Answer:
The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.
A pipe is closed at one end and the air column vibrate with a fundamental note of 550Hz calculate the length of the resonating air column neglecting end correction
Answer:
The length of the resonating air column is 15 cm.
Explanation:
Given;
fundamental frequency of the air column, F₀ = 550 Hz
Assume speed of sound in air, v = 330 m/s
The wavelength of the air column is calculated as;
[tex]\lambda = \frac{V}{F_o} \\\\\lambda = \frac{330}{550} \\\\\lambda = 0.6 \ m[/tex]
The length of the resonating air column is calculated as;
In fundamental mode for closed pipe;
[tex]L = \frac{\lambda }{4} \\\\L = \frac{0.6}{4} \\\\L = 0.15 \ m\\\\L = 15 \ cm[/tex]
Therefore, the length of the resonating air column is 15 cm.
a. Using the ideas of electric field and force, explain what would happen to an electron if released from rest at r=2.0m?
b. Would the electron released from rest move to a region of higher electrical potential or lower electrical potential?
c. Would the electron released from rest move such that the system would have higher potential energy or lower potential energy?
231 91 Pa has__neutrons
Answer:
140
Explanation:
1. A block with mass 20 kg is
sliding up a plane (Ukinetic=0.3,
inclined at 10°) at a speed of
2 m/s to the right (positive
X-direction). How far does it
go up along the plane before
it comes to rest momentarily?
Answer: 0.435 m
Explanation:
Given
mass m=20 kg
initial speed u=2 m/s
coefficient of kinetic friction [tex]\mu_k=0.3[/tex]
deceleration which opposes the motion is given by
[tex]\Rightarrow a=g\sin \theta+\mu_kg\cos \theta\\\Rightarrow a=g(\sin \theta +\mu_k\cos \theta)[/tex]
[tex]\Rightarrow a=9.8(\sin 10^{\circ}+0.3\times \cos 10^{\circ})\\\Rightarrow a=4.59\ m/s^2[/tex]
using [tex]v^2-u^2=2as[/tex]
[tex]\Rightarrow s=\dfrac{2^2}{2\times 4.59}=0.435\ m[/tex]
man is walking due east at the rate of of 4kmph and the rain is falling 30° east of vertical with a velocity of 6kmph the velocity of rain relative to the man will be?
Answer:
No answer
Explanation:
no explanation
What force is left out of the Quantum Mechanics theory?
Answer:
Quantum mechanics is a key hypothesis in material science that gives a portrayal of the actual properties of nature at the size of iotas and subatomic particles. It is the establishment of all quantum physical science including quantum science, quantum field hypothesis, quantum innovation, and quantum data science.
Explanation:
It is the greatest of issues, it is the littlest of issues. At present physicists have two separate rule books clarifying how nature functions. There is general relativity, which perfectly represents gravity and everything it overwhelms: circling planets, impacting worlds, the elements of the growing universe all in all. That is enormous. At that point there is quantum mechanics, which handles the other three powers – electromagnetism and the two atomic powers. Quantum hypothesis is very proficient at portraying what happens when a uranium molecule rots, or when singular particles of light hit a sun based cell. That is little.
A makeshift sign hangs by a wire that is extended over an ideal pulley and is wrapped around a large potted plant on the
roof as shown in the figure below. When first set up by the shopkeeper on a sunny and dry day, the sign and the pot are in
equilibrium. The mass of the sign is 27.5 kg, and the mass of the potted plant is 67.5 kg.
Plant
sale
today!
(a) Assuming the objects are in equilibrium, determine the magnitude of the static friction force experienced by the
potted plant.
N
(b) What is the maximum value of the static friction force if the coefficient of static friction between the pot and the
roof is 0.707?
N
Answer:I know the answer for B cus I’m doing the same problem. For B, you would only take the coefficient of friction given and then multiply it by the Normal Force, which in this case is the same as the Gravitational Force.
Explanation:
A 69.0 kg ice skater moving to the right with a velocity of 2.61 m/s throws a 0.22 kg snowball to the right with a velocity of 25.2 m/s relative to the ground. (a) What is the velocity of the ice skater after throwing the snowball
Answer:
0.08m/s
Explanation:
Given data
M1= 69kg
v1= 2.61m/s
M2= 0.22kg
v2= 25.2m/s
Before snowball is thrown:
Total mass of skater + snowball = 69+ 0.22 = 69.22kg
Total Momentum of skater + snowball = mv = 69.22 x 2.61 = 180.7 kgm/s
After snowball is thrown:
Let's call the velocity of the skater V.
Total momentum = momentum of skater + momentum of snowball
=69.22V + (5.544)
= 69.22V + 5.544
So:
180.7 = 69.22V+5.544
180.7- 5.544= 69.22V
175.156= 69.22V
V= 175.156/69.22
V = 2.53m/s
The total momentum after catching the snowball is mV or:
(69.0 + 0.22) x V
So:
5.544= 69.22V
V= 5.544/69.22
V=0.08m/s
The velocity of the ice skater after throwing the snowball is 0.08m/s
A airplane accelerated from 59 m/s to 95 m/s in a distance of 123 meters what was its acceleration in m/s^2, assumed constant?
Answer:
22.54 m/s^2
Explanation:
vf = final velocity = 95 m/s
vi = initial velocity = 59 m/s
d = displacement = 123 m
a = acceleration, unknown
Use this kinematics equation to find a:
vf^2 = vi^2 + 2a*d
95^2 = 59^2 + 2*a*123
22.54 m/s^2 = a
Hope this helps!! :)
The masses of astronauts are monitored during long stays in orbit, such as when visiting a space station. The astronaut is strapped into a chair that is attached to the space station by springs and the period of oscillation of the chair in a friction-less track is measured.
(a) The period of oscillation of the 10.0 kg chair when empty is 0.750 s. What is the effective force constant of the springs?
(b) What is the mass of an astronaut who has an oscillation period of 2.00 s when in the chair?
(c) The movement of the space station should be negligible. Find the maximum displacement of the 100,000 kg sace station if the astronaut's motion has an amplitude of 0.100 m.
Answer:
a) k = 701.8 N / m, b) m_{ast} = 61.1 kg, c) v ’= -1.3 10⁻⁴ m / s
Explanation:
a) For this exercise let's use the relationship of the angular velocity
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
k = w² m
the angular velocity is related to the period
w = 2π / T
we substitute
k = 4 π² [tex]\frac{m}{T^2}[/tex]
let's calculate
k = 4 π² 10 /0.75²
k = 701.8 N / m
b) now repeat the measurement with an astronaut on the chair
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
where the mass Month the mass of the chair plus the mass of the astronaut
M = m + [tex]m_{ast}[/tex]
M = k / w²
w = 2π / T
let's calculate
w = 2π / 2
w = π rad / s
M = 701.8 /π²
M = 71,111 kg
now we use that
M = m + m_{ast}
m_{ast} = M - m
m_{ast} = 71.111 - 10.0
m_{ast} = 61.1 kg
c) if the astronaut's movement is simple harmonic
x = A cos wt
therefore the speed is
v = [tex]\frac{dx}{dt}[/tex]
v = -Aw sin wt
maximum speed is
v = - Aw
v = 0.100 π
v = 0.31416 m / s
we can suppose that the movement of the space station and the astronaut is equivalent to division of the same
initial instant. Before the move
p₀ = 0
final instant. When the astronaut is moving
p_f = M_station v’+ m_{ast} v
the moment is preserved
p₀ = pf
0 = M__{station} v ’+ m_{ast} v
v ’= - [tex]\frac{m_{ast} }{M_{station} } \ v[/tex]
we substitute
v ’= [tex]\frac{61.1 }{ 100000 } \ 0.31416[/tex]
v ’= -1.3 10⁻⁴ m / s
the negative sign indicates that the station is moving in the opposite direction from the astronaut
What is the ratio of the intensities of an earthquake P wave passing through the Earth and detected at two points 14 km and 44 km from the source?
Answer:
Explanation:
Intensity of a wave is inversely proportional to distance from source
I ∝ 1 / r
I is intensity , r is distance
I₁ / I₂ = r₂² / r₁²
ratio of intensity = 44² / 14²
= 9.87
intensity of earthquake of first place will be 9.87 times that of second place .
A cylinder containing the air comprises the systemm. Cycle is completed as follows : (i) 82000 N-m of work is done by the piston on the air during compression stroke and 45 kJ of heat are rejected to the surroundings. (ii) During expansion stroke 100000 N-m of work is done by the air on the piston. Calculate the quantity of heat added to the system؟?
Answer & Explanation:
1 N-m = 1 Joule
So 82 kJ of energy put into the system during (i).
45 kJ of heat leaves the system, so 82 kJ - 45 kJ = 37 kJ is remaining.
(ii) requires 100 kJ of energy but only 37 kJ is available, so 100 kJ - 37 kJ = 63 kJ of heat energy must be added to the system.
The heat given would be equal to the heat emitted from the system and by providing some external source of energy the volume or temperature of the system may increase.
The amount of heat added to the system is 63kJ.
The energy can be estimated as:
Given,
Work done by piston = 82000 NmHeat rejected in surrounding = 45 kJWork done during expansion stroke = 100000 NmQuantity of added heat = ?During compression stroke:
Work done by the piston [tex]\rm (W_{1-2})[/tex] = - 82000Nm or - 82kJHeat rejected to the system [tex]\rm (Q_{1-2})[/tex] = - 45kJWe know that,
[tex]\rm Q_{1-2} = (U_{2} - U_{1}) + W[/tex]
Therefore,
[tex]\begin{aligned}-45 &= \rm (U_{2} - \rm U_{1}) + (-82)\\\\\rm (U_{2}-U_{1}) &= 37\rm (U_{2} - U_{1}) = 37\; kJ \end{aligned}[/tex] (equation 1 )
During Expansion system:
Work done by the piston [tex]\rm (W_{2-1})[/tex] = 100000 Nm or 100 kJ
Now putting values in the equation:
[tex]\begin{aligned}\rm Q_{2-1} &= \rm U_{1} - U_{2} + W\\\\&=\rm (U_{1} - U_{2}) + W\end{aligned}[/tex]
Substituting value from equation 1:
[tex]\begin{aligned}\rm Q_{2-1} &= - 37+100\rm \;kJ\\&= 63\rm \; kJ\end{aligned}[/tex]
Therefore, 63kJ of energy is added to the system.
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Consider the following scenario. A car for which friction is not negligible accelerates from rest down a hill, running out of gasoline after a short distance (see below). The driver lets the car coast farther down the hill, then up and over a small crest. He then coasts down that hill into a gas station, where he brakes to a stop and fills the tank with gasoline. Identify the forms of energy the car has, and how they are changed and transferred in this series of events.
Answer:
1) U-> K +W
2) K -> W
Explanation:
In this exercise, care must be taken as they indicate that the friction force (rubbing) is not negligible.
1 part at the top of the hill the car has gravitational potential energy, which is transformed in a part into kinetic energy and another part into heat by the work of the friction force that opposes the movement.
2 part when the other hill rises it loses kinetic energy that is transformed into gravitational potential energy and part in heat due to the work of the friction force on this hill.
3rd part in the last descent all the gravitational potential energy is transformed into kinetic energy and the application of the brakes is transformed into heat due to the negative lock of the friction force and filled with gasoline that has chemical energy that can be transformed in the engine.
if the
gravitational force f=GM1M2/r2 derive the dimention of G
Answer:
M-1 L3 T-2
Explanation:
Where,
M = Mass
L = Length
T = Time
Derivation
From Newton’s law of gravitation,
Force (F) = [GM1M2] × r-2
Gravitational Constant (G) = F × r2 × [Mm]-1 . . . . (1)
Since, Force (F) = Mass × Acceleration = M × [LT-2]
∴ The dimensional formula of force = M1 L1 T-2 . . . . (2)
On substituting equation (2) in equation (1) we get,
Gravitational Constant (G) = F × r2 × [Mm]-1
Or, G = [M1 L1 T-2] × [L]2 × [M]-2 = [M-1 L3 T-2].
Therefore, the gravitational constant is dimensionally represented as M-1 L3 T-2.
reason why the center of gravity must not be at 50cm
Answer:
hope this helps
hope this is what u want
4. Draw conclusions: What is the minimum energy required to break the egg?
.
Answer:
0.25 J
Explanation:
The strength of the egg shell, the size of the egg, and the force used to break it are just some of the variables that affect how much energy is needed to crack an egg. When an object hits the egg with an impact energy of 12–26 mJ, cracks occur.
What is energy?The capacity of a system or object to do work is called energy, which is a fundamental term in physics. Kinetic energy, potential energy, heat energy, electromagnetic energy, and nuclear energy are just a few of the different forms of energy.
While potential energy is the energy possessed by an object as a result of its position or position, kinetic energy is the energy of motion. While electromagnetic energy is energy carried by electromagnetic waves like light, thermal energy is energy related to the temperature of a substance. The energy stored in the nucleus of an atom is called nuclear energy.
Therefore, when an object hits the egg with an impact energy of 12–26 mJ, cracks occur.
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On the computer, we can only approximate the true Fourier transform because of the need for both time and frequency sampling. Because of the limited frequency resolution, the frequency of a cosine may not line up exactly at a frequency sample, but vou'll still get a peak around that frequency. Assuming that vou sample the signal with fs-8000 Hz, and take its Fourier transform using an FFT with 4096 points, what is the value (in Hz) of the FFT frequency bins closest to the two frequencies of the sinusoids (941 and 1336 Hz)
Answer:
hello your question is incomplete below is the missing part
Consider the sum of sinusoids de (t) = sin(24(941)t) + sin(27(1336)t)
answer : Value of DFT frequency = 684 Hz
Explanation:
Given data:
sample signal ( Fs ) = 8000 Hz
Fourier transform using an FFT with 4096 points
Determine the value of DFT frequency bins closest to the two frequencies of the sinusoids ( 941 and 1336 Hz ) ( in HZ )
fs = 8000 Hz
4096 parts are divided with a gap of Fs/N
attached below is the detailed solution
A 3.0-kilogram mass is traveling in a circle of
0.20-meter radius with a speed of 2.0 meters per
second. What is its centripetal acceleration?
(1) 10. m/s
(3) 60. m/s2
(2) 20. m/s2
(4) 6.0 m/s2
Answer:
[tex]a=20\ m/s^2[/tex]
Explanation:
Given that,
The mass of an object, m = 3 kg
The radius of a circle, r = 0.2 m
The speed of the object, v = 2 m/s
We need to find the centripetal acceleration. Its formula is given by :
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{0.2}\\\\a=20\ m/s^2[/tex]
So, the centripetal acceleration is [tex]20\ m/s^2[/tex].
A liquid fueled rocket is red on a test stand. The rocket nozzle has an exit diameter of 30 cm and the combustion gases leave the nozzle at a velocity of 3800 m/s and a pressure of 100 kPa, which is the same as the ambient pressure. The temperature of the gases in the combustion area is 2400 C. Find (a) the temperature of the gases at the nozzle exit plane, (b) the pressure in the combustion area, and (c) the thrust developed. Assume that the gases have a speci c heat ratio of 1.3, and a molar mass of 9. Assume that the ow in the nozzle is isentropic.
Answer:
1. Temperature= 869.35 K
2. Pressure of combustion = 12994.043 kpa
3. Thrust = 127x10⁶N
Explanation:
this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.
1.
The temperature = (273+2400k) - (3800)²/2(4003)
= 2673 - 14440000/8006
= 2673 - 1803.65
= 869.35 K
Approximately 869.4K
2. We first get mach number
= 3800/√1.3(923.8)(869.35)
= 3800/1021.78
= 3.719
Pressure = 100kpa[1+2.07464415]^1.3/0.3
= 12995.043kpa
C. Thrust
Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)
= 12678.621
= 126.781 kN
Thrust is approximately 127kN = 127x10⁶N
if a body of mass m is placed on earth ,what is the amount of potential energy possessed by it (g:-9.8m/s
Answer:
mgh
Explanation:
Assume the height of the body is 1.8m.
The gravity?of the body is G=mg
the height of the gravity center is about 0.9m
E=mgh
=m*9.8m/s*0.9m
= 8.82mJ
Write your paragraph explaining why astronomers' minds are more important than the instruments they use? It has to be a paragraph and please make it quick I’m being timed!!
Answer:
Ill just give you some quick pointers
Explanation:
-in order to create the instruments, they need to use their mind
-their minds and creativity are the base and foundation of anything really
-the brain is the most complex part of your body so much more complicated than any intrument
How could being mindful in conversation be helpful?
You can see if the person your talking to and being mindfull they will see respect towards you or they see you as a mindful person.It shows a person that you care.
Explanation:
Ocean waves crash on the beach at a velocity of 3.5 m/s. If the distance between the crests of each wave is 4 m, find the frequency of the waves.
a. 0.0088 Hz
b. 14.0 Hz
c. 1.14 Hz
d. 0.88 Hz
Answer:
d
Explanation:
velocity=frequency × wavelength
frequency=speed/wavelength
frequency=3.5÷4
=0.875~0.88
The frequency of the waves is (d) 0.88 Hz. So, correct answer is option (d).
What is frequency of wave?The frequency of a sinusoidal wave is the number of full oscillations performed by any wave constituent in a unit of time. According to the definition of frequency, if a body is moving periodically, it has completed one cycle after going through a number of situations or postures and then returning to its initial position. Therefore, frequency is a quantity that describes the rate of oscillation and vibration.
Given parameter,
Velocity of the waves = 3.5 m/s.
distance between the crests of each wave, that is, wavelength of the waves = 4 m.
We know that, for a wave transmission,
velocity of wave =frequency of wave × wavelength of wave
⇒ frequency of wave=speed of wave/wavelength of wave
⇒ frequency of wave =3.5 m/s ÷4m
=0.875 Hz
≈ 0.88 Hz
Hence, the frequency of the waves is 0.88 Hz.
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1. If a wave has a wavelength of 5.5m and a frequency of 45hz, what is its speed?
Answer:
By using the most simple velocity equation, velocity = distance / time, meaning the speed would be 247.5 meters per second.
You are heating an iron to iron the shirt you're going to wear to school tomorrow. Which of the following best describes the transformation
of energy in this example?
Electrical to radiant
Thermal to electrical
Chemical to radiant
Electrical to thermal
2.
3
4 5 6 7
8
9
10
Next
Answer:
electrical to the thermal
Explanation:
You are designing a new roller-coaster. The main feature of this particular design is to be a vertical circular loop-the-loop where riders will feel like they are being squished into their seats even when they are in fact upside-down (at the top of the loop). The coaster starts at rest a height of 80m above the ground, speeds up as it descends to ground level, and then enters the loop which has a radius of 20m. Suppose a rider is sitting on a bathroom scale that initially reads W (when the coaster is horizontal and at rest). What will the scale read when the coaster is moving past the top of the loop
Answer:
The reading on the scale is N = 9W
Explanation:
Since the roller coaster drops from a height, h of 80 m, the potential energy lost equals the kinetic energy gained as it enters the loop.
So, mgh = 1/2mv² where m = mass of rider, g = acceleration due to gravity = 9.8 m/s², h = initial height of roller coaster above ground level = 80 m and v = speed of roller coaster as it enters the loop.
So, mgh = 1/2mv²
v² = 2gh
v = √(2gh)
v = √(2 × 9.8 m/s² × 80 m)
v = √(1568 m²/s²)
v = 39.6 m/s
Now, as the roller coaster gets to the top of the vertical loop, the centripetal force, F and the weight W acts downwards. For the passenger not to fall off, this must equal the normal force, N
So, F = mv²/r where v = speed of roller coaster = 39.6 m/s and r = radius of vertical loop = 20 m and m = mass of rider = W/g
F = Wv²/gr
F = W(39.6 m/s)²/(9.8 m/s² × 20 m)
F = (1568 m²/s²)W/196 m²/s²
F = 8W
Since F + W = N
8W + W = N
9W = N
So, N = 9W
So, the reading on the scale is N = 9W
Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1, the ball bounces off a cement floor, and in Case 2, the ball bounces off a piece of stretchy rubber. Two cases of a ball dropping and bouncing off of a surface. In case one, the ball approaches a cement floor, comes in contact, momentarily comes to rest, and then bounces off with a velocity smaller than the approach velocity. In case 2, the ball approaches a piece of stretchy rubber. When the ball comes in contact with the rubber, it deforms the rubber and comes momentarily to rest. The ball bounces off again with a velocity smaller than the approach velocity, and the rubber regains its original shape. In both cases, the balls are dropped from the same initial height and reach the same final height. In which case is the magnitude of the ball's change in momentum the greatest
Answer:
the impulse must be the same in these two cases F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
Explanation:
For this exercise we use the relationship between momentum and momentum
I = Δp
F t = m v_f - m v₀
To know the speed we use the conservation of energy
starting point. Highest point
Em₀ = U = m g h
fincla point. Just before the crash
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
we substitute in the impulse relation
F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases
PLEASE HELP ASAP WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST!!!
What 2 forces would be responsible for exerting forces to the left on a cyclist that is already in motion moving to the right?
(Select 2 of the choices below)
A. gravity
B. normal force
C. air resistance
D. friction
Answer:
Friction and Air resistence
Explanation:
i already passed this grade years ago...
The forces responsible for exerting forces to the left on a cyclist are -
Air resistanceFrictionWhat is air resistance?Air resistance describes the forces that are in opposition to the relative motion of an object as it passes through the air. We can write air resistance as -
[tex]$F_{D}=\frac{1}{2} \rho v^{2} C_{D} A[/tex]
where -
F{D} = drag
{ρ} = density of fluid
{v} = speed of the object relative to the fluid
C{D} = drag coefficient
{A} = cross sectional area
Given is to find what 2 forces would be responsible for exerting forces to the left on a cyclist that is already in motion moving to the right.
The forces responsible for exerting forces to the left on a cyclist are -
Air resistanceFrictionTherefore, the forces responsible for exerting forces to the left on a cyclist are -
Air resistanceFrictionTo solve more questions on forces, visit the link -
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A 10-kg rock falls from a height of 8-m above the ground. What is the kinetic energy of the rock just before it hits the ground?
Answer: 800
Explanation:
1/2 x m x v^2 = m x g x h
KE = 10 x 10 x 8
KE= 800
The energy of the body by the virtue of its motion is known as the kinetic energy of the body. The kinetic energy of the rock just before it hits the ground will be 784.8 J.
What is kinetic energy?The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of the velocity.
According to the law of conservation of energy, energy can not be created nor be destroyed can be transferred from one form to another form.
Kinetic energy= potential energy
Kinetic energy= mgh
Kinetic energy= 10×9.81×8
Kinetic energy=784.8 J
Hence the kinetic energy of the rock just before it hits the ground will be 784.8 J.
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If a car is moving to the left with constant velocity, one can conclude that :
a. there must be no forces exerted on the car.
b. the net force exerted on the car is directed to the left.
c. the net force exerted on the car is zero.
d. there is exactly one force exerted on the car.
Answer:
b. the net force exerted on the car is directed to the left.
Explanation:
Applying Newton's second law of motion, the car will move in the direction of the applied force. If the applied forces are on different directions, the car will move in the direction of the greater force (net force).
Therefore, if a car is moving to the left with constant velocity, one can conclude that the net force exerted on the car is directed to the left.
The correct option is "b. the net force exerted on the car is directed to the left"
