Answer:
The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.
Explanation:
We can calculate the speed of the train using the Doppler equation:
[tex] f = f_{0}\frac{v + v_{o}}{v - v_{s}} [/tex]
Where:
f₀: is the emitted frequency
f: is the frequency heard by the observer
v: is the speed of the sound = 343 m/s
[tex] v_{o}[/tex]: is the speed of the observer = 0 (it is heard in the town)
[tex] v_{s}[/tex]: is the speed of the source =?
The frequency of the train before slowing down is given by:
[tex] f_{b} = f_{0}\frac{v}{v - v_{s_{b}}} [/tex] (1)
Now, the frequency of the train after slowing down is:
[tex] f_{a} = f_{0}\frac{v}{v - v_{s_{a}}} [/tex] (2)
Dividing equation (1) by (2) we have:
[tex] \frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}} [/tex]
[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}} [/tex] (3)
Also, we know that the speed of the train when it is slowing down is half the initial speed so:
[tex] v_{s_{b}} = 2v_{s_{a}} [/tex] (4)
Now, by entering equation (4) into (3) we have:
[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}} [/tex]
[tex] \frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}} [/tex]
By solving the above equation for [tex]v_{s_{a}}[/tex] we can find the speed of the train after slowing down:
[tex] v_{s_{a}} = 11.06 m/s [/tex]
Finally, the speed of the train before slowing down is:
[tex] v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s [/tex]
Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.
I hope it helps you!
A bag contains lenses with focal lengths 10 cm, 20 cm and 25 cm which are not marked with their focal length. Describe a simple activity to identify the three types lenses
pls give the answer ASAP!!!!!
Explanation:
ehb-pynw-ayo
joi n fast
An iron nail floats in mercury and sinks in water. explain why?
Answer:
because density of iron is more than that of water but less then that of Mercury
hope it's helpful
A car starts from rest and accelerates uniformly at 3.0 m/s2 toward the north. A second car starts from rest 6.0 s later at the same point and accelerates uniformly at 5.0 m/s2 towar the north. How long after the second car starts does it overtake the irst car?
a. 12 s
b. 19 s
c. 21 s
d. 24 s
A certain sound is recorded by a microphone. The same microphone then detects a second sound, which is identical to the first one except that the amplitude of the pressure fluctuations is larger. In addition to the larger amplitude, what distinguishes the second sound from the first one
Answer:
Loudness of the second sound is more than the first one.
Explanation:
There are two sounds, the second sound is identical to first but the loudness of second is more than the first one.
As the frequency is same so the itch is same for both the sounds.
As the loudness depends on the amplitude of the sound so the loudness of the second sound is more than the first sound.
Two small silver spheres, each of mass m=6.2 g, are separated by distance d=1.2 m. As a result of transfer of some fraction of electrons from one sphere to the other, there is an attractive force F=900 KN between the spheres. Calculate the fraction of electrons transferred from one of the spheres: __________
To evaluate the total number of electrons in a silver sphere, you will need to invoke Avogadro's number, the molar mass of silver equal to 107.87 g/mol and the fact that silver has 47 electrons per atom.
Answer:
4.60 × 10⁻⁸
Explanation:
From the given information;
Assuming that q charges are transferred, then:
[tex]F = \dfrac{kq^2}{d^2}[/tex]
where;
k = 9 ×10⁹
[tex]900000 = \dfrac{9*10^9 \times q^2}{1.2^2}[/tex]
[tex]q = \sqrt{\dfrac{900000\times 1.2^2 }{9*10^9}}[/tex]
q = 0.012 C
No of the electrons transferred is:
[tex]= \dfrac{0.012}{1.6\times 10^{-19}} C[/tex]
[tex]= 7.5 \times 10^{16} \ C[/tex]
Initial number of electrons = N × 47 × no of moles
here;
[tex]\text{ no of moles }= \dfrac{6.2}{107.87}[/tex]
no of moles = 0.0575 mol
∴
Initial number of electrons = [tex]6.023\times 10^{23} \times 47 \times 0.0575 mol[/tex]
= 1.63 × 10²⁴
The fraction of electrons transferred [tex]=\dfrac{7.5\times 10^{16} }{1.6 3\times 10^{24}}[/tex]
= 4.60 × 10⁻⁸
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.
Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?
Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
Choose the FALSE statements. In Simple harmonic motion,
I. The velocity of the object does not change at all position
II. The acceleration of the object does not change at all position.
Ill. When velocity is zero, acceleration is also zero
IV. The velocity has maximum magnitude at the equilibrium position.
V. When the net force is maximum, the velocity is zero.
A. I and II
B. III and IV
C. IV and V
D. I, II and III
E. I, II, III and IV
F. I, II, III and V
G. All the above statements are false.
Answer:
b is correct.
Explanation:
because of the question you have given
A heat engine with 0.100 mol of a monatomic ideal gas initially fills a 3000 cm3 cylinder at 800 K. The gas goes through the following closed cycle Isothermal expansion to 5000 cm3 ?
Part A How much work does this engine do per cycle? Express your answer with the appropriate units. sochoric cooling to 200 K -Isothermal compression to 3000 cm3. - Isochoric heating to 800 K Value Units
Part B What is its thermal efficiency? Express your answer with the appropriate units.
Answer:
below
Explanation:
Part A) This engine works per cycle is 254.9 J.
Part B) The thermal efficiency is 23.42%
What is the thermal efficiency?The thermal efficiency of any heat engine is represented in percentage of heat energy converted into work.
For isothermal expansion, work done is
W₁ =nRT₁ x ln(V₂/V₁)
W₁ = 0.1 x 8.314 x 800 x ln(5000/3000)
W₁ = 339.8 J =Q₁
For isochoric cooling ,
W₂ =0
Q₂ =nCvdT = 0.1 x 3R/2 x (T₂-T₁)
Q₂ = -748.3 J
For isothermal compression,
W₃ =nRT₂ ln (V₄/V₃)
W₃ = 0.1 x 8.314 x 200 x ln(3000/5000)
W₃ = -84.9J
For isochoric heating
W₄ =0
Q₄ =nCvdT = 0.1 x 3R/2 x (800-200)
Q₄ = -748.3 J
Total work done in all the process W = W₁ +W₂ +W₃ +W₄
W =254.9 J
Thus, the work done is 254.9 J
Thermal efficiency = Work done/Heat taken
η = W/ Q₁ +Q₄
η = [254.9 / 339.8 +748.3 ] x 100 %
η = 0.2342 x 100 %
η = 23.42%
Thus, the thermal efficiency is 23.42%
Learn more about thermal efficiency.
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If energy is transferred spontaneously as heat from a substance with a temperature of T1 to a substance with a temperature of T2, which of the following statements must be true?
1-T1 < T2
2-T1 = T2
3-T1 > T2
4-more information is needed
Answer: The statement [tex]T_{1} > T_{2}[/tex] must be true.
Explanation:
As it is given that heat is being transferred from a substance with temperature [tex]T_{1}[/tex] to a substance with temperature [tex]T_{2}[/tex].
It is known that heat will always being transferred from a higher temperature to a lower temperature. Because at higher temperature the molecules of a substance acquire more energy and when they lose energy then a decrease in temperature occurs.
Hence, in the given situation [tex]T_{1} > T_{2}[/tex].
Thus, we can conclude that the statement [tex]T_{1} > T_{2}[/tex] must be true.
Select the correct answer.
Which figure shows a correct pattern of field lines?
A. Figure A
B. Figure B
C. Figure C
D. Figure D
Answer the following using equations, number substitution and keep units. 1. What is the speed of an object that travels 5m in 10s. 2. What force is on a 10kg mass that accelerates at 3m/s/s. 3. What is the potential energy of a 7kg object 4m off the ground *
show all your work please
Explanation:
1. Distance, d = 5 m
Time, t = 10 s
Speed = distance/time
[tex]v=\dfrac{5}{10}=0.5\ m/s[/tex]
2. Mass, m = 10 kg
Acceleration, a = 3 m/s³
Force, F = mass (m) × acceleration (a)
F = 10 × 3
= 20 N
3. Mass, m = 7 kg
Height, h = 4 m
Potential energy, E = mgh
E = 7 × 9.8 × 4
E = 274.4 J
Hence, this is the required solution.
which unit would be most suitable for its scale?
A mm
B
с
crn?
D
cm
[0625_504_9p_1].
8
A piece of cotton is measured between two points on a ruler.
1
coton
BAS
2
4
5
6
7
8
9
10
11
12
13
14
15 16
when the lenge of coton is wound closely around a pen, goes round six times.
pen
six turns of coton
दे-
What is the distance onde round the pen?
4 2.2 m
B 26 cm
с
13.2 cm
D 15.6 cm
Answer:
Mm, thats the answer trust me men
When you release some of the paint from a spray paint can and the can remains at the same temperature, which gas law does this represent?
Answer:
Boyle's law.
Explanation:
Robert Boyle was an Irish chemist and is famously referred to as the first modern chemist. He was born on the 25th of January, 1627 in Lismore, Ireland and died on the 31st, December 1691, London, United Kingdom.
Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.
Mathematically, Boyles law is given by;
[tex] PV = K[/tex]
[tex] P_{1}V_{1} = P_{2}V_{2} [/tex]
Where;
P1 is the original pressure.
P2 is the final pressure.
V1 is the original volume.
V2 is the final volume.
Hence, when you release some of the paint from a spray paint can by applying an amount of pressure and the can remains at the same temperature, the gas law which this represent is Boyle's law.
When using the process of evaporation to separate a mixture what is left behind to an evaporating dish
A. The mixture does not separate in the entire mixture remains in the dish
B. The liquid evaporates in the solid is left in the dish
C. The mixture does not separate in the entire mixture evaporates
D. None of these
Answer:
B
Explanation:
The liquid evaporates in the solid is left in the dish..
A car is travelling at a speed of 30m/s on a straight road. what would be the speed of the car in km
Answer:
[tex] = \frac{30 \times {10}^{ - 3} }{1} \\ = 0.03 \: km \: per \: second[/tex]
Answer:
108 km/hr or 0.03 km/s
Explanation:
conversion factor for m/s to km/hr is 5/18
conversion factor for m/s to km/s is 1/1000
3. What is electric current?
The flow of moving electrons
electrons that move one time
Answer:
An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space. It is measured as the net rate of flow of electric charge through a surface or into a control volume. ... In electric circuits the charge carriers are often electrons moving through a wire.
Answer:
The flow of moving electrons
Which describes a characteristic of metallic bonds?
Answer:
arge number of electrons free to move between the charged ions in the lattice.
Explanation:
The metallic bond occurs when an atom with few electrons is united in its last level, therefore the best way to decrease the total energy of the system is to lose all its electrons to remain with the configuration of a noble gas. The electrons that it loses cannot be acquired by other atoms since they all have few electrons, thus leaving a large number of electrons free to move between the charged ions in the lattice.
Some important characteristics emerge from this description of the metallic bond:
* It has many free electrons therefore its electrical conductivity is high
* As the charged ions are fixed, the material can be malleable, bent without breaking since the free electrons create the bond that keeps the system stable.
* As the electrons are free when heating a part of the material, these electrons acquire energy and rapidly propagate it to the other side, giving a high thermal conductivity
* As the temperature increases, the electrons acquire more kinetic energy, which is why there are more collisions between them and consequently the resistivity of the material increases.
please helpp!
convert 1N into dyne
In the given relation F=ma a stands for write there SI unit
Answer:
a. 1 Newton = 100000 Dyne
b. a represents acceleration.
Explanation:
Newton is the standard unit (S.I) of measurement of force. Converting 1 Newton to dyne we have;
1 Newton = 10⁵ Dyne
1 Newton = 100000 Dyne
Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.
Mathematically, it is given by the formula;
Force = mass * acceleration
[tex] F = ma[/tex]
Hence, we can deduce that a represents the acceleration of an object and it's measured in meters per seconds square.
two blocks are held together with a compressed spring between them on the surface of a slippery table .one block has three times the inertia of the other .when the blocks are released ,the spring pushes them away from each other .what is the ratio of their kinetic energies after the release?
Explanation:
The initial kinetic energy [tex]KE_0[/tex] for both blocks is zero. Let [tex]m_1= m[/tex] and [tex]m_2 =3m[/tex]. So using the conservation law of linear momentum, we can write
[tex]0 = m_1v_1 - m_2v_2[/tex]
or
[tex]v_2 = \dfrac{m_1}{m_2}v_1 = \dfrac{m}{3m}v_1 = \dfrac{1}{3}v_1[/tex]
The final kinetic energies for the two masses are
[tex]KE_1 = \frac{1}{2}m_1v_1^2 = \frac{1}{2}mv_1^2[/tex]
[tex]KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(3m)(\frac{1}{3}v_1)^2 = \frac{1}{2}m(\frac{1}{3}v_1^2)[/tex]
Therefore, the ratio of their kinetic energies is
[tex]\dfrac{\Delta KE_2}{\Delta KE_1} = \dfrac{\frac{1}{2}(\frac{1}{3}v_1^2)}{\frac{1}{2}v_1^2} = \dfrac{1}{3}[/tex]
A van tire contacts the ground on a rectangular area of (10 cm) by (15cm). If the bus's mass is (900 kg), what pressure does the car exert on the ground as it rests on all four tires? (g= 9.8 m/s³)
3.80×10^5
14.7×10^4
6.67×10^3
58.8×10^3
Answer: Hmmm im not sure but i'd go with 3.80x10^5
Explanation: Like i mentioned im not very good at physics...sorry if its wrong
what affects our utility
Answer:
Energy Bill fluctuations are inevitable and depend on a variety of different factors. Two of the most important are the current weather your home is experiencing and the current price per Kilowatt Hour (which fluctuates more than you might think).
With what speed must a ball be thrown directly upward so that it remains in the air for 10 seconds?
a) What will be its speed when it hits the ground?
b) How high does the ball rise?
Answer:
◆ See the attachment photo.
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◆ Mark as brainlist.
Diwn unscramble the word
Answer:
wind
Explanation:
just a possible answer.
A power plant generates 150 MW of electrical power. It uses a supply of 1000 MW from a geothermal source and rejects energy to the atmosphere. Find the power to the air and how much air should be flowed to the cooling tower (kg/s) if its temperature cannot be increased more than 10oC.
Answer:
- the power to the air is 850 MW
- mass flow rate of the air is 84577.11 kg/s
Explanation:
Given the data in the question;
Net power generated; [tex]W_{net[/tex] = 150 MW
Heat input; [tex]Q_k[/tex] = 1000 MW
Power to air = ?
For closed cycles
Power to air Q₀ = Heat input; [tex]Q_k[/tex] - Net power generated; [tex]W_{net[/tex]
we substitute
Power to air Q₀ = 1000 - 150
Q₀ = 850 MW
Therefore, the power to the air is 850 MW
given that ΔT = 10 °C
mass flow rate of air required will be;
⇒ Q₀ / CpΔT
we know that specific heat of air at p=c ; Cp = 1.005 kJ/kg.K
we substitute
⇒ ( 850 × 10³ ) / [ 1.005 × 10 ]
⇒ ( 850 × 10³ ) / 10.05
⇒ 84577.11 kg/s
Therefore, mass flow rate of the air is 84577.11 kg/s
Jason throws a basketball straight downward, letting it bounce once before catching it. We can ignore air
resistance
What is true about the acceleration and velocity of the ball on its way up?
Answer:
The acceleration is a negative as the ball is now moving in the opposite direction.
The velocity would decrease as the ball moves upward
Acceleration remains constant and velocity is negative and decreasing as per the given scenario. The correct option is B.
What is acceleration?In mechanics, acceleration is defined as the rate of change of an object's velocity with respect to time.
Vector quantities are accelerations. The orientation of an object's acceleration is determined by the orientation of its net force.
Velocity is the directional speed of a moving object as an indication of its rate of change in position as observed from a specific frame of reference and measured by a specific time standard.
The rate of displacement of an entity known as its velocity. It is measured in meters per second. The rate of change in velocity of an object is defined as acceleration.
According to the scenario, acceleration remains constant while velocity decreases and is negative.
Thus, the correct option is B.
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Your question seems incomplete, the missing options are:
Acceleration increases and velocity is negative and decreasing.Acceleration remains constant and velocity is negative and decreasing,Acceleration decreases and velocity is positive and increasing.Acceleration remains constant and velocity is negative and increasing.A ball is thrown straight up in the air at an initial speed of 30 m/s. At the same time the ball is thrown, a person standing 70 m away begins to run toward the spot where the ball will land.How fast will the person have to run to catch the ball just before it hits the ground?Vperson= m/s
Answer:
Explanation:
Here's what we know and in which dimension:
y dimension:
[tex]v_0=30[/tex] m/s
v = 0 (I'll get to that injust a second)
a = -9.8 m/s/s
The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.
x dimension:
Δx = 70 m
v = ??
Velocity is our unknown.
Solving for the time in the y dimension:
[tex]v=v_0+at[/tex] and filling in:
0 = 30 + (-9.8)t and
-30 = -9.8t so
t = 3.1 seconds
We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.
In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.
Δx = vt and
70 = v(6.2) so
v = 11.3 m/s
A basketball player is getting ready to jump, pushing off the ground and accelerating upward.
A) Draw a force identification diagram.
B) Draw a free body diagram.
Answer:
B
Explanation:
B
Could you please explain the step by step process of setting up this problem?
A 10 kg, 4 m long plank of wood is going to be used as a teeter-totter for a brother and sister. The brother has a mass of 30 kg and the sister a mass of 20 kg. If the brother and sister sit at opposite ends of the plank, how far from the brother should the fulcrum be in order for the teeter-totter to be balanced?
A. 1.33 m
B. 1.60 m
C. 1.67 m
D. 2 m
Which of the following phenomenon odd called photoelectric effect?
A. High energy electrons impinge on a metallic Annie which emits electrons
B. A high energy photon emits photons as it slows down
C. A metal absorbs a quanta of light and then emits electrons
D. Two electrons are created from a quanta of light.
two electrons are created from a quanta of kight
Moving the probe 1 cm towards the non-grounded electrode changes the value the potential from about 0.90 V to about 1.2 V. Explain how you can get the magnitude of the average electric field between these two points on the paper, and give the value of this field in Newtons/Coulomb. Show your calculations.
Answer:
-30 N/C
Explanation:
Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m
Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V
Since E = -ΔV/Δx
substituting the values of the variables into the equation, we have
E = -ΔV/Δx
E = -0.30 V/0.01 m
E = -30 V/m
Since 1 V/m = 1 N/C.
E = -30 N/C
So, the average electric field is -30 N/C