Answer:
[tex]80\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
The frequency [tex]f[/tex] of a wave is the number of cycles completed in unit time ([tex]1\; {\rm s}[/tex] in this example.) In this question, [tex]f = 40\; {\rm s^{-1}}[/tex] ([tex]1\; {\rm Hz} = 1\; {\rm s^{-1}}[/tex]) means that the wave would complete [tex]40[/tex] cycles in every [tex]1\; {\rm s}[/tex].
The wavelength [tex]\lambda[/tex] of a wave is the distance the wave travels in each cycle. It is given that [tex]\lambda = 2\; {\rm m}[/tex].
The goal is to find the wave speed, which is the distance that this wave travels in unit time ([tex]1\; {\rm s}[/tex].)
In this question, it is given that [tex]\lambda = 2\; {\rm m}[/tex] and [tex]f = 40\; {\rm s^{-1}}[/tex]. Thus, this wave would travel a total of [tex]40\, (2\; {\rm m}) = 80\; {\rm m}[/tex] for the [tex]40[/tex] cycles completed in each unit time of [tex]1\; {\rm s}[/tex] ([tex]\lambda = 2\; {\rm m}[/tex] for each cycle.) The speed of this wave would be [tex]80\; {\rm m\cdot s^{-1}}[/tex].
Formally, the speed [tex]v[/tex] of this wave can be found by multiplying the wavelength [tex]\lambda[/tex] of this wave by its frequency [tex]f[/tex]:
[tex]\begin{aligned}v &= \lambda\, f \\ &= (2\; {\rm m})\, (40\; {\rm s^{-1}) \\ &= 80\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
A 0.80kg block of carbon (solid) is dropped into 1.4kg of water. If the carbon starts at -20C, the water starts at 92C, and they have equal final temperatures, what is the final temperature of the system?
The system's final temperature is roughly 16.7°C.
What is a system's final temperature?You may determine your substance's final heat by multiplying the temperature change by the initial temperature. Your water's final temperature would be 24 + 6, or 30 degrees Celsius, for instance, if it started off at 24 degrees Celsius.
The following is the formula for energy conservation:
Q1 + Q2 = 0
Q = mcΔT
Q1 + Q2 = 0
568.8
Simplifying and solving for
6394.4 - 106768 = 0
= 16.7°C
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What would be the intensity of a sound wave produced by a 150 Watt speaker from a distance of 5.8 meters? (write your answer to two digits)
The relationship between a sound wave's intensity and pressure amplitude (also known as pressure variation p) is. I is equal to (p) 2 2 v w, where is the thickness of the substance that the sound is contained in.
Describe a sound wave.
Hence, a sound wave is made up of periodically occurring compressions and compression and rarefaction, or areas of high and low pressure, that are travelling at a specific pace. In other words, it consists of a regular (i.e., oscillating or vibrating) change in pressure that takes place around the optimum pressure that is present at a specific time and location.
A sound wave is created by a speaker in what way?
A speaker creates a sound by vibrating a cone, which causes air molecules to vibrate. A speaker in Figure 17.2. 2 vibrates with a consistent frequency and amplitude, causing motions in the molecules of the surrounding air. The speaker transfers power to the air as it vibrates back and forth, primarily as thermal energy.
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According to this graph, the acceleration
is approximately:
A. 12 m/s²
C. 4 m/s²
Velocity (m/s)
14
12
10
12 2 3 4
Time t (s)
B. 1.5 m/s2
D. 3 m/s2
Help please
Answer:
Explanation:
Because you have velocity along the y axis and time along the x axis, this is a velocity v time graph which is an acceleration graph. The slope of the line in this graph IS the acceleration. We can use 2 points and the slope formula to solve for the acceleration:
(0, 0) and (1, 3):
[tex]m=\frac{3-0}{1-0}=3[/tex] m/s squared, choice D.
How long does it take for radiation from a cesuim-133 atom to complete 1.5 million cycles
A cesium-133 atom's radiation goes through 1.5 million cycles in around 0.1633 microseconds (or 163.3 nanoseconds).
What frequency does one kind of radiation that cesium-133 emits have?9,192,631,770 hertz (cycles per second) is the frequency of the microwave spectral line that the isotope cesium-133 emits. The basic unit of time is provided by this. Cesium clocks have an accuracy and stability of 1 second in 1.4 million years.
The radiation emitted by cesium-133 has a frequency of 9,192,631,770 cycles per second, or 9.192631770 109 Hz.
The following formula may be used to determine how long 1.5 million radiation cycles take to complete:
Time is equal to the frequency of cycles.
Plugging in the numbers, we get:
time = 1.5 million / 9.192631770 × 10^9 Hz
time = 1.632995101 × 10^-7 seconds
So it takes approximately 0.1633 microseconds (or 163.3 nanoseconds) for radiation from a cesium-133 atom to complete 1.5 million cycles.
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A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water rises part-way along the straight sides, and then some oil with a density of is poured into one end. This causes the water surface on that side of the tube to go down by and the surface on the other side to go up by the same amount. How much higher is the top surface of the oil on that side of the tube compared with the surface of the water on the other side of the tube?
The top surface of the oil on that side of the tube is 0.6 times higher than the surface of the water on the other side of the tube.
Describe principle of hydrostatics?The principle of hydrostatics, also known as Pascal's principle, states that when an external pressure is applied to a fluid in a container, that pressure is transmitted uniformly in all directions within the fluid, regardless of the shape or volume of the container. In other words, the pressure applied to a confined fluid will be distributed evenly throughout the fluid and will not change in magnitude at any point within the fluid. This principle is important in a number of applications, such as hydraulic systems, which use fluids to transmit force and pressure from one point to another. It is also used to explain how liquids exert pressure on the walls of their container and how objects can float or sink in fluids.
We can use the principles of hydrostatics to solve this problem. Let's call the height difference between the two water surfaces h. We can assume that the oil completely covers the water on one side of the tube and does not mix with it, so the oil and water form two separate liquid columns with a common interface. Let's call the height difference between the oil and water surfaces on the same side of the tube H.
The pressure at any given point in a fluid depends only on the depth of that point below the surface of the fluid and the density of the fluid. Since the two water columns are at the same height, they experience the same pressure from the atmosphere. Similarly, the two oil columns experience the same pressure from the atmosphere.
Now consider a point on the interface between the oil and water on the same side of the tube. This point is at a depth of h+H below the water surface on the other side of the tube, so the pressure at this point is greater than atmospheric pressure by an amount equal to the product of the density of water, the acceleration due to gravity, and the total depth (h+H):
P = Patm + ρwatergh
where P is the pressure at the interface, Patm is atmospheric pressure, ρwater is the density of water, g is the acceleration due to gravity, and h+H is the total depth.
Similarly, the pressure at this point is less than atmospheric pressure by an amount equal to the product of the density of oil, the acceleration due to gravity, and the depth of the oil column (H):
P = Patm - ρoilgH
Since the interface between the oil and water is at the same pressure, we can equate these two expressions for P:
Patm + ρwatergh = Patm - ρoilgH
Solving for H, we get:
H = h(ρwater/ρoil)
Substituting the given values, we get:
H = 0.6h
Therefore, the top surface of the oil on that side of the tube is 0.6 times higher than the surface of the water on the other side of the tube.
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If the sun were more massive, what would happen to Earth’s gravity with the sun?
A. decrease
B. would be infinite
C. would be 0
D. increase
Answer: d. increase
Explanation:
If the sun were more massive, the gravitational force between the sun and Earth would increase. This means that Earth's gravity with the sun would also increase. Therefore, the correct answer is (D) increase.
The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. So, if the mass of one of the objects increases, the gravitational force between them will also increase. In this case, if the mass of the sun were to increase, the gravitational force between the sun and Earth would become stronger, and hence, Earth's gravity with the sun would also increase.
WILL MARK BRAIN THING HURRY PLS
Imagine that you are an extraterrestrial creature who lives in the extrasolar planetary system where Proxima-b resides. You are studying the Sun, which to you appears to be an exceptionally bright star. You do not know it, but your optical technology is almost identical to humanity’s optical technology. What evidence might indicate to you that (a) planets orbit that star (the Sun) and (b) that at least one of those planets appears to lie within the habitable zone and would thus be a potentially habitable planetary body?
a.) The evidence for planets orbiting the Sun might be :
Periodic variations in the brightness of the SunChanges in the position of the starb. Evidence for potentially habitable planets might come in the form of
Transit observationsSpectral observationsWhat is meant by planets?A planet is described as a celestial body that is in orbit around the Sun, has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and has cleared the neighborhood around its orbit.
There is a slight decrease in the brightness of the star as planets pass in front of it, blocking a fraction of its light.
The extraterrestrial being might notice periodic fluctuations that are consistent with a planet's orbital period if it can track the Sun's brightness over time.
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The length of the river span of a bridge is 2799.0 ft. The total length of the bridge is 6998ft. Convert the length of the river span of the bridge to meters.
According to the question the length of the river span of the bridge in meters is 853.3232 m.
What is Length?Length is a physical quantity that measures the distance between two points. It is one of the fundamental units in the International System of Units (SI). It is usually measured in meters, although it can also be measured in other units such as centimeters, kilometers, feet, yards, miles, and so on.
The length of the river span of the bridge is 2799.0 ft. To convert this length to meters, we need to use a conversion factor. There are 0.3048 meters in one foot, so the conversion factor we will use is 1 ft
= 0.3048 m.
To convert 2799.0 ft to meters, we multiply by the conversion factor:
2799.0 ft * 0.3048 m/ft
= 853.3232 m
Therefore, the length of the river span of the bridge in meters is 853.3232 m.
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please help me in this exercise
a. We can actually see here that the girl have kinetic energy which is respect to the escalator.
b. The kinetic energy does not depend on the chosen reference.
What is kinetic energy?Kinetic energy is a form of energy that an object possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its current velocity. Kinetic energy is a scalar quantity, meaning it only has magnitude and no direction. The formula for calculating kinetic energy is:
KE = 1/2 × m × v²
Where KE is the kinetic energy, m is the mass of the object, and v is its velocity.
The concept of kinetic energy was first introduced by the French mathematician Gaspard-Gustave de Coriolis in 1829. It was later developed by other scientists such as James Prescott Joule and Hermann von Helmholtz.
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How loud in Decibels would a sound be with an intensity of 7.8x10^-4 W/m2? (write your answer to one decimal space)
A sound that is 7.8x10-4 W/m2 in intensity is equal to (10 dB)log3.2106 W/m21012 W/m2=185 dB.
How can you determine the relative volume of a sound?The decibel, often known as the db or 0.1 bel, is the standard measurement unit. Hence, b = 10 log10 (I/I0) can be used to express the relationship between relative intensities, or b, in decibels. This equation can be used to determine that one decibel equals a 26 percent intensity variations.
What does physics mean by relative intensity?The "decibel level" of a sound is a less formal term for relative intensity level. It is not the same as energy; relative intensity level reflects loudness more faithfully by using a logarithmic scale.
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A rock climber stands on top of a 59 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s apart and observes that they cause a single splash. The initial speed of the first stone was 1.7 m/s . Include value and units.
a) How long after the release of the first stone does the second stone hit the water?
b) What was the initial speed of the second stone?
c) What is the speed of the first stone as it hits the water?
d) What is the speed of the second stone as it hits the water?
a) The time after the release of the first stone that the second stone hits the water is 2.0 s.
b) 15.7 m/s is the initial speed of the second stone.
c) The speed of the first stone as it hits the water is 15.7 m/s.
d) The speed of the second stone as it hits the water is 28.2 m/s.
What is velocity?Velocity is a vector quantity that measures both the speed and direction of an object's motion. It is equal to the rate of change of an object's position with respect to time. Velocity is usually represented by the symbol v and is measured in meters per second (m/s).
a) The time between first and second stone's release is 1.0 s. Since the time of release of first stone and the time of splash of both stones are same, the time between the release of second stone and the splash of both stones is 1.0 s.
Thus, the time after the release of the first stone that the second stone hits the water is 2.0 s.
b) The initial speed of the second stone can be calculated using the equation of motion,
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.
Substituting the values,
v² = (1.7)² + 2(9.8) * 59
v = 15.7 m/s
c) The speed of the first stone as it hits the water can be calculated using the equation of motion,
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.
Substituting the values,
v² = (1.7)² + 2(9.8) * 59
v = 15.7 m/s
d) The speed of the second stone as it hits the water can be calculated using the equation of motion,
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.
Substituting the values,
v² = (15.7)² + 2(9.8) * 59
v = 28.2 m/s
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