As the speed of the object increases, the fluid resistance increases.
What is fluid resistance?The fluid resistance is what could otherwise be referred to as the drag force. This is the force that inhibits the object from moving through the fluid. Thus it drags the object and prevents its motion.
This fluid resistance is caused by the layers of the fluid as the object moves through the fluid. As the speed of the object increases, the fluid resistance increases.
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A radio wave transmits 25.0W/m² of power per unit area. A flat surface of area A is perpendicular to the direction of propagation of the wave. Assuming the surface is a perfect absorber, calculate the radiation pressure on it.
The radiation pressure is 83.3 nPa.
What is radiation pressure?The mechanical pressure that is applied to any surface as a result of the exchange of momentum between an item and an electromagnetic field is known as radiation pressure.
It is the pressure that electromagnetic radiation exerts on a surface as a function of the momentum it carries; radiation pressure doubles if the radiation is reflected as opposed to absorbed.
The radiation pressure will be:
P = I/C
P = 25 / 3 / 10^8
P = 83.3 nPa.
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The Sun is lower in the sky during the winter than it is during the summer. (b) How does this change affect the weather?
When the Sun is lower in the sky during the winter than it is during the summer, reduced flux results, on the average, causes colder weather while increased flux causes hot weather.
The sun's rays strike the Earth at an acute angle during the summer. Because the light does not spread as much, the quantity of energy striking any specific location is increased. Furthermore, the lengthy daytime hours give the Earth plenty of time to warm up or become hot. In contrast, during the winter, the sun's rays strike the Earth at a shallow angle. These beams are more dispersed, reducing the amount of energy that impacts any specific place. Furthermore, the long nights and short days keep the Earth from overheating. As a result, we have cold weather.
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Is it possible for the gravitational force between 2 50kg objects to be less than the gravitational between a 50kg object and a 5kg object?
Yes, if the two 50 kg objects are significantly farther apart, the gravitational force between them may be less than the gravitational force between a 50 kg and a 5 kg object.
The gravitational force increases as the mass of an object increases (also called the gravity force). Since gravitational force is inversely proportional to the square of the distance between the two interacting objects, greater separation distance will result in less gravitational forces.
Therefore, the gravitational attraction between two things weakens as they become farther apart. The size of this force is determined by the mass of each object and the separation of their centers.
According to mathematics, the force of gravity is proportional to the square of the distance between the objects and directly related to the masses of the items.
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Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a downward current of 4.00 A . A third wire, wire 3 , is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force. (b) the position of wire 3 and
The position of the wire is at a distance of 12 cm, in the direction left of the wire.
The force per unit length between two parallel thin current-carrying [tex]I_1[/tex] and [tex]I_2[/tex] wires at distance ' r ' is given by [tex]f=\frac{u_0I_1I_2}{2\pi r}[/tex] ....(1) . If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.It is given that I₁ = 1.50 A and I₂ = 4.00 A and the distance between wire 1 and wire 2 = 20cm = 0.2m
Let the distance of wire 3 from wire 2 be " d " .
We have to place the 3 wire such that each wire experiences no net force which means force on wire 2 due to wire 3 = force on wire 1 due to wire 3
[tex]F_2=F_1[/tex]
Using equation (1) , we get
[tex]\frac{u_0I_2I_3}{0.2+d}=\frac{u_0I_1I_3}{d} \\\\\frac{I_2}{0.2+d} =\frac{I_1}{d} \\\\\frac{4.00}{0.2+d} =\frac{1.50}{d} \\\\4.00d=1.50(0.2+d)\\\\4.00d=0.3+1.50d\\\\\2.5d=0.3\\\\d=0.12m[/tex]
d= 12 cm in the direction left of the wire.
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The complete question is given below .
A schematic of the information provided in the question can be seen in the image attached below.Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a downward current of 4.00 A . A third wire, wire 3 , is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force. (b) the position of wire 3 and
What is the mass of an object that requires 100N (kg-m/s2) of force in order to accelerate it at 10m/s2 (Please use G-R-E-S-A)
Answer:
10kg
Explanation:
(I'm not super familiar with the GRESA method so apologies for any inaccuracies)
Given: We are given values for Force: 100N, and Acceleration: 10m/s2.
Required: We are trying to find Mass (m)
Equation: The best equation to use to solve this problem is F=ma,
Force = Mass x Acceleration. We can rearrange this for mass: m = F/a.
Solution: By substituting in the values we have: [tex]m = \frac{100}{10}[/tex]
Answer: Mass = 10kg
Hope this helped!
What is the wavelength of the photon needed to excite an electron from e1 to e4?.
The wavelength needed in transition energy is λ = 9.73 x 10¯⁸ m.
We need to know about transition energy to solve this problem. Electrons in a hydrogen atom can move to another level of energy. It is also called a transition. In this process, electrons will absorb or release their energy to move. The transition energy can be determined as
ΔE = E2 - E1
where ΔE is transition energy, E2 is the final state of energy and E1 is the initial state of energy.
The state energy of a hydrogen atom can be calculated by
E = -(13.6) / n² eV
E = h.c / λ
where n is the number of energy states, h is Planck constant (4.136 x 10¯¹⁵ eV/Hz), c is speed of light (3 x 10⁸ m/s) and λ is wavelength.
From the question above, we know that
n2 = 4
n1 = 1
By substituting the following parameter, we get
ΔE = E2 - E1
ΔE = -(13.6) / n2² - (-(13.6) / n1²)
ΔE = -(13.6) / 4² - (-(13.6) / 1²)
ΔE = 13.6 - 0.85
ΔE = 12.75 eV
Calculate the wavelength needed
ΔE = 12.75 eV
h.c / λ = 12.75
4.136 x 10¯¹⁵ . 3 x 10⁸ / λ = 12.75
λ = 9.73 x 10¯⁸ m
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You exert a force of 100N on a rope that pulls a sled across a flat icy surface. The rope is oriented 30 degrees above the horizontal. The sled has a mass of 10kg and it is carrying a 30kg child. The sled starts at rest. You can assume friction is negligible.
You reach a hill and start going up, still applying a constant 100N force at 30 degrees from the slope. The sled is now moving at constant speed. What is the angle of the slope?
The angle of slope if the sled carrying a 30 kg child which is moving at a constant speed is 12.7°
Given that,
Force, F = 100 N
Mass, m = 10 + 30 = 40 kg
Angle at which the force is applied, θ = 30°
Resolving the force applied into its horizontal and vertical components,
cos θ = [tex]F_{x}[/tex] / F
[tex]F_{x}[/tex] = 100 * cos 30° = 100 * 0.866
[tex]F_{x}[/tex] = 86.6 N
A force due to the weight of the cart is applied on the ground which is directly vertical downwards. But the slope is at an angle. So resolving this force into its horizontal and vertical components.
sin θ = [tex]w_{x}[/tex] / w
w = m g
w = 40 * 9.8
w = 392 N
Given that the velocity when travelling up the slope is constant. So the acceleration is zero.
Σ [tex]F_{x}[/tex] = m a = 0
[tex]F_{x}[/tex] - [tex]w_{x}[/tex] = 0
[tex]w_{x}[/tex] = 86.6 N
θ = [tex]sin^{-1}[/tex] ( 86.6 / 392 ) = [tex]sin^{-1}[/tex] ( 0.22 )
θ = 12.7°
Therefore, the angle of slope if the sled carrying a 30 kg child which is moving at a constant speed is 12.7°
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Table of current and potential differences for resistors in series
Answer:
use ohms law for tabling.
Potential (V) = current (I) × R
Let R = 1 ohm.
table:-
V. (voltage difference) current (I)
1. 1 Volt 1A
2. 2 volt 2 A
3. 4 volt 4 A
4. 6 volt 6A
.
.
.
.
the value, v(m), of a comic book m months after publication has an average rate of change of –0.04 between m
The value of the comic book decreased by an average of $0.04 each month between m = 36 and m = 60.
The average rate of change for the function f(x) from [tex]x_{1} =a[/tex] and [tex]x_{2}=b[/tex] is [tex]\frac{f(b)-f(a)}{b-a}[/tex]
It is the average amount by which the function changed per unit throughout that time period.
Actually, this is the angle of the line on the function's graph that passes through two points.
The value of the comic book fell by an average of $0.04 per month (a unit in this case is a month) between m = 36 and m = 60 if the average rate of change of the function f(x) between m = 36 and m = 60 is -0,04.
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Celia and jane are on the roof of their school. Each woman has an identical ball that they release from the roof at the same time. However, while celia drops her ball from rest, jane releases her ball with a small horizontal velocity. Assuming that air resistance is negligible, which ball will hit the ground first?.
Answer:
Jane's ball will hit the ground first
describe and explain how energy if thermally transferred from the heating element to the food in a conventional oven
Answer:
The thermal transfers via "Conduction and Radiation"
Explanation:
Because, Convection is the transfer of thermal energy by particles moving through a fluid. Thermal energy is always transferred from an area with a higher temperature to an area with a lower temperature.
Moving particles transfers the thermal energy through a fluid by forming convection currents.
It transfers ( thermal ) heat energy to a food's surface via convection and radiation. So that, the heat energy on the food's surface, then it spreads via "conduction" towards the food's center.
two like charges of the same magnitude are 5.0 mm apart. if the force of repulsion they exert upon each other is 2.0 n, what is the magnitude of each charge? (the constant of proportionality for the coulombic force is 9.0 × 109 n·m2/c2.)
The magnitude of charge of each particle is 5.55 μC
Given that the two like charges are of same magnitude. Let Q be the magnitude of the charges.
Distance between the particles, r = 5 mm = [tex]5\times 10^{-3}[/tex] m
The force of repulsion exerted upon each other by the charges = 2 N
Coulomb's law states that the force of interaction between the charges is directly proportional to the product of the magnitude of charges and is inversely proportional to the square of the distance between the charges.
I.e., [tex]F = \frac{kQ^2}{r^2}[/tex] , k = [tex]9\times 10^9 Nm^2C^{-2}[/tex] is the constant of proportionality for the Coulombic force.
⇒ 2 = [tex]\frac{9\times 10^9\times Q^2}{5^2\times 10^{-6}}[/tex]
Therefore, [tex]Q^2 =\frac{50\times 10^{-6}}{9\times10^9}[/tex]
Therefore, Q = 5.55 x [tex]10^{-15}[/tex] C = 5.55 μC is the magnitude of the charge of each particle.
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On a velocity-time graph, when is the object not moving?
when the slope is a straight line rising to the right
when the slope is a straight line rising to the right
at the point in which the line crosses the x-axis
at the point in which the line crosses the x -axis
when the slope is a straight line falling to the right
when the slope is a straight line falling to the right
when the slope is a line curving upward to the right
when the slope is a line curving upward to the right
The time when the object is not moving on the velocity time graph is at the point in which the line crosses the x - axis.
What is the velocity time graph?The velocity time graph can be used to observe the movement of a particle. The graph consist of three main parts;
Uniform accelerationConstant velocityUniform decelerationWe can say that the time when the object is not moving on the velocity time graph is at the point in which the line crosses the x - axis.
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Structures of the skeletal system
Bones, Cartilage, Ligaments, Tendons, and Muscles.
An engineer at NASA is investigating the impact of meteorites on planets. He performs experiments in a vacuum by dropping metal spheres of different sizes and different heights above a horizontal bed of sand. In one experiment, a metal sphere of mass 0.1 kg is dropped from a height of 1.2 m. The sphere makes a depression in the sand of 2 cm. [Gravitational field strength, g = 10 Nkg -1 ] i) Calculate the potential energy of the sphere at a height of 1.2 m [2 marks]
The potential energy of the sphere at a height of 1.2 m is 1.2 J.
The gravitational potential energy of an object is defined as the energy stored inside a body due to virtue of its position.
The equation for calculating the potential energy(PE) is as follows:
PE = mgh
Here m is the mass of the body
g is the acceleration due to gravity
h is the height of the body above the ground level/reference level
Putting m = 0.1 kg, h = 1.2 m and g = 10 m/s^2
PE = 0.1 x 1.2 x 10 = 1.2 J.
Thus, the potential energy of the sphere at a height of 1.2 m is 1.2 J.
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Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a downward current of 4.00 A . A third wire, wire 3 , is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force. (a) Is this situation possible? Is it possible in more than one way? Describe
Yes, this situation is possible. No, It is not possible in more than one way.
The force per unit length between two parallel thin current-carrying [tex]I_1[/tex] and [tex]I_2[/tex] wires at distance ' r ' is given by [tex]f=\frac{u_0I_1I_2}{2\pi r}[/tex] ....(1) If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.A schematic of the information provided in the question can be seen in the image attached below.
From the picture: Assuming that the forces exerted on the wire by the other two wires are equal and opposite, the third wire exerts no net force. So you can say: Yes the situation is possible.
The forces acting on the other two wires will be in opposite directions, so they cannot have multiple passes, but they will not be of the same magnitude.
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A long, vertical, metallic wire carries downward electric current. (ii) What would be the direction of the field if the current consisted of positive charges moving downward instead of electrons moving upward? Choose from the same possibilities as in part (i).
The direction of the conventional current that generates the magnetic field at a specified place determines its direction. The direction of the field if the current consisted of positive charges moving downward instead of electrons moving upward is South.
What is the electric current's direction?In terms of circuit analysis, the direction of electric current from positive to negative is usually considered as well. Positive charges moving in the opposite direction are mathematically similar to negative charges moving in the same way. Therefore, it does not have any effect. When examining a circuit, the flow of current from positive to negative or vice versa can be taken into account. In actuality, negatively charged electrons can attract positively charged ions. Conventionally, a positive charge would flow in the same direction as an electric current. As a result, the battery's positive terminal receives less current in the external circuit than its negative counterpart. Actually, electrons would flow in the reverse direction across the wires.
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put the types of electromagnetic radiation in order of decreasing frequency. view available hint(s)for part a put the types of electromagnetic radiation in order of decreasing frequency. microwave > radio > infrared > ultraviolet radio > microwave > infrared > ultraviolet ultraviolet > infrared > radio > microwave ultraviolet > infrared > microwave > radio
The types of electromagnetic radiation in order of decreasing frequency will be D. ultraviolet > infrared > microwave > radio.
What is electromagnetic radiation?
This is the radiation that travels in waves and possesses both an electric and magnetic field. Both natural and artificial sources contribute to it. The energy levels of electromagnetic radiation can range from low to high. It consists of x-rays, gamma rays, infrared light, visible light, ultraviolet light, radio waves, and microwaves.
When a charged particle, like an electron, changes its velocity—that is, when it is accelerated or decelerated—electromagnetic radiation is created. As a result, the charged particle loses the energy of the electromagnetic radiation that is thus created.
It should be noted that radio has highest wavelength and ultraviolet has lowest wavelength.
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Acceleration = 2.8 m/s^2
If after take-off, the jet continues to accelerate at the same rate for another
15 s, how fast will it be going at that time?
Speed and velocity are similar in term of their unit. the speed of the jet is 42 m/s
What is Speed ?
Speed can simply be defined as how fast an object moves. It is the distance covered per time taken.
Given that acceleration = 2.8 m/s² and If after take-off, the jet continues to accelerate at the same rate for another 15 s, to know how fast it will be going at that time, we will use acceleration formula.
Acceleration = Velocity / time
speed = acceleration × time
speed = 2.8 × 15
Speed = 42 m/s
Therefore, the speed or the velocity of the jet is 42 m/s
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Why is the following situation impossible? An experimenter is accelerating electrons for use in probing a material. She finds that when she accelerates them through a potential difference of 84.0 kV , the electrons have half the speed she wishes. She quadruples the potential difference to 336 kV , and the electrons accelerated through this potential difference have her desired speed.
When the electrons accelerated through a potential difference of 84 kV, the electrons have half the speed that she wishes.
When the potential is quadrupled to 336 kV, the electrons acquire the desired speed.
Let u be the speed of the electrons after accelerating through a potential difference ΔV.
Now,
K = eΔV = (γ - 1)mc² where K is the kinetic energy.
K = (γ - 1)mc² =[ 1/ ( √1 - (u/c)² ) - 1]mc²
Therefore,
[ 1/ ( √1 - (u/c)² ) - 1] = (eΔv / mc² ) + 1 = (eΔV + mc² ) / (mc²)
1 - (u/c)² = [ (mc²) / (eΔV + mc² ) ]²
u/c = √[ 1 - ( m / {(eΔV/c²) + m})² ]
u / c = √[ 1 - ( 9.11 × 10⁻¹³ / {(1.6 × 10⁻¹⁹ × 8.4 × 10⁴ /(3 × 10⁸)² + 9.11 × 10⁻¹³ })² ]
u = 0.512c
Despite the increased accelerating voltage, this speed cannot be doubled because it is greater than half that of light. The electrons move at u = 0.798c faster if the accelerating voltage is quadrupled to 336 kV.
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(a) a spring stretches by 0.015 m when a 1.75 kg object is suspended from its end. what is its spring constant? (b) how much mass should be attached to the spring so that its frequency of vibration is f
(a) Total force along y axis,
∑F =ma=0
kx−mg=0 ⇒k= [tex]\frac{mg}{x}[/tex]
The spring constant is, k is
k=(1.5)×(9.8)/(0.020)=14.7/0.020 =735N/m
(b) Frequency f = [tex]\frac{1}{2} \pi[/tex] [tex]\sqrt{\frac{k}{m} }[/tex] ⇒ 4[tex]\pi ^{2}[/tex][tex]f^{2}[/tex] = [tex]\frac{k}{m}[/tex]
[tex]\frac{k}{4\pi ^{2}f^{2} }[/tex]= m ⇒ m = (735)4[tex]\pi ^{2}[/tex](3.0)(0.3)
=(735)×4×(3.75)×(3.75)×(9.0)
=735×4×9.85×9.0=7354×9.85×9.0
=735354.6 N
=2.07kg
The spring constant is the force required to stretch or compress the spring divided by the distance the spring is lengthened or shortened. It is used to determine the stability or instability of a spring and thus the system for which it is intended. It is denoted by k.
Therefore the spring constant is 735 N/m and the mass should be 2.07kg.
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Few meteors make it through the thick air of this planet without disintegrating. True or false?.
Few meteors make it through the thick air of this planet without disintegrating. This statement is false.
Meteors refer to heavily sized metallic objects or rock fragments that are formed as a result of a collision between asteroids.
These rock lumps revolve around the sun and are smaller in size than asteroids.
Their size ranges from grain-sized particles to about 1 meter wide objects
As they enter the atmosphere of earth they crash forming streaks of light and disintegrate into small-sized dust particles and never get through the thick air of the earth.
However, they impact the earth in several ways such as heating the earth's surface.
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Give a physical argument that shows it is impossible to accelerate an object of mass m to the speed of light, even with a continuous force acting on it.
It is not possible to accelerate an object to the speed of light.
What is the speed of light?The speed of light is generally regarded as the fastest speed on earth. We can tell from the Newton's law that the force that acts on a body is the product of the mass and the acceleration of the body.
It then follows that the force that is required to accelerate an object to the speed of light is a very high force that s somewhat unattainable and as the object approaches the speed of light, the mass of the object has to become infinitely small.
Hence, it is not possible to accelerate an object to the speed of light.
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A speeding car passes a highway patrol checkpoint, then decelerates at a constant rate. After 5 s, the car is 225 m from the checkpoint, and its speed is then 30 m/s. What was the car’s velocity when it passed the checkpoint?.
Answer:
Car velocity when crossing checkpoint = 60 m/s
Explanation:
For an object traveling at an acceleration of a m/s², with an initial velocity of u m/s the displacement at time = t secs is given in meters by the equation
[tex]\displaystyle d=ut+\frac{1}{2}at^{2}[/tex]
Here we are given displacement, and time but not acceleration and initial velocity : d = 225 meters, t = 5 seconds
Let's find an equation relating u and a in terms of d and using data given
Switching sides we get
[tex]ut+\frac{1}{2}at^{2}=d[/tex]
Substituting values for t = 5, d = 225 we get
[tex]5u+\frac{1}{2}a.25=225[/tex]
Multiplying both sides by 2 yields
[tex]10u+25a=450\;\;\; ...... (1)[/tex]
We also have the formula:
[tex]\displaystyle a=\frac{{v-u}}{t}[/tex]
where v is the current velocity and u the initial velocity
So
[tex]\displaystyle a=\frac{30-u}{5}[/tex]
[tex]u+5a=30\;\;...... (2)[/tex]
Multiply equation (2) by 5 and subtract from (1) to eliminate the a terms and solve for u
[tex]\displaystyle 10u+25a-5u-25a=450-150\\\\5u=300\\\\u=60m/s\\\\\textsf{which is the speed at which the car passes the checkpoint}\\[/tex]
A parallel-plate capacitor is disconnected from a battery, and the plates are pulled a small distance further apart. Do the following quantities increase, decrease, or stay the same?.
When the parallel-plate capacitors are disconnected from a battery and plates are pulled a small distance farther apart then Q remains the same, Change in potential increases, C decreases, E stays the same and Energy stored increases.
Solution:
Let's say the capacitance of parallel plates is C
Charge stored = Q
Distance between a plate of capacitor = d
The capacitance of a parallel plate capacitor equals when a parallel- plate capacitor is disconnected.
Then, charge Q remains the same i.e.
Q = εA / d = Q / Ed = Q/V
Here Q=Charge stored in a capacitor
E = Electric field, V is the potential difference, and 'd' is the separation between plate
Both potential difference and distance are inversely proportional to the capacitance between a plate of the capacitor.
Similarly, the potential difference increases when capacitance decreases.
E = V/d
While the value of V and d decrease.
E remains constant, therefore energy stored in the capacitor will be E = 1/2CV^2
The decrease in C causes an increase in V, as a result of which energy stored in the capacitor increases.
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suppose your bathroom scale reads your mass is 55 kg, with a 1% uncertainty. what is the uncertainty in your mass in kilograms?
Answer:
± .55 kg
Explanation:
55 * 1% = .55 kg
± .55 kg
As the people sing in church, the sound level everywhere inside is 101 dB . No sound is transmitted through the massive walls, but all the windows and doors are open on a summer morning. Their total area is 22.0 m² . (b) Suppose the ground is a good reflector and sound radiates from the church uniformly in all horizontal and upward directions. Find the sound level 1.00 km away.
The sound level from 1 km away is 46.4 dB and the sound energy radiated through the windows and doors in 20 min is 332.6 J
What do you mean by sound radiates?Multiple plate modes participate in the forced vibration of a baffled plate that produces sound. The emitted sound power depends on the activation of each plate mode by the external pressures as well as the radiation characteristics of the individual plate modes and their mutual interaction via their radiated sound, as shown in eq (20) and (24). The net contribution of the mutual interaction between plate modes to the overall sound power may be disregarded if a plate is activated in a manner that results in a plate vibration field that can be defined as reverberant with an equal distribution of energy over all modes.
(Lw) = 10·log (W/Wo) dB
Given:
sound level, [tex]\beta= 101 dB[/tex]
Area, A = [tex]22\;m^{2}[/tex]
Time, [tex]\triangle t = 20\;min=1200\;s[/tex]
Intensity, [tex]I=1\times 10^{-12}\;W/m^{2}[/tex]
[tex]r=1\;km=1000\;m[/tex]
(a)
We know that, Sound level is,
[tex]\beta=10\times log(\frac{I}{I_{o} } )[/tex]
Solving the above equation for sound intensity,
[tex]I=I_{o} \times 10^{\frac{\beta}{10} }[/tex]
[tex]I=1 \times 10^{-12} \times 10^{\frac{101}{10} }[/tex]
[tex]I=0.0126\;W/m^{2}[/tex]
Therefore, The sound energy is,
[tex]E=P\times \triangle t[/tex]
Substitute [tex]P=I \times A[/tex] in the above equation,
[tex]E=I \times A \times \triangle t[/tex]
[tex]E=0.0126 \times 22 \times 1200[/tex]
[tex]E=332.6\;J[/tex]
(b)
Half of a sphere area with 1 km radius is equal to Area of the sound at 1 km distance,
[tex]A_{hemisphere} = \frac{1}{2} \times 4 r^{2} \pi[/tex]
Substitute the known value in the above equation ,
[tex]A_{hemisphere} = \frac{1}{2} \times 4 (1000)^{2} \pi[/tex]
[tex]A_{hemisphere} = 6283185\;m^{2}[/tex]
Sound Intensity is,
[tex]I = \frac{P}{A_{hemisphere}}[/tex]
Substitute [tex]P=I \times A[/tex] in the above equation,
[tex]I = \frac{I \times A}{A_{hemisphere}}[/tex]
Substitute the known value in the above equation,
[tex]I = \frac{0.0126 \times 22}{6283185}[/tex]
[tex]I = 4.4 \times 10^{-8}\;W/m^{2}[/tex]
Sound level is,
[tex]\beta=10\times log(\frac{I}{I_{o} } )[/tex]
Substitute the known value in the above equation,
[tex]\beta=10\times log(\frac{4.4 \times 10^{-8} }{1 \times 10^{-12} } )[/tex]
[tex]\beta=46.4\;dB[/tex]
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A proton moves at 0.950 c . Calculate its (a) rest energy,
The rest energy of the proton is 900 MeV.
We need to know about relativistic energy to solve this problem. The rest energy of the object can be determined by
Eo = m₀ . c²
where Eo is rest energy, m₀ is rest mass and c is speed of light (3 x 10⁸ m/s).
From the question above, we know that :
m₀ = 1.6 x 10¯²⁷ kg
c = 3 x 10⁸ m/s
By substituting the following parameter, we get
Eo = m₀ . c²
Eo = 1.6 x 10¯²⁷ . (3 x 10⁸)²
Eo = 1.44 x 10¯¹⁰ joule
Eo = 1.44 x 10¯¹⁰ / (1.6 x 10¯¹⁹) eV
Eo = 900 x 10⁶ eV
Eo = 900 MeV
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9. Which parts of the ear have a solid medium?
Middle part of the ear have a solid medium. The ossicles, which are bones that amplify sound waves, are located in the middle ear, therefore the medium is solid.
The outer ear is the first place that sound waves pass through, followed by the middle ear and lastly the inner ear. The first medium is gas because the external auditory canal, which is part of the outer ear, is filled with ambient air. The ossicles, which are bones that amplify sound waves, are located in the middle ear; therefore the second medium is solid. The last medium is liquid because the inner ear is loaded with fluids like endolymph and perilymph that vibrate when sound waves pass through them.
The middle ear amplifies sound after being directed toward it by the outer ear, which is in charge of catching it. Amplified sound waves are sent from the middle ear to the inner ear, where hair cells detect the sound waves and transmit the data to the brain.
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(i) What happens to the magnitude of the magnetic field inside a long solenoid if the current is doubled? (a) It becomes four times larger. (b) It becomes twice as large. (c) It is unchanged. (d) It becomes one-half as large. (e) It becomes one-fourth as large.
The magnetic field inside the solenoid is
B = μo nI
Here n = N/l
Here I be the current
N be the number of turns
l be the length of solenoid
Because the magnetic field in this instance is independent of radius, "c) it is unaffected."
What is magnetic field made of?Electric charges in motion create magnetic fields. Everything is constructed of atoms, and each atom contains a nucleus that is composed of protons and neutrons, with orbiting electrons. Each atom creates a tiny magnetic field because the orbiting electrons are tiny moving charges.
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