Answer:
Explanation:
When the distance is decreasing, the frequency of the received wave form will be higher than the source wave form. Besides sound and radio waves, the Doppler effect also affects the light emitted by other bodies in space. If a body in space is "blue shifted," its light waves are compacted and it is coming towards us.
The Doppler effect is a phenomenon that describes how the perceived frequency and wavelength of waves change due to relative motion. It can result in a blue shift or a red shift, depending on the direction of motion. The Doppler effect has applications in astronomy and medicine, among other fields.
Explanation:The Doppler effect is a phenomenon in physics that describes how the perceived frequency and wavelength of waves change when there is relative motion between the source of the waves and the observer. This effect can alter the way in which we perceive radiation, such as light or sound.
When an object emitting waves moves towards an observer, the waves are compressed, resulting in a higher frequency and shorter wavelength. This is known as a blue shift. On the other hand, when an object moves away from an observer, the waves are stretched, resulting in a lower frequency and longer wavelength, which is called a red shift.
The Doppler effect has several applications in various fields. For example, it is used in astronomy to determine the speed and direction of celestial objects based on their red or blue shifts. In medicine, it is used in ultrasound technology to measure blood flow velocity. Overall, the Doppler effect allows us to understand and interpret the motion of objects emitting waves.
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On a ring road, 12 trams are spaced at regular intervals and travel at a constant speed. How many trams need to be added to the circuit so that, maintaining the same speed, the intervals between them will decrease by 1 5
3 trams must be added
Explanation:
In this problem, there are 12 trams along the ring road, spaced at regular intervals.
Calling L the length of the ring road, this means that the space between two consecutive trams is
[tex]d=\frac{L}{12}[/tex] (1)
In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become
[tex]d'=(1-\frac{1}{5})d=\frac{4}{5}d[/tex]
And the number of trams will become
[tex]12+n[/tex]
So eq.(1) will become
[tex]\frac{4}{5}d=\frac{L}{n+12}[/tex] (2)
And substituting eq.(1) into eq.(2), we find:
[tex]\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3[/tex]
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Answer: 3
Explanation:
We need to do (1/(5-1))x12
A sperm whale can accelerate at about 0.120 m/s 2 0.120 m/s2 when swimming on the surface of the ocean. How far will a sperm whale travel if it starts at a speed of 1.10 m/s 1.10 m/s and accelerates to a speed of 2.05 m/s 2.05 m/s ? Assume the whale travels in a straight line.
Answer:
The distance traveled by the sperm whale is 12.46 meters.
Explanation:
Given that,
Acceleration of a sperm whale, [tex]a=0.12\ m/s^2[/tex]
Initial speed of the whale, u = 1.1 m/s
Final speed of the whale, v = 2.05 m/s
We need to find the distance traveled by the sperm whale. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2ad[/tex]
d is the distance covered
[tex]d=\dfrac{v^2-u^2}{2a}[/tex]
[tex]d=\dfrac{(2.05)^2-(1.1)^2}{2\times 0.12}[/tex]
d = 12.46 meters
So, the distance traveled by the sperm whale is 12.46 meters. Hence, this is the required solution.
Design at least two independent experiments to determine the wavelength of a microwave source. You have a microwave source, microwave detector, a barrier with different regions, and a goniometer to measure angles. (To avoid over-ranging the detector, please use the largest meter multiplier setting that allows you to see what you’re doing. Nevertheless, if you are seeing nothing at any angle, do go to a smaller multiplier.) One of your experiments should involve setting up a standing wave.
One of your experiments should involve Include the following in your report:
a. Design experiments to solve the problem and discuss how you will use the available equipment to make measurements.
b. Describe the mathematical procedures you will use.
c. List the assumptions are you making. Explain how each could affect the outcome.
d. What are the sources of experimental uncertainty? How could you minimize the uncertainties?
Answer:
The Double Slits Experiment and Michelson Interferometer
The questions will be answered for the double slits experiment.
(b) Mathematically, the double slits experiment equation can be given as: d sin 0 = m(Wavelength). d is the separation distance, 0 is the diffraction angle, m = 1,2,3,...
(c) assumptions
The width of the slits is lesser than the microwave's wavelength. This is to set up a standing wave between the microwave source and detector.
m is an positive integer. To obtain a constructive interference of the E-M wave(microwave)
(d) The uncertainties are:
(i)The zero error in the reading of the multiplier will disrupt the value of the wavelength by small percentage. This can be adjusted to obtain more accurate result.
(ii) The angle as obtained by the gionometer can not be measured to highest level of accuracy, as there are some approximations. High sensitive equipment should be used to obtain accurate result
Explanation:
Double Slits Experiment.
The double slits is a pair of 2cm wide slits with a 6cm separation, cut in a metal foil, with the double slits at the centre of turnable with the microwave source and microwave detector through the slits. The shunt on the microammeter is adjusted to
so that a large scale deflection is obtained. The interference pattern may then be explored by moving the receiver arm, whilst keeping the transmitter arm fixed. A graph of intensity (current) versus θ(Measured through the gionometer) should be plotted, and an estimate made of the microwave wavelength by measuring the angular separation of adjacent maxima in the interference pattern and the separation of the two slits.
Michelson Interferometer
Setting up the interferometer, the mirrors are metal sheets and the beamsplitter is a hardboard sheet. The metal sheets should be carefully positioned to be perpendicular to the microwave beam, and the beamsplitter positioned at the centre of the turntable and at 45° to the beam. Any slight change in the path length of either of the beams leaving the beamsplitter changes the interference pattern at the receiver. One of the metal sheets is mounted in a frame which slides along rails. If this sheet is slowly moved, maxima and minima of current will be recorded by the receiver and a plot of “intensity” against distance can be made. The distance between successive maxima (or minima) is one half of the microwave wavelength which can thus be measured.
A 1500 kg car traveling at 90 km/hr East collides with a 3000 kg truck traveling at 60 km/hr South. They stick together and move which direction?
Answer:
They both move towards South direction.
Explanation:
NB: The car traveling due east must be backing west and the car traveling due south must be backing north.
From their momentum in calculation below:
Momentum of the car traveling east is MU= 1500x90x1000/3600= 37,500kgm/s.
Momentum of car traveling due South =MU = 3000x60x1000/3600= 50,000. So, from here we see the the momentum of the car moving south has the high momemtum so they stick together and move towards the direction of south.
A vertical, piston-cylinder device containing a gas is allowed to expand from 1 m3 to 3 m3. The heat added to the system during the constant pressure expansion was 200 kJ and the decrease in the energy of the system is 340 kJ. Calculate the gas pressure, in kPa.
Answer:
Explanation:
Given
Initial Volume [tex]v_1=1\ m^3[/tex]
final Volume [tex]v_2=3\ m^3[/tex]
Heat added at constant Pressure [tex]Q=200\ kJ[/tex]
Decrease in Energy of System [tex]\Delta U=-340\ kJ[/tex]
According to First law of thermodynamics
[tex]Q=\Delta U+W[/tex]
[tex]W=Q-\Delta U[/tex]
[tex]W=200-(-340)[/tex]
[tex]W=540\ kJ[/tex]
Work done in a constant Pressure Process is given by
[tex]W=P\Delta V[/tex]
where P is the constant Pressure
[tex]540=P\times (3-1)[/tex]
[tex]P=270\ kPa[/tex]
Consider a bicycle that has wheels with a circumference of 1 m. What is the linear speed of the bicycle when the wheels rotate at 2 revolutions per second?
Answer:
V= 12.56 m/s
Explanation:
Given that
R= 1 m
The angular speed ,ω = 2 rev/s
We know that 1 rev = 2 π rad
Therefore
ω = 4 π rev/s
The linear speed V is given as
V= ω x R
V = 4 π x 1 m/s
V= 4 π m/s
Therefore the speed of the bicycle will be 4 π m/s
We know that , π = 3.14
V= 12.56 m/s
The linear speed will be 12.56 m/s
What is the maximum number of 7.00 μF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored within the capacitor array under 953 μC ?
Final answer:
To maintain the total charge below 953 µC for capacitors connected in parallel to a 3.00 V battery, a maximum of 45 capacitors, each with a capacitance of 7.00 µF, can be used. This is calculated by dividing the total charge by the charge per capacitor (953 µC / 21 µC per capacitor).
Explanation:
To find out the maximum number of 7.00 µF capacitors that can be connected in parallel while keeping the total charge under 953 µC, we use the formula for charge on a capacitor, Q=CV, where C is the capacitance and V is the voltage. Since the capacitors would be connected in parallel, the voltage across each capacitor would be the same as the voltage of the battery, which is 3.00 V.
The total charge Q for one capacitor is given by:
Q = C × V = 7.00 µF × 3.00 V = 21.00 µC per capacitor.
To find out how many such capacitors we could connect in parallel without exceeding the total charge of 953 µC, we divide the total permissible charge by the charge per capacitor:
Number of capacitors = Total charge / Charge per capacitor = 953 µC / 21.00 µC per capacitor ≈ 45.38.
Since we can't have a fraction of a capacitor, the maximum number you can use is 45 capacitors.
The maximum number of 7.00 µF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored under 953 μC is 45 capacitors.
To determine the maximum number of 7.00 µF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored within the capacitor array under 953 μC, follow these steps:
First, recall the formula for the charge stored in a capacitor: Q = C × V, where Q is the charge, C is the capacitance, and V is the voltage.Each 7.00 µF capacitor connected to a 3.00 V battery will store a charge of: Q = 7.00 µF × 3.00 V = 21.00 μCTo find the maximum number of capacitors (n) that can be connected while keeping the charge under 953 μC, set up the inequality. n × 21.00 μC < 953 μCSolving for n, we get: n < 953 μC / 21.00 μC ≈ 45.38Since the number of capacitors must be an integer, the maximum number of 7.00 µF capacitors that can be connected in parallel is 45 capacitors.
Find the magnitude of the net electric force exerted on this charge. Express your answer in terms of some or all of the variables q, R, and appropriate constants.
Answer:
Th steps is as shown in the attachment
Explanation:
from the diagram, its indicates that twelve identical charges are distributed evenly on the circumference of the circle. assuming one of gthe charge is shifted to the centre of he circle alomng the x axis, as such the charge is unbalanced and there is need ot balanced all the identical charges for the net force to be equal to zero.
The mathematical interpretation is as shown in the attachment.
Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the effect of temperature on reaction rate and the effect of pressure on boiling point.)
Answer:
In New York is at ocean level while Denver is at an altitude of 1 mile from ocean level. In this way, the breaking point of water is lower in Denver than in New York, that is, the water will boil at lower temperatures in Denver than in New York. If the breaking point of the water decreases, at this point it will put aside more effort to cook an egg. Now the time required to cook the egg is higher in Denver than in New York.
An egg cooks slower in Denver because water boils at a lower temperature due to the city's higher altitude and lower atmospheric pressure. This lower boiling point decreases the rate of the cooking chemical reactions.
Explanation:An egg cooks more slowly in boiling water in Denver than in New York City due to differences in atmospheric pressure and the effect this has on boiling point temperatures. Denver, known as the Mile-High City, is approximately one mile above sea level. This altitude results in a lower atmospheric pressure than New York City which is virtually at sea level. Because of this, water boils at a slightly lower temperature in Denver (approximately 202 degrees Fahrenheit) than it does in New York City (approximately 212 degrees Fahrenheit).
Since cooking is essentially a series of chemical reactions, and these reactions occur faster at higher temperatures, an egg will cook more slowly in Denver’s boiling water than in New York's. This is because the boiling water in Denver is at a lower temperature due to the effect of pressure on boiling point.
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5. An automobile cooling system holds 16 L of water. How much heat does it absorb if its temperature rises from 20o C to 90o C?
Answer:
[tex]Q=4704000J\\Q=4.7MJ[/tex]
Explanation:
Given data
Heat capacity c=4.2 J/gC
Water weigh m=1000g
Temperature Risen T=20°C to 90°C
To find
Heat Q
Solution
The formula to find Heat is:
Q=mcΔT
Where
m is mass
c is heat capacity
ΔT is change in temperature
So
[tex]Q=4.2(16*1000)*(90-70)\\Q=4704000J\\Q=4.7MJ[/tex]
The automobile cooling system would absorb approximately 4,699,840 Joules of heat when its temperature rises from 20°C to 90°C, according to the formula for heat absorption Q = mcΔT.
Explanation:To answer this question, we apply the formula for heat absorption: Q = mcΔT.
Here, 'Q' is the total heat absorbed, 'm' is the mass of the substance (water in this case), 'c' is the specific heat capacity of the substance, and 'ΔT' is the change in temperature. The mass can be defined as the volume of water multiplied by the density of water (1 g/cm^3). So for 16L, m = 16000g. The specific heat capacity of water is approximately 4.186 J/g °C, and the change in temperature, ΔT, is 70°C (90°C - 20°C).
Plugging in these values, we get Q = 16000g× 4.186 J/g °C × 70°C = 4,699,840 Joules. Therefore, it's estimated that the automobile cooling system would absorb around 4,699,840 Joules of heat.
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If the balloon takes 0.19 s to cross the 1.6-m-high window, from what height above the top of the window was it dropped?
Answer:
[tex]heigth=2.86m[/tex]
Explanation:
Given data
time=0.19 s
distance=1.6 m
To find
height
Solution
First we need to find average velocity
[tex]V_{avg}=\frac{distance}{time}\\V_{avg}=\frac{1.6m}{0.19s}\\V_{avg}=8.42m/s[/tex]
Also we know that average velocity
[tex]V_{avg}=(V_{i}+V_{f})/2\\[/tex]
Where
Vi is top of window speed
Vf is bottom of window speed
Also we now that
[tex]V_{f}=V_{i}+gt\\V_{f}=V_{i}+(9.8)(0.19)\\V_{f}=V_{i}+1.862[/tex]
Substitute value of Vf in average velocity
So
[tex]V_{avg}=(V_{i}+V_{f})/2\\where\\V_{f}=V_{i}+1.862\\and\\V_{avg}=8.42m/s\\So\\8.42m/s=(V_{i}+V_{i}+1.862)/2\\2V_{i}+1.862=16.84\\V_{i}=(16.84-1.862)/2\\V_{i}=7.489m/s\\[/tex]
Vi is speed of balloon at top of the window
Now we need to find time
So
[tex]V_{i}=gt\\t=V_{i}/g\\t=7.489/9.8\\t=0.764s[/tex]
So the distance can be found as
[tex]distance=(1/2)gt^{2}\\ distance=(1/2)(9.8)(0.764)^{2}\\ distance=2.86m[/tex]
To determine the height from which the balloon was dropped, we can use the equation for vertical motion: h = 0.5*g*t^2, where h is the height, g is the acceleration due to gravity, and t is the time of flight. Given that the balloon takes 0.19 s to cross the 1.6 m high window, we can solve for the initial height to be approximately 0.01485 m or 14.85 cm above the top of the window.
Explanation:To determine the height from which the balloon was dropped, we can use the equation for vertical motion: h = 0.5*g*t^2, where h is the height, g is the acceleration due to gravity, and t is the time of flight.
Given that the balloon takes 0.19 s to cross the 1.6 m high window, we can plug in these values into the equation to solve for the initial height:
1.6 = 0.5*9.8*(0.19)^2
Simplifying the equation, we find that the balloon was dropped from a height of approximately 0.01485 m or 14.85 cm above the top of the window.
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A strong lightning bolt transfers an electric charge of about 16 C to Earth (or vice versa). How many electrons are transferred? Avogadro’s number is 6.022 × 1023 /mol, and the elemental charge is 1.602 × 10−19 C.
Answer:
Number of electrons, [tex]n=9.98\times 10^{19}[/tex]
Explanation:
A strong lightning bolt transfers an electric charge of about 16 C to Earth, q = 16 C
We need to find the number of electrons that transferred. Let there are n electrons transferred. It is given by using quantization of electric charge as :
q = ne
[tex]n=\dfrac{q}{e}[/tex]
e is elemental charge
[tex]n=\dfrac{16}{1.602\times 10^{-19}}[/tex]
[tex]n=9.98\times 10^{19}[/tex]
So, there are [tex]9.98\times 10^{19}[/tex] electrons that gets transferred. Hence, this is the required solution.
A strong lightning bolt that transfers an electric charge of about 16 C to Earth, transfers 1.0 × 10²⁰ electrons (1.7 × 10⁻⁴ moles of electrons).
A strong lightning bolt transfers an electric charge of about 16 C to Earth (or vice versa).
What is the electric charge?Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field.
We want to calculate the number of electrons that have a charge of 16 C. We have to consider the following relationships:
The charge of 1 electron is 1.602 × 10⁻¹⁹ C.There are 6.022 × 10²³ electrons in 1 mole (Avogadro's number).16 C × 1 electron/1.602 × 10⁻¹⁹ C = 1.0 × 10²⁰ electron
1.0 × 10²⁰ electron × 1 mol/6.022 × 10²³ electron = 1.7 × 10⁻⁴ mol
A strong lightning bolt that transfers an electric charge of about 16 C to Earth, transfers 1.0 × 10²⁰ electrons (1.7 × 10⁻⁴ moles of electrons).
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If you find two stars with the same Right Ascension, are they necessarily close together in the sky? Why or why not?
In space, spatial coordinates can be roughly divided into measures of Right ascension and declination. The declination is measured in degrees while the ascent is measured in hours, minutes, seconds. When you have objects in space such as those of the characteristics presented we will have to they are not necessarily close together in the sky because we can find two stars on the same right ascension but on different declination lines (Which means they can be very far apart from each other)
Two 1.4 g spheres are charged equally and placed 1.6 cm apart. When released, they begin to accelerate at 110 m/s2 . Part A What is the magnitude of the charge on each sphere? Express your answer to two significant figures and include the appropriate units.
Answer:
[tex]q = 2.17\times 10^{-15}~{\rm C}[/tex]
Explanation:
The relation between the acceleration and the electrical force between the spheres can be found by Newton's Second Law.
[tex]\vec{F} = m\vec{a}\\\vec{F} = (1.4\times 10^{-3})(110) = 0.154~N[/tex]
This is equal to the Coulomb's Force.
[tex]F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{2q}{(1.6 \times 10^{-2})^2} = 0.154\\q = 2.17\times 10^{-15}~C[/tex]
A particle moves so that its position (in meters) as a function of time (in seconds) is . Write expressions (in unit vector notation) for (a) its velocity and (b) its acceleration as functions of time.
Answer:
a.Velocity=[tex]\vec{v}=(6t)\hat{j}+8\hat{k}[/tex]
b.[tex]\vec{a}=6\hat{j}[/tex]
Explanation:
We are given that
A particle moves so that its position (in m) as function of time is
[tex]\vec{r}=2\hat{i}+(3t^2)\hat{j}+(8t)\hat{k}[/tex]
a.We have to find its velocity
We know that
Velocity,[tex]v=\frac{dr}{dt}[/tex]
Using the formula
Velocity,v=[tex]\frac{d(2i+3t^2j+8tk)}{dt}[/tex]
Velocity=[tex]\vec{v}=(6t)\hat{j}+8\hat{k}[/tex]
b.Acceleration[tex]=\vec{a}=\frac{d\vec{v}}{dt}[/tex]
[tex]\vec{a}=\frac{d((6t)\hat{j}+8\hat{k})}{dt}[/tex]
[tex]\vec{a}=6\hat{j}[/tex]
The velocity and acceleration of a particle can be calculated from its position function using differentiation in respect to time. The first derivative gives the velocity, while the second derivative gives the acceleration. Unit vector notation is used to display the results.
Explanation:The question involves the calculations of a particle's velocity and acceleration over time, represented in unit vector notation. This scenario falls within the realm of physics, specifically kinematics. Given a particle's motion described by a position function, we can calculate its velocity and acceleration as functions of time.
V(t), the velocity of the particle, is the first derivative of the position function. Hence, to find the velocity, we simply differentiate the position function with respect to time. Now, to calculate A(t), the acceleration of the particle, we take the first derivative of the velocity function, or equivalently, the second derivative of the position function. This gives us the rate of change of velocity, which represents acceleration.
For example, suppose the particle's position function r(t) is given by r(t) = 3t^2 + 2 squares. Differentiating the position function once gives us v(t) = 6t + 2, the velocity function. Differentiating v(t) gives us a(t) = 6, the acceleration function. Both velocity and acceleration are given in unit vector notation.
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At the instant that you fire a bullet horizontally from a rifle, you drop a bullet from the height of the gun barrel. If there is no air resistance, which bullet hits the level ground first? Explain.
Answer:
Both hit at the same time
Explanation:
If air resistant is ignored, then gravitational acceleration g is the only thing that affect the bullets vertical motion, no matter what their horizontal motion is. Since both bullets are starting from rest, vertically speaking their speeds are 0 initially, they are both subjected to the same acceleration g, then they travel at the same rate and would reach the ground at the same time.
A resultant vector is 8.00 units long and makes an angle of 43.0 degrees measured ������� – ��������� with respect to the positive � − ����. What are the magnitude and angle (measured ������� – ��������� with respect to the positive � − ����) of the equilibrant vector? Please show all steps in your calculations
Answer:
223 degree
Explanation:
We are given that
Magnitude of resultant vector= 8 units
Resultant vector makes an angle with positive -x in counter clockwise direction
[tex]\theta=43^{\circ}[/tex]
We have to find the magnitude and angle of the equilibrium vector.
We know that equilibrium vector is equal in magnitude and in opposite direction to the given vector.
Therefore, magnitude of equilibrium vector=8 units
x-component of a vector=[tex]v_x=vcos\theta[/tex]
Where v=Magnitude of vector
Using the formula
x-component of resultant vector=[tex]v_x=8cos43=5.85[/tex]
y-component of resultant vector=[tex]v_y=vsin\theta=8sin43=5.46[/tex]
x-component of equilibrium vector=[tex]v_x=-5.85[/tex]
y-component of equilibrium vector=[tex]-v_y=-5.46[/tex]
Because equilibrium vector lies in III quadrant
[tex]\theta=tan^{-1}(\frac{v_x}{v_y})=tan^{-1}(\frac{-5.46}{-5.85})=43^{\circ}[/tex]
The angle [tex]\theta'[/tex]lies in III quadrant
In III quadrant ,angle =[tex]\theta'+180^{\circ}[/tex]
Angle of equilibrium vector measured from positive x in counter clock wise direction=180+43=223 degree
(5 pts) A sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz. How far doesthe needlemove in one period?A.0.0127 mB.0.0254 mC.0.0508 m
Answer:
B. 0.0254m
Explanation:
A is the amplitude of the oscillation, i.e. the
maximum displacement of the object from
equilibrium, either in the positive or negative x-direction. Simple harmonic motion is repetitive.
The period T is the time it takes the object tocomplete one oscillation and return to the startingposition.
d = 2A = 2×0.0127
A water droplet of mass ‘m’ and net charge ‘-q’ remains stationary in the air due to Earth’s Electric field.
(a) What must be the direction of the Earth’s electric field?
(b) Find an expression for the Earth’s electric field in terms of the mass and charge of the droplet.
Answer:
(a) the electric field of the Earth will be directed towards the negatively charged water droplet.
(b) E = (9.8*m)/q
Explanation:
Part (a) the direction of the Earth’s electric field
Electric field is always directed towards negatively charged objects.
The water droplet has a negative charge ‘-q’, therefore the electric field of the Earth will be directed towards the negatively charged water droplet.
Part (b) an expression for the Earth’s electric field in terms of the mass and charge of the droplet
The magnitude of Electric field is given as;
E = F/q
where;
f is force and q is charge
Also from Newton's law, F = mg
where;
m is mass and g is acceleration due to gravity = 9.8 m/s²
E = F/q = mg/q
E = (9.8*m)/q, Electric field in terms of mass and charge of the droplet.
Two cars, one in front of the other, are travelling down the highway at 25 m/s. The car behind sounds its horn, which has a frequency of 640 Hz. What is the frequency heard by the driver of the lead car? (Vsound=340 m/s).
The answer choices are:
A) 463 Hz
B) 640 Hz
C)579 Hz
D) 425 Hz
E) 500 Hz
Answer:
[tex] f_s = 640 Hz[/tex]
Explanation:
For this case we know that the speed of the sound is given by:
[tex] V_s = 340 m/s[/tex]
And we have the following info provided:
[tex] v_c = 25 m/s [/tex] represent the car leading
[tex] v_s= 25 m/s[/tex] represent the car behind with the source
[tex] f_o = 640 Hz[/tex] is the frequency for the observer
And we can find the frequency of the source [tex] f_s[/tex] with the following formula:
[tex] f_s = \frac{v-v_o}{v-v_s} f_o [/tex]
And replacing we got:
[tex] f_s = \frac{340-25}{340-25} *640 Hz = 640 Hz[/tex]
So then the frequency for the source would be the same since the both objects are travelling at the same speed.
[tex] f_s = 640 Hz[/tex]
Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it 2.20 m above the ground. Ignore air resistance. (a) What is the speed of the shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?
Answer:
a. 6.69m/s
b. y=4.48m
c. t=1.43secs
Explanation:
Data given, acceleration,a=35m/s^2
distance covered,d=64cm=0.64m,
a. to determine the speed, we use the equation of motion
initial velocity,u=0m/s
if we substitute values we arrive at
[tex]v^{2}=u^{2}+2as\\v^{2}=0+2*35*0.64\\v^{2}=44.8m/s\\v=\sqrt{44.8}\\ v=6.69m/s\\[/tex]
b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2
and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.
Hence we can write the equation above again
[tex]v^{2}=u^{2}-2a(y-2.2)\\[/tex]
if we substitute values we have
[tex]v^{2}=u^{2}-2a(y-2.2)\\0=6.69^{2}-2*9.81(y-2.2)\\y-2.2=\frac{44.76}{19.62} \\y=2.28+2.2\\y=4.48m[/tex]
c. the time it takes to arrive at 1.83m is obtain by using the equation below
[tex]1.83-2.2=6.69t-\frac{1}{2} *9.81t^{2}\\4.9t^{2}-6.69t-0.37\\using \\t= \frac{-b±\sqrt{b^{2}-4ac} }{2a}\\ where \\a=4.9, b=-6.69, c=-0.37[/tex]
if we insert the values, we solve for t , hence t=1.43secs
(a) The speed of the shot when Sam releases it 6.75 m/s.
(b) The height risen by the shot above the ground is 4.52 m.
(c) The time taken for the shot to return to 1.8 m above the ground is 1.44 s.
The given parameters;
constant acceleration, a = 35 m/s²height above the ground, h₀ = 64 cm = 2.2 mheight traveled, Δh = 64 cm = 0.64 mThe speed of the shot when Sam releases it is calculated as;
[tex]v^2 = u^2 + 2as\\\\v^2 = 0 + 2a(\Delta h)\\\\v = \sqrt{2a(\Delta h)} \\\\v = \sqrt{2\times 35(0.65)} \\\\v = 6.75 \ m/s[/tex]
The height risen by the shot is calculated as follows;
[tex]v^2 = u^2 + 2gh\\\\at \ maximum \ height , v = 0\\\\0 = (6.75)^2 + 2(-9.8)h\\\\19.6h = 45.56 \\\\h = \frac{45.56}{19.6} \\\\h =2.32 \ m[/tex]
The total height above the ground = 2.20 m + 2.32 m = 4.52 m.
The time taken for the shot to return to 1.8 m above the ground is calculated as follows;
the time taken to reach the maximum height is calculated as;
[tex]h = vt - \frac{1}{2} gt^2\\\\2.32 = 6.75t - (0.5\times 9.8)t^2\\\\2.32 = 6.75t -4.9t^2\\\\4.9t^2 -6.75t + 2.32=0\\\\a = 4.9, \ b = -6.75, \ c = 2.32\\\\t = \frac{-b \ \ + /- \ \ \sqrt{b^2-4ac} }{2a} \\\\t = \frac{-(-6.75) \ +/- \ \ \sqrt{(-6.75)^2 - 4(4.9\times 2.32)} }{2(4.9)} \\\\t = 0.72 \ s\ \ or \ \ 0.66 \ s[/tex]
[tex]t \approx 0.7 \ s[/tex]
height traveled downwards from the maximum height reached = 4.52 m - 1.8 m = 2.72 m
[tex]h = vt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2(2.72)}{9.8} } \\\\t = 0.74 \ s[/tex]
The total time spent in air;
[tex]t = 0.7 \ s \ + \ 0.74 \ s\\\\t = 1.44 \ s[/tex]
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An experiment is performed in the lab, where the mass and the volume of an object are measured to determine its density. Two completely different valid methods are used. Each experimental method to measure the density is performed, two considerable sets of data are taken on each and the results are compared. The results of the density measurement by each method should be:______.
a. dependent on the mass and the volume of the object
b. completely different, because it was measured different ways.
c. correct on the most accurate method and wrong in the other
d. the same within the uncertainty of each measurement method
Answer:
d. the same within the uncertainty of each measurement method
Explanation:
The density of an object and in general any physical property, has the same value regardless of the method used to measure it, either directly or indirectly. Since two completely different valid methods are used, the results must be the same, taking into account the level of precision of each of the methods.
You are going to an outdoor concert, and you'll be standing near a speaker that emits 60 W of acoustic power as a spherical wave. What minimum distance should you be from the speaker to keep the sound intensity level below 94 dB?
Answer:
[tex]r=44m[/tex]
Explanation:
β is calculated as:
[tex]\beta =(10dB)log_{10}(I/I_{o} )\\ I=I_{o}10^{\frac{\beta }{10dB} }\\ I=(1.0*10^{-12}W/m^{2} )10^{\frac{\(94dB }{10dB} }\\I=2.51mW/m^{2}[/tex]
The distance r is defined as the radius of spherical wave.solve for r
We have
[tex]I=\frac{P_{source} }{4\pi r^{2} }\\ r=\sqrt{\frac{P_{source}}{4\pi I} }\\ r=\sqrt{\frac{60W}{4\pi (2.51mW/m^{2} )} }\\r=44m[/tex]
A tank with a volume of 0.150 m3 contains 27.0oC helium gas at a pressure of 100 atm. How many balloons can be blown up if each filled balloon is a sphere 30.0 cm in diameter at 27.0oC and absolute pressure of 1.20 atm? Assume all the helium is transferred to the balloons.
Answer:
884 balloons
Explanation:
Assume ideal gas, since temperature is constant, then the product of pressure and volume is constant.
So if pressures reduces from 100 to 1.2, the new volume would be
[tex]V_2 = \frac{P_1V_1}{P_2} = \frac{100*0.15}{1.2} = 12.5 m^2[/tex]
The spherical volume of each of the balloon of 30cm diameter (15 cm or 0.15 m in radius) is
[tex]V_b = \frac{4}{3}\pir^3 = \frac{4}{3}\pi 0.15^3 = 0.014 m^3[/tex]
The number of balloons that 12.5 m3 can fill in is
[tex]V_2/V_b = 12.5 / 0.014 = 884[/tex]
Final answer:
Based on the volume of the tank, temperature, and pressure, we can calculate the number of moles of helium. By dividing this number by the number of moles of helium in one balloon, we can determine how many balloons can be blown up.
Explanation:
To find out how many balloons can be blown up with the given amount of helium, we need to calculate the total volume of helium in the tank and divide it by the volume of each balloon.
Given:
The volume of the tank is 0.150 m³
The temperature of the helium in the tank is 27.0°C
The pressure of the helium in the tank is 100 atm
The volume of each balloon is 30.0 cm in diameter, which is equivalent to a radius of 15.0 cm or 0.15 m
The temperature of the filled balloon is 27.0°C
The absolute pressure of the filled balloon is 1.20 atm
First, we need to convert the volume of the tank to liters:
0.150 m³ * 1000 L/m³ = 150 L
Next, we need to calculate the number of moles of helium in the tank using the ideal gas law:
P * V = n * R * T
n = (P * V) / (R * T)
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Using the given values:
Pressure (P) = 100 atm
Volume (V) = 150 L
Ideal gas constant (R) = 0.0821 L·atm/K·mol
Temperature (T) = 27.0°C + 273.15 = 300.15 K
Calculating the number of moles:
n = (100 atm * 150 L) / (0.0821 L·atm/K·mol * 300.15 K) = 6.19 moles
Now we can calculate the number of balloons that can be blown up:
Number of balloons = (Number of moles of helium) / (Number of moles of helium in one balloon)
The number of moles of helium in one balloon can be found using the ideal gas law:
P * V = n * R * T
n = (P * V) / (R * T)
Using the given values:
Pressure (P) = 1.20 atm
Volume (V) = (4/3) * π * (0.15 m)³ = 0.141 m³
Ideal gas constant (R) = 0.0821 L·atm/K·mol
Temperature (T) = 27.0°C + 273.15 = 300.15 K
Calculating the number of moles:
n = (1.20 atm * 0.141 m3) / (0.0821 L·atm/K·mol * 300.15 K) = 0.006 moles
Finally, calculating the number of balloons:
Number of balloons = 6.19 moles / 0.006 moles = 1031
Therefore, 1031 balloons can be blown up with the given amount of helium.
A uniformly charged ring of radius 10.0 cm has a total charge of 78.0 μC. Find the electric field on the axis of the ring at the following distances from the center of the ring. (Choose the x-axis to point along the axis of the ring.)(a) 1.00 cm(b) 5.00 cm(c) 30.0 cm(d) 100 cm
Answer:
6908316.619 N/C
25087609.3949 N/C
6652357.02259 N/C
690831.6619 N/C
Explanation:
x = Distance from the ring
R = Radius of ring = 10 cm
q = Charge = 78 μC
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
Electric field at a point x is from a ring given by
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}[/tex]
For 1 cm
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 0.01}{(0.01^2+0.1^2)^{1.5}}\\\Rightarrow E=6908316.619\ N/C[/tex]
The electric field is 6908316.619 N/C
For 5 cm
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 0.05}{(0.05^2+0.1^2)^{1.5}}\\\Rightarrow E=25087609.3949\ N/C[/tex]
The electric field is 25087609.3949 N/C
For 30 cm
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 0.3}{(0.3^2+0.1^2)^{1.5}}\\\Rightarrow E=6652357.02259\ N/C[/tex]
The electric field is 6652357.02259 N/C
For 100 cm
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 1}{(1^2+0.1^2)^{1.5}}\\\Rightarrow E=690831.6619\ N/C[/tex]
The electric field is 690831.6619 N/C
A battery has emf E and internal resistance r = 1.50 Ω. A 14.0 Ω resistor is connected to the battery, and the resistor consumes electrical power at a rate of 92.0 J/s. What is the emf of the battery?
Answer:
E = 39.68 V
Explanation:
EMF ( Electromotive force) : This is defined as the magnitude of the potential difference of both the external circuit and the inside of the cell. The S.I unit is Volt.
The Expression for E.M.F is given as,
E = I(R+r) ................... Equation 1
Where E = EMF, I = current, R = External Resistance, r = internal resistance.
Also,
P = I²R
I = √(P/R) ..................... Equation 2
Where P = power, R = External resistance.
Given: P = 92 J/s, R = 14 Ω.
Substitute into equation 2
I = √(92/14)
I = √(6.57)
I = 2.56 A.
Also Given: r = 1.5 Ω.
Substitute into equation 1
E = 2.56(1.5+14)
E = 2.56(15.5)
E = 39.68 V.
A battery has emf E and internal resistance r = 2.00 Ω. A 11.5 Ω resistor is connected to the battery, and the resistor consumes electrical power at a rate of 96.0 J/s.
Part APart complete
What is the emf of the battery?
Express your answer with the appropriate units.
E =
39.0 V
A plane flies 1.4hours at 110mph on a bearing of 40degrees.It then turns and flies 1.7hours at the same speed on a bearing of 130degrees.How far is the plane from its starting point?
Answer:
Distance from starting point= 242.249871 miles
Explanation:
Distance covered on bearing of 40 degree=a= [tex]1.4*110\ mph[/tex]
Distance covered on bearing of 40 degree=a=154 miles
Distance covered on bearing of 130 degree=b=[tex]1.7*110\ mph[/tex]
Distance covered on bearing of 40 degree=b=187 miles
Angle between bearing=[tex]130-40[/tex]
Angle between bearing=90 degree
With the angle of 90 degree, Distance from starting point can be calculated from Pythagoras theorem.
Distance from starting point=[tex]\sqrt{a^2+b^2}[/tex]
Distance from starting point=[tex]\sqrt{154^2+187^2}[/tex]
Distance from starting point= 242.249871 miles
Answer:
Distance from starting point= 242.249871 miles
Distance covered on bearing of 40 degree=a=
Distance covered on bearing of 40 degree=a=154 miles
Distance covered on bearing of 130 degree=b=
Distance covered on bearing of 40 degree=b=187 miles
Angle between bearing=
Angle between bearing=90 degree
With the angle of 90 degree, Distance from starting point can be calculated from Pythagoras theorem.
Distance from starting point=
Distance from starting point=
Distance from starting point= 242.249871 miles
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a skydiver jumps out of a plane waering a suit and decleration due to air i s 0.45 m/s^2 for each 1 m/s, skydyvers initial value problems that models the skydiver velocity is v(t). temrinal speed assuming that accelration due to gravity is 9.8m/s
Answer:
[tex]v = 21.77\ m/s[/tex]
Explanation:
Given,
Air resistance, = 0.45 m/s² for each 1 m/s
Air resistance for velocity v = 0.45 v.
Terminal velocity = ?
acceleration due to gravity= g = 9.8 m/s²
now,
a = 0.45 v
we know,
[tex]a =\dfrac{dv}{dt}[/tex]
[tex]\dfrac{dv}{dt}=0.45 v[/tex]
[tex]\dfrac{dv}{v} = 0.45 dt[/tex]
integrating both side
[tex]\int \dfrac{dv}{v} = 0.45\int dt[/tex]
[tex] ln(v) = 0.45 t[/tex]
[tex]t = e^{0.45 t}[/tex]
When a body is moving at terminal velocity, Force due to gravity is equal to force due to air resistance.
m a = m g
0.45 v = 9.8
[tex]v = \dfrac{9.8}{0.45}[/tex]
[tex]v = 21.77\ m/s[/tex]
Hence, the terminal velocity of the skydiver is equal to 21.77 m/s.
The main waterline into a tall building has a pressure of 90 psia at 16 ft elevation below ground level. Howmuch extra pressure does a pump need to add to ensure a waterline pressure of 30 psia at the top floor 450 ft above ground?
Answer:
Explanation:
Given
initial Pressure [tex]P_1=90\ psia[/tex]
elevation [tex]z_1=16\ ft[/tex]
Final Pressure [tex]P_2=30\ psia[/tex]
elevation [tex]z_2=450\ ft[/tex]
Pressure after Pumping(pump inlet pressure ) is given by
[tex]P_{after\ pump}=P_{top}+\Delta P[/tex]
[tex]\Delta P=\rho gh[/tex]
where [tex]\rho [/tex]=density of water[tex](62.2\ lbm\ft^3)[/tex]
g=acceleration due to gravity[tex](32.2\ ft/s^2)[/tex]
h=elevation
[tex]\Delta P=62.2\times 32.2\times (16+450)\times \frac{1\ lbf}{32.174\ lbm\ ft}[/tex]
[tex]\Delta P=28,985\ lbf/ft^2[/tex]
[tex]\Delta P=201.3\ lbf/in.^2[/tex]
[tex]P_{after\ pump}=30+201.3=231.3\ lbf/in.^2[/tex]
Pressure required to be applied
[tex]\Delta P_{pump}=P_{after\ pump}-P_{bottom}[/tex]
[tex]\Delta P_{pump}=231.3-90=141.3\ psi[/tex]
A ball is thrown at an angle of to the ground. If the ball lands 90 m away, what was the initial speed of the ball?
Answer:
The initial speed of the ball is 30 m/s.
Explanation:
It can be assumed that the ball is thrown at an angle of 45 degrees to the ground. The ball lands 90 m away. We need to find the initial speed of the ball. We know that the horizontal distance covered by the projectile is called its range. It is given by :
[tex]R=\dfrac{u^2\ sin2\theta}{g}[/tex]
u is the initial speed of the ball.
[tex]v=\sqrt{\dfrac{Rg}{sin2\theta}}[/tex]
[tex]v=\sqrt{\dfrac{90\times 9.8}{sin2(45)}}[/tex]
v = 29.69 m/s
or
v = 30 m/s
So, the initial speed of the ball is 30 m/s. Hence, this is the required solution.
The exact initial speed requires the angle [tex]\( \theta \)[/tex]; assuming [tex]\( \theta = 45^\circ \)[/tex], the initial speed is 29.71 m/s.
To solve this problem, we need to use the kinematic equations for projectile motion. The horizontal distance (range) \( R \) that the ball travels can be found using the following equation:
[tex]\[ R = \frac{{v_0^2 \sin(2\theta)}}{g} \][/tex]
Given that the ball lands 90 m away and assuming that the angle of projection [tex]\( \theta \)[/tex] is known (but not provided in the question), we can rearrange the equation to solve for [tex]\( v_0 \)[/tex]:
[tex]\[ v_0^2 = \frac{R \cdot g}{\sin(2\theta)} \][/tex]
[tex]\[ v_0 = \sqrt{\frac{R \cdot g}{\sin(2\theta)}} \][/tex]
However, since the angle [tex]\( \theta \)[/tex] is not given, we cannot calculate the exact initial speed [tex]\( v_0 \)[/tex] without additional information. The problem as stated is incomplete because it requires the value of [tex]\( \theta \)[/tex] to find [tex]\( v_0 \)[/tex].
If we assume that the angle [tex]\( \theta \)[/tex] is such that [tex]\( \sin(2\theta) \)[/tex] is maximized (which occurs at [tex]\( \theta = 45^\circ \)[/tex], then [tex]\( \sin(2 \cdot 45^\circ) = \sin(90^\circ) = 1 \)[/tex], and the equation simplifies to:
[tex]\[ v_0 = \sqrt{R \cdot g} \][/tex]
[tex]\[ v_0 = \sqrt{90 \, \text{m} \cdot 9.81 \, \text{m/s}^2} \][/tex]
[tex]\[ v_0 = \sqrt{882.9 \, \text{m}^2/\text{s}^2} \][/tex]
[tex]\[ v_0 =29.71 \, \text{m/s} \][/tex]
Therefore, under the assumption that [tex]\( \theta = 45^\circ \)[/tex], the initial speed of the ball would be [tex]\( 29.71 \, \text{m/s} \)[/tex].
In conclusion, without the value of [tex]\( \theta \)[/tex], we cannot determine the exact initial speed of the ball. However, if we assume the most favorable condition for maximum range, where [tex]\( \theta = 45^\circ \)[/tex], the initial speed [tex]\( v_0 \)[/tex] would be [tex]\( 29.71 \, \text{m/s} \)[/tex].