Calculate wavelength of tube of length 15, 18, 20 cm.
Answer:In this lesson, the mathematical relationship between the tube's length, the speed of sound through air, and the ... Thus, the length of the air column is equal to one-half of the wavelength for the first harmonic. ... Determine the fundamental frequency (1st harmonic) of an open-end air column that has a length of 67.5 cm.
Explanation:
Which of the following is an example of kinetic mechanical energy?
Immersive Reader
(2 Points)
A. A bike rolling down a hill
B. An elevated wrecking ball
C. A compressed spring
D. A loaded gun
E. A set mouse trap
Answer:
A
Explanation:
Kinetic energy must be moving. Potential energy has the ability to move but is not doing so at the moment.
A is likely the answer. But there's lots involved in that kind of motion.
B If the ball is elevated, it implies it is not moving yet. It has potential energy.
C Again, the spring is compressed. It will push something when it moves, but it is not moving yet.
D The load gun's bullet is not moving. It's still potential energy.
E. The mouse trap is set, but it is not moving. When the mouse eats the bait then it's potential energy will transform into kinetic energy.
An enormous thunderstorm covers Dallas-Ft. Worth. Your best friend Clark is a storm chaser and heads to the center of the storm to take some readings while you stay dry at home. While Clark is at the center of the storm, he sees and hears lightning strike a tree that is 190 m from where he is standing. You are 140 km from the tree. How long does it take for the sound to reach Clark
Answer:
407.61 seconds
Explanation:
Given that an enormous thunderstorm covers Dallas-Ft. Worth. Your best friend Clark is a storm chaser and heads to the center of the storm to take some readings while you stay dry at home. While Clark is at the center of the storm, he sees and hears lightning strike a tree that is 190 m from where he is standing. You are 140 km from the tree. How long does it take for the sound to reach Clark
Solution.
The distance between the Clark and you will be:
Distance = 140km - 190m
Distance = 140000 - 190
Distance = 139810 m
The radius of the earth is also considered but it will surely be insignificant as you subtracted the two distance values.
Using the formula
Speed = distance/time
Where
The speed of sound = 343 m/s
343 = 139810 / t
Make time the subject of formula
t = 139810 / 343
t = 407.61 s
Therefore, it will take 407.61 seconds for the sound to reach Clark
Compare and contrast Albert Einstein and Karl Popper regarding their views on the nature and philosophy of science. Use the key concepts of each to explain their views. Consider how they begin their accounts, the arguments they make to support their views, and their key conclusions.
Answer:
Well Albert Einstein is more smarter PERIOD
Explanation:
You push a 1.5 kg ball across a desk for 2 seconds, so that 10 J of work have been done. How much power was produced?
Answer
Answer:
i need help with the same question
Explanation:
Each vertical line on the graph is 1 millisecond (0.001 s) of time. What is the period and
frequency of the sound waves?
Help me!!!!!
Answer:
1000 Hz
Explanation:
please mark me as brainliest
PLS HELP
SHOW WORK
A car starting from rest accelerates at a rate of 1.5 m/s ^ 2 What is its final speed at the end of 18.0 seconds ?
Answer:
27 ms^-1
Explanation:
by using v= u + at
u = 0 ( because the object id starting from rest)
v= 0 + 1.5 x 18
v = 27 ms^-1
The sun warms Earth through the process of _________
conduction
convection
insulation
radiation
A ball is thrown straight up with an initial velocity of 128 ft/sec, so that its height (in feet) after t sec is given by s(t) = 128t-16t2. (a) What is the average velocity of the ball over the following time intervals? [3,4] ft/sec [3,3.5] ft/sec [3,3.1] ft/sec (b) What is the instantaneous velocity at time t = 3? ft/sec (c) What is the instantaneous velocity at time t = 6? ft/sec Is the ball rising or falling at this time? rising falling (d) When will the ball will hit the ground? t = sec
Answer:
one sec let me think
Explanation:
(a)The average velocity of the ball over the following time intervals will be [3,4] ft/sec.
(b)The instantaneous velocity at time t = 3 will be32 ft/sec.
(c)The instantaneous velocity at time t = 6 will be -64 ft/sec.
(d)The ball will hit the ground at 13.4 sec.
What is velocity?The change of displacement with respect to time is defined as the velocity. velocity is a vector quantity. it is a time-based component.
The given data in the question will be ,
u is the initial velocity by which ball thrown=128 ft/sec.
V₃ is the instantaneous velocity at time t=3 sec.
V₆ is the instantaneous velocity at time t=6 sec.
t is the time when ball hits the ground=?
(a) Given equation for the displacement
s(t) =128t-16t² (on differenting got the velocity )
v(t) = 128-32t
Time when velocity is zero will be
[tex]\rm{ t=\frac{128}{32}[/tex]
[tex]\rm{ t=4 sec[/tex]
If the velocity got in the equation is 128 and 32 ft /sec. it can be only when the average velocity is [3,4] ft/sec .
Hence the average velocity obtained from the problem will be [3,4] ft/sec
(b)
s(t) =128t-16t² (on differenting got the velocity )
v(t) = 128-32t
At time( t=3 sec)
v(3) = 128-32×3
v(3) =32 m/sec.
Hence the instantaneous velocity at time t = 3 will be32 ft/sec.
(c)
s(t) =128t-16t² (on differenting got the velocity )
v(t) = 128-32t
At time( t=6 sec)
v(6) = 128-32×6
v(6) = -64 m/sec.
Hence the instantaneous velocity at time t = 6 will be -64 ft/sec.
(d)
According to Newtons third equation of motion we got
v=u+gt
If the body returens from a certain height at max height its velocity must be zero; ( u=0)
[tex]\rm t=\frac{(v-u)}{g} \\\\\ \rm t=\frac{(128-0)}{9.81}\\\\\rm t=13.04 sec.[/tex]
Hence the ball will hit the ground at 13.4 sec.
To learn more about the velocity refer to the link ;
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Astronomers observe the motion of four planets that orbit a star similar to the Sun. Each planet follows an elliptical orbit around its star. The astronomers measure each planet's orbital period, as shown in the table.
Planet; Orbital Period (Earth days)
Planet W; 10
Planet X; 640
Planet Y; 80
Planet Z; 270
To determine the distance each planet is from the star, astronomers applied one of Kepler's three laws.
Kepler's first law: The path of each planet around a star is an ellipse, with the star at one focus. Kepler's second law: A planet sweeps out equal areas in equal amounts of time as it revolves around the star.
Kepler's third law: The square of the time for one revolution of a planet is proportional to the cube of the radius of its orbit.
Based on the table, identify the planet that is the farthest distance from the star, and indicate which of Kepler's three laws can be used to justify your answer. Enter your answer in the box provided.
Answer:
planet that is farthest away is planet X
kepler's third law
Explanation:
For this exercise we can use Kepler's third law which is an application of Newton's second law to the case of the orbits of the planets
T² = ([tex]\frac{4\pi ^2}{ G M_s}[/tex] a³ = K_s a³
Let's apply this equation to our case
a = [tex]\sqrt[3]{ \frac{T^2}{K_s} }[/tex]
for this particular exercise it is not necessary to reduce the period to seconds
Plant W
10² = K_s [tex]a_{w}^3[/tex]
a_w = [tex]\sqrt[3]{ \frac{100}{ K_s} }[/tex]
a_w = [tex]\frac{1}{ \sqrt[3]{K_s} }[/tex] 4.64
Planet X
a_x = [tex]\sqrt[3]{ \frac{640^3}{K_s} }[/tex]
a_x = \frac{1}{ \sqrt[3]{K_s} } 74.3
Planet Y
a_y = [tex]\sqrt[3]{ \frac{80^2}{K_s} }[/tex]
a_y = \frac{1}{ \sqrt[3]{K_s} } 18.6
Planet z
a_z = [tex]\sqrt[3]{ \frac{270^2}{K_s} }[/tex]
a_z = \frac{1}{ \sqrt[3]{K_s} } 41.8
From the previous results we see that planet that is farthest away is planet X
where we have used kepler's third law
What is the weight of a 25 kg object on Earth with an acceleration due to gravity of 9.8m/s/s?
2.45 n
24.5 n
245 n
2450 n
A turtle crawls at 4.32 m/s to cover the short 3.84 m distance to his food bowl. How long does it take?
10POINTS!!
A satellite orbits Earth 350 km above Earth's surface. Calculate the free-fall acceleration at this altitude.
Answer:
8.82 m/s²
Explanation:
Formula for the free fall or gravitational acceleration is;
a = GM/r²
Where;
G is gravitational constant = 6.67 × 10^(-11) m³/kg.s²
M is mass of earth = 5.972 × 10^(24) kg
r is radius of earth = 6371 km
We are given that the satellite orbits Earth 350 km above Earth's surface.
Thus, new radius = 6371 + 350 = 6721 km = 6721000 m
Thus;
a = (6.67 × 10^(-11) × 5.972 × 10^(24))/(6721000²)
a = 8.82 m/s²
Who watching all star draft? Luka better get picked first ong
pls help its already late lol
Answer:
Temperature
Explanation:
Answer:
Gravity
Explanation:
Gravity has nothing to do with weather. Temperature means hot or cold, water is rain and thunder, and wind is if it's in the winter or something like that
Physics help, thank you guys so much!
Answer:
Δt = 5.85 s
Explanation:
For this exercise let's use Faraday's Law
emf = [tex]- \frac{d \phi}{dt}[/tex] - d fi / dt
[tex]\phi[/tex] = B. A
\phi = B A cos θ
The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.
suppose a linear change of the magnetic field
emf = - A [tex]\frac{B_f - B_o}{ \Delta t}[/tex]
Dt = - A [tex]\frac{B_f - B_o}{emf}[/tex]
the final field before a fault is zero
let's calculate
Δt = - 0.046 (0- 1.4) / 0.011
Δt = 5.85 s
We intend to measure the open-loop gain (LaTeX: A_{open}A o p e n ) of an actual operational amplifier. The magnitude of LaTeX: A_{open}A o p e n is in the range of 106 V/V. However, the signal generator in measurement setup can supply minimal voltage of 1 mV, and the oscilloscope used at amplifier output can measure maximal voltage level of 10 V. Can you design a simple measurement setup using this signal generator and oscilloscope, and accurately measure the LaTeX: A_{open}A o p e n
Answer:
voltage divider, R₂ = 1000 R₁
measuring the output in the resistance R₁
Explanation:
Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V
in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.
To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.
If we use two resistors whose relationship is
R₂ / R₁ = 10³
R₂ = 1000 R₁
When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator
When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how fast is the angle of elevation of the rocket increasing, as seen by an observer on the ground 5 kilometers from the launching pad
Answer:
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.
Explanation:
Geometrically speaking, the distance between the rocket and the observer ([tex]r[/tex]), measured in kilometers, can be represented by a right triangle:
[tex]r = \sqrt{x^{2}+y^{2}}[/tex] (1)
Where:
[tex]x[/tex] - Horizontal distance between the rocket and the observer, measured in kilometers.
[tex]y[/tex] - Vertical distance between the rocket and the observer, measured in kilometers.
The angle of elevation of the rocket ([tex]\theta[/tex]), measured in sexagesimal degrees, is defined by the following trigonometric relation:
[tex]\tan \theta = \frac{y}{x}[/tex] (2)
If we know that [tex]x = 5\,km[/tex], then the expression is:
[tex]\tan \theta = \frac{y}{5}[/tex]
And the rate of change of this angle is determined by derivatives:
[tex]\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y[/tex]
[tex]\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}[/tex]
[tex]\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}[/tex]
[tex]\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}[/tex]
Where:
[tex]\dot \theta[/tex] - Rate of change of the angle of elevation, measured in sexagesimal degrees.
[tex]\dot y[/tex] - Vertical speed of the rocket, measured in kilometers per hour.
If we know that [tex]y = 4\,km[/tex] and [tex]\dot y = 400\,\frac{km}{h}[/tex], then the rate of change of the angle of elevation is:
[tex]\dot \theta = 48.780\,\frac{\circ}{s}[/tex]
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.
Find the current if 20C of charge pass a particular point in a circuit in 10 seconds.
Answer:
2 A
Explanation:
From the question,
Q = it..................... Equation 1
Where Q = Quantity of charge, i = cudrrent, t = time.
Make i the subject of the equation
i = Q/t.......................... Equation 2
Given: Q = 20 C, t = 10 seconds.
Substitute these values into equation equation 2
i = 20/10
i = 2 A.
Hence the current is 2A
Given values are:
Charge, Q = 20 CTime, t = 10 secondsAs we know,
→ [tex]Current = \frac{Charge}{Time}[/tex]
or,
→ [tex]i = \frac{Q}{t}[/tex]
BY substituting the values, we get
[tex]= \frac{20}{10}[/tex]
[tex]= 2 \ A[/tex]
Thus the answer above is right.
Learn more about current here:
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what is impulse in physical science
Answer:
In classical mechanics, impulse (symbolized by J or Imp) is the integral of a force, F, over the time interval, t, for which it acts. Since force is a vector quantity, impulse is also a vector quantity. ... A resultant force causes acceleration and a change in the velocity of the body for as long as it acts.
What happens when a moving object experiences no net force?
Answer:
An object with no net forces acting on it which is initially at rest will remain at rest. If it is moving, it will continue to move in a straight line with constant velocity. Forces are "pushes" or "pulls" on the object, and forces, like velocity and acceleration are vector quantities.
Define speed and what is it’s SI unit.
what is dimensional analysis
Answer:
The formula used to convert from the metric system outside of it. So like converting kilograms into pounds. The Formula is as follows:
# unit x (#unit/#unit) = # unit
^ ^ ^
I I I
given Conversion Answer
factor
*Note: Italic "units" are the same. Bold "units" are the same.
Example:
One thousand eighty kilometers is how many miles? Set it up dimensionally.
1080 km (1 mi/1.61 km) = 670.81 mi
*This is because 1080 x 1 = 1080, but then you divide 1080 by 1.61
A ray of light is incident on a plane mirror at an angle of 30 degree state the angle of reflection?
Answer:
30°
Explanation:
The law of reflection states that when a ray of light reflects off a surface, the angle of incidence is equal to the angle of reflection.
∠i = ∠r
so as ∠i = 30°
∠r will also be 30°
1. Odysseus traveled from Troy to Ithaca. What
was the acceleration of Odysseus' ship if its mass
was 900,000 kg and it moves across the water with
a force of 300,000 N?
Answer: 0.33 m/s^2
Explanation:
The acceleration of Odysseus' ship as it moves across the water from Troy to Ithaca is 0.33m/s²
Given the data in the question;
Mass of Odysseus' ship; [tex]m= 900000kg[/tex]Force with which Odysseus' ship moves across the water; [tex]F = 300000N[/tex]Acceleration; [tex]a = ?[/tex]To determine the acceleration of the ship, We the equation from Newton's Second Law of Motion:
[tex]F = m\ *\ a[/tex]
Where F is the force, m is the mass and a is the acceleration
Lets make acceleration ''a'', the subject of the formula
[tex]a = \frac{F}{m}[/tex]
Now, we substitute our given values into the equation
[tex]a = \frac{300000N}{900000kg}[/tex]
We know that, A newton is defined as [tex]1 kg.m/s^2[/tex]
[tex]a = \frac{300000 kg.m/s^2}{900000kg} \\\\a = 0.33m/s^2[/tex]
Therefore, the acceleration of Odysseus' ship as it moves across the water from Troy to Ithaca is 0.33m/s²
Learn more, https://brainly.com/question/2842540
8. What is the mass of a toy car with a speed of 12.5 m/s and 47500 J of kinetic energy?
Answer:
608kg
Explanation:
Formula : Kinetic energy
½ ×mass x speed²
47500
½×12.5²
=608 Kg
A truck of mass 1600 kg is traveling at 15 m/s. In order to avoid running over a squirrel in the middle of the road, the truck driver begins to brake. What must the braking force on the truck be if the truck comes to stop in 2 s? *
a)12000n
b) 8000n
c)7000n
d)6000n
The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 100 mg. Find a formula for the mass of the sample that remains after t years. (b) Find the mass after 1000 years correct to the nearest milligram. (c) When will the mass be reduced to 40 mg? SOLUTION (a) Let m(t) be the mass of radium-226 (in milligrams) that remains after t years. Then dm/dt = km and m(0) = 100, so this theorem gives m(t) = m(0)ekt = ekt. In order to determine the value of k, we use the fact that m(1590) = 1 2 . Thus e1590k = so e1590k = and 1590k = ln 1 2 = − ln(2) k = . Therefore m(t) = . We could use the fact that eln(2) = 2 to write the expression for m(t) in the alternative form m(t) = . (b) The mass after 1000 years is as follows. (Round your final answer to the nearest milligram.) m(1000) = ≈ mg
Answer:
See explanation
Explanation:
a) Formula for the mass of the sample that remains after t years = N= Noe^-kt
Where;
N = mass at time t years
No = mass at time t= 0
k = decay constant
t = time taken
So,
N = 100e^-kt
b) First,
t1/2 = -ln(1/2)/k
t1/2 = 0.693/k
t1/2 = half life of radium-226 =1590 years
1590 = 0.693/k
k = 0.693/1590
k = 4.36 * 10^-4
So,
N= 100e^-(4.36 * 10^-4 * 1000)
N= 65 mg
c) From
N = 100e^-kt
40 = 100e^-(4.36 * 10^-4t)
40/100 = e^-(4.36 * 10^-4t)
0.4 = e^-(4.36 * 10^-4t)
ln(0.4) = ln(e^-(4.36 * 10^-4t))
-0.9163 = -4.36 * 10^-4t
t = 0.9163/4.36 * 10^-4
t = 2102 years
Which statement about oceans is incorrect?
A Evaporation occurs when water is warmed by the sun.
B Most evaporation and precipitation occur over the ocean.
C 97 percent of Earth's water is fresh water from the ocean.
D Water leaves the ocean by the process of evaporation.
What is the difference in the speed of the generator with a small magnet and a generator with a large magnet?
Answer:
uADHTiszA?kv
Explanation: