The term that is unnecessary to include when accurately drawing ray diagrams is light speed.
When drawing ray diagrams accurately, it is not necessary to include the speed of light. Ray diagrams are used to show how light travels through an optical system, such as a lens or mirror. The diagrams show the path of light rays as they reflect or refract through the system, and are used to predict the behavior of the system.
For example, a ray diagram can be used to show how an image is formed by a lens. By tracing the path of several light rays through the lens, it is possible to predict the size and position of the image. This is done by using the laws of reflection and refraction, which describe how light behaves when it passes through different media. Therefore, the main answer in 30 words is Light speed. It is not necessary to include the speed of light when drawing ray diagrams.
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The focal length of a converging lens is 10 cm. What is the angular magnification of this lens if the image is viewed by a relaxed eye with a near point of 25 cm?.
The angular magnification of this lens if the image is viewed by a relaxed eye with a near point of 25 cm is 0.056.
The given values are: The focal length of a converging lens is 10 cm and the near point of the relaxed eye is 25 cm. To determine the angular magnification of this lens if the image is viewed by a relaxed eye with a near point of 25 cm, we need to use the formula given below:Angular magnification, m = (-) v / uwhere v = the distance of the image from the eyeand u = the distance of the object from the eye
The near point of the relaxed eye is 25 cm, so the distance of the object from the eye, u = 25 cm.Using the formula of the lens equation, we can find the value of v:1/f = 1/v - 1/u1/10 = 1/v - 1/25v = 1/10 + 1/25v = (5 + 2) / 50v = 7/50 cmNow we can substitute the values of u and v to calculate the angular magnification:m = (-) v / um = (-) (7/50) / 25m = (-) 7 / 125m = -0.056 or 0.056 (taking magnitudes)
Therefore, the angular magnification of this lens if the image is viewed by a relaxed eye with a near point of 25 cm is 0.056.
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Please help me!! Time = 98 seconds speed=7.5 meters/second what is the distsnce in meters
Answer:
735 mExplanation:
The distance covered by an object given it's speed and time taken can be found by using the formula
distance = speed × time
From the question we have
distance = 7.5 × 98 = 735 m
We have the final answer as
735 mHope this helps you
Express 48 m/s in terms of
1.km/h
2.m/min
3.km/s
4.km/minutes
48 m/s in terms of km/h is 720.8 km/h. In terms of m/min is 2880 m/min, in terms of km/s is 0.048 km/s and in terms of km/min is 2.88 km/min.
To solve this question, we need to understand some terms. The unit of velocity is measured in m/s. It can be expressed in different units of velocity.
1 km (kilometer) = 1000 meter
1 h (hour) = 3600 seconds
1 minutes = 60 seconds
To convert m/s into km/h,
48 m/s * 3600/1000 = 172.8 km/h
To convert m/s into m/min,
48 m/s * 60 = 2880 m/min
To convert m/s into km/s,
48 m/s ÷ 1000 = 0.048 km/s
To convert m/s into km/minutes,
48 m/s * 60 / 1000 = 2.88 km/min
Therefore, the 48 m/s expressed is 172.8 km/h, 2880 m/min, 0.048 km/s and 2.88 km/min.
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48 m/s is equivalent to 172.8 km/h, 2880 m/min, 0.048 km/s, and 2.88 km/minute.
To express 48 m/s in different units of velocity:
km/h (kilometers per hour):
To convert m/s to km/h, we can use the conversion factor of 3.6 since 1 m/s is equal to 3.6 km/h.
48 m/s * (3.6 km/h / 1 m/s) = 172.8 km/h
Therefore, 48 m/s is equivalent to 172.8 km/h.
m/min (meters per minute):
To convert m/s to m/min, we can use the conversion factor of 60 since there are 60 seconds in a minute.
48 m/s * (60 m/min / 1 s) = 2880 m/min
Therefore, 48 m/s is equivalent to 2880 m/min.
km/s (kilometers per second):
Since 1 kilometer is equal to 1000 meters, to convert m/s to km/s, we divide the value by 1000.
48 m/s / 1000 = 0.048 km/s
Therefore, 48 m/s is equivalent to 0.048 km/s.
km/minute (kilometers per minute):
To convert m/s to km/minute, we first need to convert m/s to km/s (as calculated in the previous step) and then multiply by 60 to convert seconds to minutes.
0.048 km/s * 60 = 2.88 km/minute
So, 48 m/s is equivalent to 2.88 km/minute.
Hence, 48 m/s is equivalent to approximately 172.8 km/h, 2880 m/min, 0.048 km/s, and 2.88 km/minute.
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The Hubble Space Telescope is observing a distant Type 1 supernova with peak apparent magnitude 24. Using the light curve in your book (21.8) estimate how long after the peak brightness the supernova will become too faint to be seen.
- 270 days
- 1 year
- 125 days
- 50 days
Based on the light curve in the book, the distant Type 1 supernova will become too faint to be seen approximately 270 days after reaching its peak apparent magnitude of 24.
According to the given information, the peak apparent magnitude of the distant Type 1 supernova is 24. By referring to the light curve in the book, we can estimate the duration for which the supernova remains visible. Typically, the brightness of a supernova decreases exponentially over time.
While specific light curves may vary, a rough estimate can be made. Considering the options provided, the closest estimate is approximately 270 days. This means that after the supernova reaches its peak brightness, it will gradually fade and become too faint to be observed after around 270 days.
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A convex mirror has a focal length of -13.0 cm. When you place a 6.00 cm tall pencil 60.0 cm in front of that mirror, what is the location of the pencil's image?
The location of the pencil's image is =16.6cm
Calculation of image locationThe focal length of the convex mirror (f)= -13.0 cm
The object distance (u) = 60cm
The image distance (v) = xcm
Using the formula,1/f= 1/v + 1/u
Make 1/v the subject of formula,
1/v = 1/f -1/u
1/v = 1/13 - 1/60
1/v = 60-13/780
1/v = 47/780
V = 780/47
V = 16.6 cm
Therefore, he location of the pencil's image is = 16.6cm
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a wire carries a 15 μa current. how many electrons pass a given point on the wire in 1.0 s ?
Given a current of 15 μA, the number of electrons that pass a given point in the wire in 1.0 s is approximately 9.36 × 10¹² electrons.
One ampere is defined as the flow of one coulomb of charge per second. Since 1 microampere = 1/1,000,000 ampere, a current of 15 μA is equal to 15 × 10⁻⁶ A.
To calculate the number of electrons passing through a point in one second, we can use the equation:
number of electrons = (current in amperes) × (time in seconds) / (charge of one electron)
The charge of one electron is approximately 1.602 × 10⁻¹⁹ C. Therefore, the number of electrons passing a given point on the wire in 1.0 s is:
(15 × 10⁻⁶A) × (1.0 s) / (1.602 × 10⁻¹⁹ C) ≈ 9.36 × 10¹² electrons.
So, approximately 9.36 × 10¹² electrons pass through the point in one second.
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What is the range of electronegativity for polar nonpolar and ionic bondings?.
The range of electronegativity for polar nonpolar and ionic bonding will be between 1.5 and 2.1
Electronegativity is a chemical property that describes the tendency of an atom or a functional group to attract electrons toward itself.
The range of electronegativity for polar nonpolar and ionic bonding . So, the difference in electronegativity is somewhere between 1.5 and 2.1, between a polar covalent bond and an ionic bond.
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The range of electronegativity for polar nonpolar and ionic bonding is less than 0.4 and greater that 0.4, respectively.
In general there are two major categories of bonds, which are polar molecules and nonpolar molecules.
In the field of chemistry, polarity can be described as the distribution of electric charge around atoms, chemical groups, or molecules.
When there is an electronegativity difference between the bonded atoms then polar molecules are formed. Water - \(H_{2}\)O, Ammonia - N\(H_{3}\) are some of the examples of polar molecules.And when electrons are shared equal between atoms of a diatomic molecule or when polar bonds in a larger molecule cancel out each other then non polar molecules are formed. Any of the noble gasses: He, Ne, Ar, Kr, Xe, Benzene, methane etc are some of the examples of the molecules which forms non polar bondAnd if we talk about ionic bond, which are also be called as electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound. These types of bond forms when the outermost electrons of one atom are transferred permanently to another atom.There is no hard and fast rule but in general, a bond is considered as non polar if the difference in electronegativities is less than about 0.4, and it will be called a polar bond molecule if the difference is greater than 0.4.
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consider the cantilevered w14×30 beam shown in (figure 1) . e = 29(10^3) ksi, i = 291 in^4.
(a) Determine the expression for the elastic curve using the coordinate x for 0 < x < 9 ft where x is in feet.
(b) Determine the maximum slope of the beam , measured counterclockwise from the positive x-axis.
(c) Determine the maximum deflection of the beam.
The cantilever beam's elastic curve equation, including the deflection at the free end and the slope at the free end
(a) For the cantilever beam AB, the elastic curve equation is
y(x) = (F/2EI)*x^2
(b) The deflection at the free end is given by
y(L) = (F/2EI)*L^2
(c) The slope at the free end is
dy/dx = (F/EI)*L
A system's behaviours when a force is applied to it is described mathematically by an elastic curve. Its formula is the product of the ratio of the change in stress to the change in strain. Elastic curves are used to gauge a material's stiffness and forecast how a material will respond to different forces. They are also employed in the calculation of a material's elastic modulus, which is a measure of stiffness. For constructing structures and for comprehending how various materials respond to forces, elastic curves are crucial. They are also employed in engineering and industrial processes to calculate the strain and force required to distort a material.
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which of the following does not influence the speed of water in a
stream?
a. gradient
b. discharge
c. channel geometry
d. flow
Pption d. flow does not independently influence the speed of water in a stream. Instead, it is an outcome determined by the combined effects of gradient, discharge, and channel geometry.
The flow itself, or the rate at which water moves in a stream, is not an independent factor that influences the speed of water in a stream. Instead, it is a result of other factors such as gradient, discharge, and channel geometry.
a. Gradient: The gradient, or slope, of the stream affects the speed of water. Steeper gradients generally lead to faster-flowing water.
b. Discharge: The discharge of a stream, which refers to the volume of water passing through a given point in a specified amount of time, directly affects the speed of water. Higher discharge results in faster flow.
c. Channel geometry: The shape and dimensions of the stream channel also influence the speed of water. Wider and deeper channels tend to have slower flow, while narrower and shallower channels often lead to faster flow.
Therefore, option d. flow does not independently influence the speed of water in a stream. Instead, it is an outcome determined by the combined effects of gradient, discharge, and channel geometry.
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The potential difference across the terminals of a battery is 8. 4v when there is a current of 1. 50 a in the ba
Answer:
The potential difference across the terminals of a battery is 8. 4v when there is a current of 1. 50 a in the ba
Explanation:
The potential difference across the terminals of a battery is `8.4 V` when there is a current of `1.50 A` in the battery from the negative to the positive terminal. When the current is `3.50 Air`, the reverse direction, the potential difference becomes `9.4 V`. <br>(a) What is the internal resistance of the battery? <br>(b) What is the emf of the battery?
If the thumb and finger each squeeze with a force Ft = Ff = 10.0 N, what force do the forceps jaws exert on the plastic rod?
Express your answer to three significant figures and include the appropriate units.
Answer:
FJ = 31.48N
Explanation:
In this case you have that the torque exerted by each, thumb and finger, over the point P, is equal and opposite by the torque exerted by each Jaw.
You equal one of the torque produced by your hand, as for example, the torque of the finger, with the torque produced by one jaw:
\(\tau_F=\tau_J\\\\F_Fd_1=F_Jd_2\) (1)
FF: finger force = 10.0N
FJ: jaw force = ?
d1: distance from finger to point P = 8.50cm = 0.085m
d2: distance form the border of the Jaws to the point P = 2.70cm = 0.027m
You solve the equation (1) for FJ and replace the values of the other parameters:
\(F_J=\frac{F_Fd_1}{d_2}=\frac{(10.0N)(0.0850m)}{0.027m}=31.48N\)
The force exerted by each jaw is 31.48N
The force that the forceps jaws exert on the plastic rod will be 31.48N.
From the information given, the following can be depicted:
d1 = 8.50cm = 0.085md2 = 2.70cm = 0.027mFf = 10NFj = UnknownTherefore, the jaw force will be calculated thus:
Fj = (Ff × d1) / d2
Fj = (10 × 0.085) / 0.027
Fj = 31.48N
Therefore, the jaw force is 31.48N.
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Question 19 (4 points) A 2-kg ball traveling in the y-direction (North) collides with a 6-kg ball, initially at rest. After the collision, the 6-kg ball is moving at 5 m/s, 53° West of North. The 2-kg ball is moving 37° East of North. 6 kg 530 5 m/s 379 VE 2 kg before 6 kg after 2 kg What was the initial speed of the small ball (before the collision)? 20 m/s 35 m/s 30 m/s 15 m/s 40 m/s 10 m/s 25 m/s
Be careful of the wire size application chart you use, because many of them allow up to a ____ voltage drop over the length of the wire, which is more than is allowed in most automotive circuits. Group of answer choices
Answer: 0.1
Explanation: The wire size is very important when making decisions on the wiring of any equipment,house and other things. It has also been found that the size of the wire determines the load that can be allowed through it.
The right wire size application chart should always be considered before selecting the wiring diagram or system for any building or equipment as it can help to determine the amount of voltage that can be effectively carried through it.
Create a hypothesis using If, then and because.
Hypothesis: If I lower the amount of prey, then the frog population would lower, because the frogs would not have any food and then they would die.
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A 2018 Tesla Model S has a horsepower of 518. How many watts of power does this car model have?
Answer:
It depends on the model but these are the watts of each motor
Explanation:
Un motor: 258 HP (262 CV) Dual motor: 154 kW (209 CV) y 197 kW (268 CV). Performance: 154 kW (209 CV) y 225 kW (306 CV).
Which is the equation of a parabola with focus (-2, -4) and vertex (-4, -4)?
The equation of the parabola is (x + 4)^2 = 8(y + 4).
What is the equation of the parabola with focus (-2, -4) and vertex (-4, -4)?The equation of a parabola with a focus and vertex can be expressed in the form: (x - h)^2 = 4p(y - k), where (h, k) represents the vertex and p represents the distance between the vertex and the focus.
Given the focus (-2, -4) and the vertex (-4, -4), we can determine that the distance between the vertex and the focus is p = 2.
Substituting the known values into the equation, we have: (x - (-4))^2 = 4(2)(y - (-4)).
Simplifying further: (x + 4)^2 = 8(y + 4).
Therefore, the equation of the parabola is (x + 4)^2 = 8(y + 4).
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to date, we've found few planets orbiting their stars at distances comparable to the distances of the jovian planets in our solar system. why do astronomers think this is the case?
The answer is (b), meaning that we haven't been looking for planets so at distances for long enough.
What are the definitions of the 12 planets?(Astronomy) The solar system is made up of the sun and the objects gravitationally attracted to it, such as the planets Mercury, Venus, Mars, Mars, Octavian, Saturn, Uranus, and Neptune, as well as the minor planets Eris, Pluto, and Ceres.
What are the planets' roles?The world sustains us and gives us the resources we require. Last but not least, experts claim that if our planet disappeared, we and everything else on the world would have died. The one and only planet capable of supporting life is Earth.
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B2-RT15: RIFLE SHOTS-TIME TO HIT GROUND The rifles in the figures are being fired horizontally (straight outward, off platforms). The bullets fired from the rifles are all identical, but the rifles propel the bullets at different speeds. The speed of each bullet and the height of each platform are given. All of the bullets miss the targets and hit the ground. 500 m/s 800 m/s 800 m/s 1200 m/s 40 m 40 m 30 m 20 m Rank the time it takes the bullets to hit the ground. OR All All the same zero determine Cannot 4 Least Greatest Explain your reasoning.
THE ANSWER BELOW THE PICTURES.
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the unit of power is derived unit.
show how
\(power= \frac{work}{time} = \frac{joule}{second} = watt\)
An airplane travels for 2.5 hours at an average rate
of 130 miles per hour. Use the distance formula, d=rt, to find how
far the plane travels.
The plane travels a distance of 325 miles if the airplane travels for 2.5 hours at an average speed of 130 miles per hour. Using the distance formula (d = rt), we can calculate the distance.
To find the distance traveled by the airplane, we can use the distance formula, which is represented as d = rt. In this formula, "d" represents the distance, "r" represents the rate or speed at which the object is traveling, and "t" represents the time taken for the travel.
Given that the airplane travels for 2.5 hours at an average rate of 130 miles per hour, we can substitute these values into the formula. The rate of the airplane is 130 miles per hour, and the time taken is 2.5 hours.
Using the formula, we can calculate the distance traveled as follows:
d = rt
d = 130 mph × 2.5 hours
Multiplying the rate (130 mph) by the time (2.5 hours) gives us:
d = 325 miles
Therefore, the airplane travels a distance of 325 miles during the 2.5 hours of travel at an average rate of 130 miles per hour.
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A sprinter starts from rest and runs with constant acceleration to a top speed of 12m/s in 4.0 seconds.
Answer:
a = 3,0 m/s²
Explanation:
En este ejercicio se pide calcular la aceleracion del cuerpo, usemos las ecuaciones de cinematica en una dimensión.
v= v₀ + a t
como el corredor parte del reposo si velocidad inicial es cero
v = at
a = v/t
calculemos
a = 12 /4,0
a = 3,0 m/s²
Un zapato de golf tiene 10 tacos cada uno con un área de 0.0 pulgadas en contacto con el piso suponga que al caminar hay un instante en que los tacos soportan el peso completo de una persona de 180 libras ¿cuál es la presión ejercida por los tacos sobre el suelo?
Pregunta completa :
Un zapato de golf tiene 10 tacos cada uno con un área de 0.01 pulgadas en contacto con el piso suponga que al caminar hay un instante en que los tacos soportan el peso completo de una persona de 180 libras ¿cuál es la presión ejercida por los tacos sobre el suelo?
Responder:
12230825.435 pascal
Explicación:
Dado que:
Peso (W) = 180 libras
Número de tacos = 10
Área por montante = 0.01 in²
Área total (10 * 0.01) = 0.1 pulg²
Usando la calculadora de conversión:
180 libras = 81.647 kg
0.1 pulg² = 6.452 × 10 ^ -5 m²
Recordar :
Presión = Fuerza (Newton) / Área (m²)
Fuerza = masa * aceleración debida a la gravedad
Aceleración por gravedad = 9,8 m / s²
Presión = (81.647 * 9.8) / (6.452 * 10 ^ -5)
Presión = 800.1406 / 6.452 * 10 ^ -5
Presión = 12230825.435 pascal
how far apart are the object and image produced by a converging lens with 35.5- cm focal length when the object is 45 cm from the lens?
To find the distance between the object and image produced by a converging lens with a 35.5 cm focal length when the object is 45 cm from the lens, you can use the lens formula:
1/f = 1/do + 1/di
Where:
f = focal length (35.5 cm)
do = object distance (45 cm)
di = image distance
Step 1: Plug in the values for f and do:
1/35.5 = 1/45 + 1/di
Step 2: Subtract 1/45 from both sides:
1/35.5 - 1/45 = 1/di
Step 3: Find a common denominator and subtract:
(45 - 35.5)/(35.5 * 45) = 1/di
9.5/(35.5 * 45) = 1/di
Step 4: Take the reciprocal of both sides:
di = (35.5 * 45)/9.5
Step 5: Calculate di:
di ≈ 168.42 cm
So, the object and image produced by the converging lens with a 35.5 cm focal length when the object is 45 cm from the lens are approximately 168.42 cm apart.
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Find a real root of the equation f(x)=x^2-2x-5=0, using bisection method in five stages.
A real root of the equation f(x) = x^2 - 2x - 5 = 0, found using the bisection method in five stages, is approximately x = -0.046875. To find a real root of the equation f(x) = x^2 - 2x - 5 = 0 using the bisection method, we can start by identifying an interval [a, b] that contains the root. Let's choose the interval [-3, 0], where f(a) = f(-3) = 4 and f(b) = f(0) = -5. Since f(a) and f(b) have opposite signs, there must be a root within this interval.
Stage 1:
- Start with interval [a, b] = [-3, 0]
- Calculate the midpoint c = (a + b) / 2 = (-3 + 0) / 2 = -1.5
- Evaluate f(c) = (-1.5)^2 - 2(-1.5) - 5 = -0.25
- Since f(c) has the same sign as f(a), replace a with c: [a, b] = [-1.5, 0]
Stage 2:
- Calculate the new midpoint c = (a + b) / 2 = (-1.5 + 0) / 2 = -0.75
- Evaluate f(c) = (-0.75)^2 - 2(-0.75) - 5 = -4.4375
- Since f(c) has the same sign as f(a), replace a with c: [a, b] = [-0.75, 0]
Stage 3:
- Calculate the new midpoint c = (a + b) / 2 = (-0.75 + 0) / 2 = -0.375
- Evaluate f(c) = (-0.375)^2 - 2(-0.375) - 5 = -2.7461
- Since f(c) has the same sign as f(a), replace a with c: [a, b] = [-0.375, 0]
Stage 4:
- Calculate the new midpoint c = (a + b) / 2 = (-0.375 + 0) / 2 = -0.1875
- Evaluate f(c) = (-0.1875)^2 - 2(-0.1875) - 5 = -1.2217
- Since f(c) has the same sign as f(a), replace a with c: [a, b] = [-0.1875, 0]
Stage 5:
- Calculate the new midpoint c = (a + b) / 2 = (-0.1875 + 0) / 2 = -0.09375
- Evaluate f(c) = (-0.09375)^2 - 2(-0.09375) - 5 = -0.6104
- Since f(c) has the same sign as f(a), replace a with c: [a, b] = [-0.09375, 0]
After five stages, the interval [a, b] has become [-0.09375, 0]. Since the interval is small, we can approximate the root as the midpoint of this interval:
Root ≈ (a + b) / 2 = (-0.09375 + 0) / 2 = -0.046875
Therefore, a real root of the equation f(x) = x^2 - 2x - 5 = 0, found using the bisection method in five stages, is approximately x = -0.046875.
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When entering an expressway, in the acceleration lane you should: A search for a gap in traffic and adjust your speed to the speed of the traffic. B set the cruise control for highway speed. C make a complete stop and check traffic for a suitable gap. D get as close to the vehicle ahead as possible so you can merge into the same gap.
Answer:A
Explanation: search for a gap in traffic and adjust your speed to the speed of the traffic.
A hummingbird is flying around and its velocity v as a function of time t is given in the graph below where
rightwards is the positive velocity direction.
v (m/s)
4
3
2
1
→t (8)
1.5 2 2.5 3 3.5
0.51
-1
-2
What is the hummingbird's displacement Ax from t = 3.0 s to 3.5 s?
Answer:
2 m
Explanation:
The following data were obtained from the question:
Time 1 = 3 s
Time 2 = 3.5 s
Velocity = 4 m/s (constant)
Displacement =?
Next, we shall determine the change in time. This can be obtained as follow:
Time 1 = 3 s
Time 2 = 3.5 s
Change in time = 3.5 – 3
Change in time = 0.5 s
Finally, we shall determine the displacement of the bird between 3 s and 3.5 s as follow:
Velocity = 4 m/s
Time = 0.5 s
Displacement =?
Velocity = Displacement /Time
4 = Displacement /0.5
Cross multiply
Displacement = 4 × 0.5
Displacement = 2 m
Therefore, the displacement of the birth between 3 and 3.5 s is 2 m.
if the intensity of a sound increases from 1 bel (10 db) to 3 bel (30 db), what is the increase in energy per square meter?
The increase in energy is 100times.
Does distance affect the loudness of a sound?The square of the distance from the source of the sound wave determines the inverse proportionality of sound intensity. The strength of a sound wave reduces with increasing distance from the source because sound waves transmit their energy via a two-dimensional or three-dimensional medium.
A sound wave's amplitude determines its intensity, and its frequency corresponds to its rate of vibration.
Acoustic intensity, also referred to as sound intensity, is the amount of energy carried by sound waves per unit of area in a direction perpendicular to that area. The SI unit for intensity, which also includes sound intensity, is the watt per square meter (W/m²).
Mathematically, the sound intensity level can be expressed as,
β(dB) = 10log₁₀(I/I₀)
β₂-β₁ = 10log₁₀(I/I₀)
30-10 = 10log₁₀(I/I₀)
20 = 10log₁₀(I/I₀)
I/I₀ = 10²(or) 100times.
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A particle starts at the origin and moves away from it in a straight line. After 1. 6 seconds it is 12. 8 meters from the origin. It then reverses direction due to a force field with a like charge. After 3. 8 seconds total , the particle is at -9. 23 m what is the average velocity of the particle over 3. 8 s? HELP PLEASE
The average velocity of the particle over 3.8 seconds is -2.43 meters/second.
The given data is as follows:Initial position (x1) = 0 meterFinal position (x2) = -9.23 meterTime taken (t) = 3.8 secondsFormula used: Average velocity formula is given by the formula: $$\text{Average velocity} = \frac{\text{displacement}}{\text{time taken}}$$ Calculation:The displacement is equal to the change in the position of the particle. Therefore, the displacement can be found as follows:Displacement = Final position - Initial position= (-9.23) - (0) = -9.23 meterWe need to substitute this value in the formula for average velocity:Average velocity = $$\frac{\text{displacement}}{\text{time taken}}$$Average velocity = $$\frac{-9.23}{3.8}$$Average velocity = -2.43 meters/second.
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a 1020-hertz sound wave travels at 340 m/s in air with a wavelength of a) 30 m. b) 3 m. c) 0.333 m. d) 1 m. e) none of the above choices are correct.
The wavelength of the sound wave is equal to 0.333 m. Therefore, option (c) is correct.
What are frequency and wavelength?The frequency of the wave can be defined as the number of oscillations that occur in one second and can be expressed in hertz. The wavelength can be defined as the distance between the two adjacent points of a wave such as two crests or troughs.
The relationship between frequency (ν), speed of sound waves (V), and wavelength (λ):
V = νλ
Given, the frequency of the sound wave, ν = 10 Hz
The speed of the sound wave,\(V = 340 m/s\)
The wavelength of the sound waves can determine as follows
λ = V/ν = 340/1020 = 0.333 m.
Therefore, the wavelength of the sound wave is 0.333 m
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A power station that is being started up for the first time generates 6120 MWh of energy over a 10 hour period. (i) If the rated power at full capacity is 660 MW, calculate how long it takes the power station to reach its full power output. (You may assume a constant increase in power from zero to full power) (ii) State what type of power station can be started up fastest and explain why the start-up times for other types of power station are slower. Explain briefly, how this is relevant to optimising the usage of windfarms. c) What is the Bremsstrahlung effect and how can it be avoided in shielding design? d) Sketch the electromagnetic field output from an antenna, describing in detail the two main regions in the output field.
(i)Therefore, it takes approximately 9.27 hours to reach its full power output.(ii)It is necessary to have quick-start power sources, this helps maintain a stable and reliable electricity supply even when wind speeds fluctuate.(c)The Bremsstrahlung effect needs to be considered to ensure proper radiation protection.(d) The near-field region is characterized by strong electric and magnetic fields while the far-field region represents the radiation zone.
(i) To calculate the time it takes for the power station to reach its full power output, we can use the formula:
Energy = Power × Time
Given that the power station generates 6120 MWh of energy over a 10-hour period and the rated power at full capacity is 660 MW, we can rearrange the formula to solve for time:
Time = Energy ÷ Power
Converting the energy to watt-hours (Wh):
Energy = 6120 MWh × 1,000,000 Wh/MWh = 6,120,000,000 Wh
Converting the power to watt-hours (Wh):
Power = 660 MW × 1,000,000 Wh/MW = 660,000,000 Wh
Now we can calculate the time:
Time = 6,120,000,000 Wh ÷ 660,000,000 Wh ≈ 9.27 hours
Therefore, it takes approximately 9.27 hours (or 9 hours and 16 minutes) for the power station to reach its full power output.
(ii) The type of power station that can be started up fastest is a gas-fired power station. Gas-fired power stations can reach full power output relatively quickly because they use natural gas combustion to produce energy.
In contrast, other types of power stations, such as coal-fired or nuclear power stations, have longer start-up times. Coal-fired power stations require time to heat up the boiler and generate steam, while nuclear power stations need to go through a complex series of procedures to ensure safe and controlled nuclear reactions.
This is relevant to optimizing the usage of windfarms because wind power is intermittent and dependent on the availability of wind. This helps maintain a stable and reliable electricity supply even when wind speeds fluctuate.
(c) The Bremsstrahlung effect is a phenomenon that occurs when charged particles, such as electrons, are decelerated or deflected by the electric fields of atomic nuclei or other charged particles. As a result, they emit electromagnetic radiation in the form of X-rays or gamma rays.
In shielding design, the Bremsstrahlung effect needs to be considered to ensure proper radiation protection. These materials effectively absorb and attenuate the emitted X-rays and gamma rays, reducing the exposure of individuals to harmful radiation.
(d) The electromagnetic field output from an antenna can be represented by two main regions:
Near-field region: This region is closest to the antenna and is also known as the reactive near-field. It extends from the antenna's surface up to a distance typically equal to one wavelength. In the near-field region, the electromagnetic field is characterized by strong electric and magnetic field components.
Far-field region: Also known as the radiating or the Fraunhofer region, this region extends beyond the near-field region.The electric and magnetic fields are perpendicular to each other and to the direction of propagation. The far-field region is further divided into the "Fresnel region," which is closer to the antenna and has some characteristics of the near field, and the "Fraunhofer region," which is farther away and exhibits the properties of the far-field.
The transition between the near-field and the far-field regions is gradual and depends on the antenna's size and operating frequency. The size of the antenna and the distance from it determine the boundary between these regions.
In summary, the near-field region is characterized by strong electric and magnetic fields, while the far-field region represents the radiation zone where the energy is radiated away as electromagnetic waves.
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