right) = \tan^{-1} \left( -0.571 \right) \approx -30^\circ \][/tex]
The direction clockwise is:
[tex]\[ 360^\circ - 30^\circ = 330^\circ \][/tex]
When you drive a car in a circle at a constant speed you are accelerating towards the center of your circular motion.
Answer:True
Explanation:
The given statement is true because when we drive a car in a circle with constant speed , the car experiences the centripetal acceleration towards the center.
But acceleration is change in velocity of object in a given time, here direction of velocity is constantly changing to give rise to acceleration.
If the magnitude of velocity is changing with time then the car would have experiences the tangential acceleration .
An object is suspended from a spring with force constant 10 N/m. (c) Find the mass suspended from this spring that would result in a period of 2.4 s on Earth. 0.142 Incorrect: Your answer is incorrect. Use the expression for the period of oscillation for a mass attached to a spring to find the mass of the object. kg (d) Find the mass suspended from this spring that would result in a period of 2.4 s on Mars. 0.142 Incorrect: Your answer is incorrect.
To solve this problem we must use the perioricity equations given as a function of the mass and spring constant. Mathematically this can be expressed as:
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
m = mass
k = Spring constant
Re-arrange to find the mass we have
[tex]m = \frac{T^2k}{4\pi^2 }[/tex]
Replacing with our values we have that
[tex]m = \frac{2.4^2*10}{4\pi^2}[/tex]
[tex]m = 1.459kg[/tex]
D) Mass is independent of acceleration due to gravity (as you can see at the equation previously given) for this reason the mass suspended on mars is given as the same found. Therefore the mass would be
m = 1.459kg
Final answer:
The mass suspended from the spring that would result in a period of 2.4 s is 0.142 kg.
Explanation:
To find the mass suspended from the spring that would result in a period of 2.4 s, we can use the formula for the period of oscillation:
T = 2π √(m/k)
Where T is the period, m is the mass, and k is the force constant of the spring.
Let's rearrange the formula to solve for m:
m = (T^2 · k) / (4π^2)
Substituting the given values:
m = (2.4^2 · 10) / (4π^2)
m = 0.142 kg
PART ONE
A jet aircraft is traveling at 281 m/s in horizontal flight. The engine takes in air at a
rate of 107 kg/s and burns fuel at a rate of
4.23 kg/s. The exhaust gases are ejected at
679 m/s relative to the aircraft.
Find the thrust of the jet engine.
Answer in units of N.
PART TWO
Find the delivered power.
Answer in units of W
Answer:
1. F = 45,458.17 N
2. P = 12,800,000 W
Explanation:
Part 1. The thrust force is the sum of the forces on the air and on the fuel.
For the air, 107 kg of air is accelerated from 281 m/s to 679 m/s in 1 second.
F = ma
F = (107 kg) (679 m/s − 281 m/s) / (1 s)
F = 42,586 N
For the fuel, 4.23 kg of fuel is accelerated from 0 m/s to 679 m/s in 1 second.
F = ma
F = (4.23 kg) (679 m/s − 0 m/s) / (1 s)
F = 2,872.17 N
So the thrust on the jet is:
F = 42,586 N + 2,872.17 N
F = 45,458.17 N
Rounded to three significant figures, the force is 45,500 N.
Part 2. Power = work / time, and work = force × distance, so:
Power = force × distance / time
Power = force × velocity
P = (45,458.17 N) (281 m/s)
P = 12,773,745.77 W
Rounded to three significant figures, the power is 12,800,000 W.
A 30 gram bullet is shot upward at a wooden block. The bullet is launched at the speed vi. It travels up 0.40 m to strike the wooden block. The wooden block is 20 cm wide and 10 cm high and its thickness gives it a mass of 500 g. The center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height.
Answer
Mass of bullet (m) = .03 kg
Mass of wooden block M = 0.5 kg
Since the center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height
finding the launch speed of bullet
Velocity of wooden block + bullet just after impact
= [\tex]\sqrt{2gh}[/tex]
=[\tex]\sqrt{2\times 9.8 \times 0.6}[/tex]
= 3.43 m/s
v₁ be the launch velocity
Applying law of conservation of momentum
0.03 x v₂ = 0.530 x 3.43
v₂ = 60.6 m /s
if v₁ be initial velocity
v₂² = v₁² + 2 g h
v₁² = v₂² - 2 gh
v₁² = 60.6 ² - 2 x (-9.8 )x 0.4
v₁ = 60.65 m /s this is launch speed
On a bet, you try to remove water from a glass by blowing across the top of a vertical straw immersed in the water. What is the minimum speed you must give the air at the top of the straw to draw water upward through a height of 1.6cm?
Answer:
v₂ = 0.56 m / s
Explanation:
This exercise can be done using Bernoulli's equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Where points 1 and 2 are on the surface of the glass and the top of the straw
The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of the straw the velocity of the surface of the vessel is almost zero v₁ = 0
The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m
We substitute in the equation
[tex]P_{atm}[/tex] + ρ g y₁ = [tex]P_{atm}[/tex] + ½ ρ v₂² + ρ g y₂
½ v₂² = g (y₂-y₁)
v₂ = √ 2 g (y₂-y₁)
Let's calculate
v₂ = √ (2 9.8 1.6 10⁻²)
v₂ = 0.56 m / s
A student stands on the edge of a merry-go-round which rotates five times a minute and has a radius of two meters one evening as the sun is setting. The student produces a shadow on the nearby building. (a) Write an equation for the position of the shadow. (b) Write an equation for the velocity of the shadow.
To solve this problem it is necessary to apply the concepts related to the simple harmonic movement, to the speed in terms of displacement and the timpo, as well as the angular frequency and the period of frequency.
PART A) According to the description given, 5 revolutions are made in one minute (or 60 seconds) that is to say that the frequency would be given by
[tex]f = \frac{1}{12s^{-1}}[/tex]
Therefore the angular velocity can be found as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 0.52rad/s[/tex]
The displacement that determines the maximum displacement based on the angular velocity time and the time in the simple harmonic movement, is equal to the radius of the circle, in other words
[tex]x = Acos(\omega t)[/tex]
Where,
A = Amplitude
[tex]\omega =[/tex] Angular velocity
t = time
If for our given values the value of the amplitude is 2m and the value of the angular velocity is 0.52rad/s
[tex]x = 2cos(0.52t)[/tex]
So the equation for the position of the shadow is of
[tex]x(t) = 2cos(0.52t)[/tex]
PART B) The equation for the velocity of the shadow is calculated as a expression of the displacement against the time, if we differenciate the previous value found, we have that,
[tex]\frac{dx}{dt} = \frac{d(2cos(0.52t))}{dt}[/tex]
[tex]v(t) = 2(-0.52)sin(0.523t)[/tex]
[tex]v(t) = -1.05sin(0.52t)[/tex]
The position of the shadow can be described using the equation x=2cos(10πt+φ) and the velocity of the shadow can be described using the equation v=-2*10π*sin(10πt+φ), using principles of simple harmonic motion.
Explanation:The subject of this question involves physics and in particular, kinematics and rotational motion, as it relates to the shadow produced by a student standing on a rotating merry-go-round.
The position and velocity of the shadow will vary as the merry-go-round rotates, and these can be represented using mathematical equations. Let's denote the angular speed of the merry-go-round as 'w', which is given by the formula w = 2πn, where n is the number of rotations per minute. For the given question, n = 5 rotations per minute. Therefore, w = 10π rad/min.
(a) Position of the shadow: We can describe the position of the shadow in terms of simple harmonic motion - as the merry-go-round rotates, the shadow produced on the nearby building would be oscillating back and forth. This kind of motion can be mathematically represented as x=Acos(wt+φ), where x is the position of the shadow, A is the amplitude (which is the radius of the merry-go-round, hence 2 meters), w is the angular frequency, t is time, and φ is the phase constant.
(b) Velocity of the shadow: The velocity of the shadow is the derivative of the position with respect to time, which can be represented as v=-Aw*sin(wt+φ). In our case, the angular speed w = 10π rad/min, A is the amplitude (radius of the merry-go-round, hence 2 meters), and 't' is the time.
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Consider two different rods. The greatest thermal conductivity will be in the rod with:
a. electrons that are freer to move from atom to atom.
b. the greater specific heat.
c. the greater cross-sectional area.
d. the greater length.
Answer:
Options A and D are correct
Explanation:
The thermal conductivity of a metal is the property of a metal to allow heat flow through it. conductivity is higher in conductors and low in insulators. Thermal conductivity is high in metals due to the metallic bonds that exist in metals and the presence of free electrons within the metal which allow easy flow of heat from one atom to another.From the problem the rod which contains freer electrons will allow more heat to flow easily hence have a higher thermal conductivity.
Thermal conductivity has the formula below;
[tex]k= \frac{QL}{AΔT}[/tex]
k is thermal conductivity,A is cross sectional area L is length of rod Q is quantity of heat transferred to material. ΔT is temperature change.From the above equation we can see that thermal conductivity is inversely proportional to A and directly proportional to L. This mean the rod with less area will have a higher thermal conductivity and the rod with a higher length will have higher k. Hence option C i wrong and option D is correct.
For specific heat, its very much different from thermal conductivity. Specific heat is the ability of a material to hold heat while thermal conductivity is the ability of heat to flow through a material.
The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation is ?y.
Part A What will happen to the fringe spacing if the wavelength of the light is decreased?
a. Δy will decrease
b. Δy will increase
c. Δy will not change
Part B What will happen to the fringe spacing if the spacing between the slits is decreased?
a. Δy will decrease
b. Δy will increase
c. Δy will not change
Part C What will happen to the fringe spacing if the distance to the screen is decreased?
a. Δy will decrease
b. Δy will increase
c. Δy will not change
Part D Suppose the wavelength of the light is 460 nm . How much farther is it from the dot on the screen in the center of fringe E to the left slit than it is from the dot to the right slit?
Answer:
Part A:
a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.
Part B:
b) If the spacing between the slits is decreased the fringe spacing Δy will increase.
Part C:
a) If the distance to the screen is decreased the fringe spacing will decrease.
Part D:
The dot in the center of fringe E is [tex]920\ x\ 10^{-9} m[/tex] farther from the left slit than from the right slit.
Explanation:
In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.
The position of bright fringes in the screen where the pattern is formed can be calculated with
[tex]\vartriangle y =\frac{m \lambda L}{d} [/tex]
[tex]m=0,\pm 1,\pm 2,\pm 3,.....[/tex]
m is the order number.[tex]\lambda[/tex] is the wavelength of the monochromatic light.L is the distance between the screen and the two slits.d is the distance between the slits.Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light [tex]\lambda[/tex] is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be [tex]\vartriangle r=m \lambda[/tex]We simply replace the values in that equation :
[tex]\vartriangle r= m \lambda =2.\ 460\ nm[/tex]
[tex]\vartriangle r= 920\ x\ 10^{-9} m[/tex]
The dot in the center of fringe E is [tex]920\ x\ 10^{-9}m[/tex] farther from the left slit than from the right slit.
By using the general equation for the double-slit maximums, we will get
A) aB) bC) aD) Can't be done completely, but below there is a approach to it.What is the distance between consecutive maximums?The equation for the constructive interference in a double-slit experiment is given by:
[tex]y = \frac{m*\lambda*D}{d} [/tex]
Where:
m is the number of the maximum.λ is the wavelength.D is the distance between the double-slit and the screen.d is the distance between the slits.Now let's answer:
A) If the wavelength is decreased, then the numerator is decreased, meaning the separation between consecutive fringes will also be decreased, so the correct option is a.
B) If d is decreased then the denominator decreases, meaning that the distance between consecutive fringes increases, so the correct option is b.
C) is the distance D is decreased, similar like in case A, the numerator decreases, meaning that the correct option is a again.
D) Sadly, as we do not know:
Which fringe is E.The value of DThe value of d.We can't answer this question.
What we should do here, is to compare the distance between the fringe and each slit, that distance will be the hypotenuse of a right triangle with one cathetus equal to D, and the other cathetus equal to y ± d/2, where each sign corresponds to each slit.
Then the difference in the distance will just be:
[tex]\sqrt{(y - d/2)^2 + D^2} - \sqrt{(y + d)^2 + D^2} [/tex]
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Isolating a variable in two equations is easiest when one of them has a coefficient of 1. Let's say we have the two equations 3A−B=5 2A+3B=−4 and want to isolate one of the variables, such that it appears by itself on one side of the equation. Which of the following is an equation with one of the above variables isolated?
Answer:
[tex]B=3A-5[/tex]
Explanation:
Variable Isolation
It's a common practice when dealing with equations that we have to isolate one variable in terms of other variables and/or constants. The isolation of the variable usually implies adding, subtracting, multiplying or dividing by constants. The following example shows how to isolate the A:
[tex]2A+3B=-4\\\\2A=-4-3B\\\\\displaystyle A=\frac{-4-3B}{2}[/tex]
We are required to find the equation where the variable has a coefficient of 1 and isolate it. The following equation fits into the description:
[tex]3A-B=5[/tex]
Isolating B:
[tex]B=3A-5[/tex]
The total volume in milliliters of a glucose-water solution is given by the equation below: V = 1001.93 + 111.5282m + 0.64698m2 where m is the molality of the solution. The partial molar volume of glucose ?glucose, is the slope of a V versus m curve, (?V/?m). Find the partial molar volume of glucose in a 0.100m solution of glucose in water
Answer:
111.657596
Explanation:
The expression of volume is given by
[tex]V=1001.93+111.5282+0.64698m^2[/tex]
Partially differentiating the term we get
[tex]\dfrac{\partial V}{\partial x}=\dfrac{\partial (1001.93+111.5282+0.64698m^2)}{\partial x}\\\Rightarrow \dfrac{\partial V}{\partial x}=111.5282+2\times 0.64698m\\\Rightarrow \dfrac{\partial V}{\partial x}=111.5282+1.29396m[/tex]
m = 0.100
[tex]\dfrac{\partial V}{\partial x}=111.5282+1.29396\times 0.100\\\Rightarrow \dfrac{\partial V}{\partial x}=111.657596[/tex]
The partial molar volume of glucose is 111.657596
Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 1.1% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth?
Answer:
3900000 m/s
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
v = Speed of the galaxy relative to the earth
Observed frequency is
[tex]f'=(1+0.013)f\\\Rightarrow \dfrac{f'}{f}=1.013[/tex]
Here the Doppler relation must be used.
So, observed frequency is given by
[tex]f'=f\dfrac{v+c}{c}\\\Rightarrow \dfrac{f'}{f}=\dfrac{v+c}{c}\\\Rightarrow v=\dfrac{f'c}{f}-c\\\Rightarrow v=(1.013\times 3\times 10^8)-3\times 10^8\\\Rightarrow v=3900000\ m/s[/tex]
The speed of the galaxy relative to the earth is 3900000 m/s
A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of the surroundings). Given these two temperatures, what is the maximum possible efficiency the car can have? (Note that actual car engine efficiencies are in the 20-25% range.) _______ %.
One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.
From the temperature the value is given as
[tex]\eta = 1-\frac{T_L}{T_H}[/tex]
Where,
T_L = Cold focus temperature
T_H = Hot spot temperature
Our values are given as,
T_L = 20\° C = (20+273) K = 293 K
T_H = 440\° C = (440+273) K = 713 K
Replacing we have,
[tex]\eta = 1-\frac{T_L}{T_H}[/tex]
[tex]\eta = 1-\frac{293}{713}[/tex]
[tex]\eta = 0.589[/tex]
Therefore the maximum possible efficiency the car can have is 58.9%
A container of water, diameter 12 cm, has a small opening near the bottom that can be unplugged so that the water can run out. If the top of the tank is open to the atmosphere, what is the exit speed of the water leaving through the hole. The water level is 15 cm above the bottom of the container. The center of the 3.0 diameter hole is 4.0 cm from the bottom. How long does it take the water to hit the ground? How far from the container will the water land?
Answer:
1.32 m
Explanation:
from the given figure we can find the velocity at the hole in the container
let it be v and we know that
then [tex]v= \sqrt{2gh}[/tex]
h= 15-4= 11 cm , g=9.81
[tex]v= \sqrt{2(9.81)(11)}[/tex]
v=14.69 m/s
now using
s=ut+0.5at^2
t= is the time required by water to reach bottom of the container
u = velocity in vertical direction is zero.
therefore,
[tex]0.04= 0\times t+\frac{1}{2}\times9.81\times t^2[/tex]
t= 0.09030 sec
let x be the distance far from the container will the water land
x=vt
x=14.69×0.09030 = 1.32 m
You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 90.3 minutes, what is the half-life of this substance?
Answer : The half-life of this substance will be, 45 minutes.
Explanation :
First we have to calculate the value of rate constant.
Expression for rate law for first order kinetics is given by:
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = ?
t = time passed by the sample = 90.3 min
a = initial amount of the reactant = 400
a - x = amount left after decay process = 100
Now put all the given values in above equation, we get
[tex]k=\frac{2.303}{90.3min}\log\frac{400}{100}[/tex]
[tex]k=1.54\times 10^{-2}\text{ min}^{-1}[/tex]
Now we have to calculate the half-life of substance, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]1.54\times 10^{-2}\text{ min}^{-1}=\frac{0.693}{t_{1/2}}[/tex]
[tex]t_{1/2}=45min[/tex]
Therefore, the half-life of this substance will be, 45 minutes.
A rocket burns fuel at a rate of 264 kg/s and
exhausts the gas at a relative speed of 8 km/s.
Find the thrust of the rocket.
Answer in units of MN.
Answer:
2.112 MN
Explanation:
Force = mass × acceleration. Each second, 264 kg of gas is accelerated from 0 to 8 km/s.
F = ma
F = m Δv / Δt
F = 264 kg × (8000 m/s − 0 m/s) / 1 s
F = 2,112,000 kg m/s²
F = 2.112 MN
Round as needed.
The thrust of the rocket, calculated using the formula - Thrust = mass rate of exhaust x exhaust velocity, is 2.112 Mega Newtons.
Explanation:The thrust produced by a rocket is typically calculated using the formula: Thrust = mass rate of exhaust x exhaust velocity. In this scenario, the rocket is burning fuel at a rate of 264 kg/s (mass rate of exhaust), and the exhaust gas is ejected at a relative speed of 8 km/s.
However, to ensure the units are consistent, we have to convert the relative speed from km/s to m/s, which results in 8000 m/s (exhaust velocity).
Substituting these values into the formula, we get: Thrust = 264 kg/s x 8000 m/s = 2,112,000 N = 2.112 MN.
Therefore, the thrust of the rocket is 2.112 Mega Newtons.
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Consider the four quantum numbers of an electron in an atom, n, l, ml, and ms. The energy of an electron in an isolated atom depends on:______a. l, ml, and msonly.b. n only.c. n and l only.d. n, l, and mlonly.e. all four quantum numbers.
Answer:
The energy of an electron in an isolated atom depends on b. n only.
Explanation:
The quantum number n, known as the principal quantum number represents the relative overall energy of each orbital.
The sets of orbitals with the same n value are often referred to as an electron shell, in an isolated atom all electrons in a subshell have exactly the same level of energy.
The principal quantum number comes from the solution of the Schrödinger wave equation, which describes energy in eigenstates [tex]E_n[/tex], and for the case of an hydrogen atom we have:
[tex]E_n=-\cfrac{13.6}{n^2}\, eV[/tex]
Thus for each value of n we can describe the orbital and the energy corresponding to each electron on such orbital.