(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².
(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².
(1.c) The relative intensity of the sound as heard by the listener is 103 dB.
(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.
(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.
Surface area being vibratedThe surface area being vibrated by the time the sound reaches the listener is calculated as follows;
A = 4πr²
A = 4π x (20)²
A = 5,026.55 m²
Intensity of the soundThe intensity of the sound is calculated as follows;
I = P/A
I = (100) / (5,026.55)
I = 0.02 W/m²
Relative intensity of the sound[tex]B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB[/tex]
Speed of sound at the given temperature[tex]v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s[/tex]
Frequency of the soundThe frequency of the sound heard is determined by applying Doppler effect.
[tex]f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )[/tex]
where;
-v₀ is velocity of the observer moving away from the source-vs is the velocity of the source moving towards the observerfs is the source frequencyfo is the observed frequencyv is speed of sound[tex]f_0 = f_s(\frac{v-v_0}{v- v_s} )[/tex]
[tex]f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz[/tex]
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What is electrical power and how is it measured?
Answer:
power is measured in watts and its the messure of work over time.
Explanation:
What is one benefit of smart meters?
Answer:
You get up-to-date readings
Explanation:
What would be the potential difference of the capacitor if it is connected to a battery of 3v?
Answer: 3V
Explanation:
If the capacitor is connected to a 3V battery:
[tex]\bold{V = \frac{Q}{C} }[/tex]
The potential difference is also 3V on the capacitor
You are pushing a 30 crate along a rough surface by applying a horizontal force. The coefficients of
frictions between the surface and the crate are 0.3 and 0.6. Sketch the situation and create a Free Body diagram
from it, then Calculate:
a. The normal force on the crate by the floor.
b. The force required to get the crate to move.
c. Determine the kinetic friction force that the floor applies on the object.
d. Determine the acceleration of the crate.
Answer:
ill say math too!
Explanation:
stion 3
Define the law of moments. It is applicable to a certain condition of the acting forces.
Name the condition.
Answer:
algebraic sum of moments of all the forces acting on the body about the axis of rotation is zero, the body is in equilibrium. According to the principle of moments, in equilibrium: Sum of anticlockwise Moments = Sum of clockwise moments.
Condition:
For an object to be in equilibrium, it must be experiencing no acceleration.
Hope it may helpful to you
Which best illustrates projectile motion?
A pictorial diagram showing a hot air balloon in four different positions, each the same height from the ground and equally spaced.
A pictorial diagram with a rocket launched straight upward. The first three positions of the rocket are shown with equal spacing between them.
A pictorial diagram with a person swimming and position shown in equal time intervals and equally spaced.
A pictorial diagram of a person hopping sideways at 5 different positions. The start and end positions are on the ground. The third image is the highest off the ground.
The pictorial diagram with a rocket launched straight upward, showing the first three positions of the rocket with equal spacing between them, best illustrates projectile motion. In this case, the rocket is projected upward into the air, and its path follows a parabolic trajectory due to the influence of gravity.
What is projectile motion?Projectile motion refers to the motion of an object that is thrown, launched or otherwise projected into the air and then moves under the influence of gravity alone. In this type of motion, the object moves along a curved path, called a parabola, due to the combined effect of its initial velocity and the force of gravity acting on it.
When a rocket is launched straight upward, it experiences a projectile motion. The rocket's initial velocity is upwards, but gravity causes it to decelerate until it eventually stops at the highest point of its trajectory. The rocket then starts to fall back down, accelerating due to gravity until it reaches the ground. If we take three equal intervals of time, we can show the rocket's positions at those times. Initially, the rocket will be moving upwards with a decreasing velocity, then it will reach its maximum height before falling back down with an increasing velocity.
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Answer: I believe it should be the final answer,
A pictorial diagram of a person hopping sideways at 5 different positions. The start and end positions are on the ground. The third image is the highest off the ground
Explanation:
Find the magnitude of the sum of these two vectors:
Answer:
Explanation:
You can decompose those vectors into their components in x and y direction. For the first vector:
[tex]\vec{r}_{1}=r_{1}\cos 30\hat{i}+r_{1}\sin 30\hat{j}=3.14(\frac{1}{2}\sqrt{3}\hat{i}+\frac{1}{2}\hat{j})=1.57\sqrt{3}\hat{i}+1.57\hat{j}[/tex]
For the second vector:
[tex]\vec{r}_{2}=r_{2}\cos 60\hat{i}-r_{2}\sin60 \hat{j}=1.355\hat{i}-1.355\sqrt{3}\hat{j}[/tex]
The sum of two vectors will be:
[tex]\vec{r}=\vec{r}_{1}+\vec{r}_{2}=(1.57\sqrt{3}+1.355)\hat{i}+(1.57-1.355\sqrt{3})\hat{j}\approx 4.0711\hat{i}-0.77415\hat{j}[/tex]
The magnitude of the sum of two vectors is:
[tex]r=\sqrt{(4.0711)^{2}+(-0.77415)^{2}}\approx 4.14[/tex] meter
Answer:
4.14m
Explanation:
A toy car, with mass of 6 Kg, is pushed with a force of 10 N. If the toy car is in the grass with a coefficient of friction of 0.1 then what is the acceleration?
How far does it get pushed in 10 s?
at 27 degrees C and 750 mmhg a sample of hydrogen occupies 5.0 L how much space will occupy at STP
Answer:
Explanation:
The number of mole of the hydrogen sample at 27 C and 750 mmHg pressure is:
[tex]n=\frac{PV}{RT}=\frac{(0.9898\times 10^{5})(5\times 10^{-3})}{(8.314)(27+273)}\approx 0.0198[/tex] mol
This number of mole will be the same although we change the pressure and temperature. For STP (Temperature = 0 C = 273 K, and Pressure = 760 mmHg = 76 cmHg = 1 Bar = 1.01 x [tex]10^{5}[/tex] Pa):
[tex]P'V'=nRT' \rightarrow V'=\frac{nRT'}{P'}=\frac{0.0198(8.314)(273)}{1.01\times 10^{5}}\approx 44.49 \times 10^{-5}[/tex] in [tex]m^{3}[/tex]. But in L, we find that V' = 0.4449 L
Which nucleus completes the following equation?
Answer:
I think that CL IS THE ONLY BEST ANSWER
Explanation:
PLEASE MARK ME BRAINLIEST IF MY ANSWER IS CORRECT PLEASE
A wheel rotates at 24 revolutions every 3 minutes. This is equivalent to
a car 150m to the east then backtracks 30m to the west. Describe the mathematical operation you will use to determine the net movement of the car
Answer:
Total displacement will be 120m
Explanation:
Since displacement is shortest distance travelled
Td= 150m east - 30m west ( Since they are opposite in direction)
Td= 120m East
How Does a Broken Bone Heal? Interactivity. Please i need the answer soon.
Answer:
The body develops a blood clot over the fractured bone in the first few days of a fracture to protect it and deliver the requisite cells for healing. Then, around the fractured bone, a field of healing tissue grows. This is referred to as a callus.
Explanation:
Answer:
Bone heals by making cartilage to temporarily plug the hole created by the break. This is then replaced by new bone. Many people think of bones as being solid, rigid, and structural. Bone is, of course, key to keeping our bodies upright, but it is also a highly dynamic and active organ
Explanation:
Friction helps your vehicle stop quickly.
A. TRUE
B. FALSE
Answer:
true
Explanation:
Friction helps your vehicle stop quickly. this statement ids true. Hence option A us correct. A vehicle can stop more rapidly because to friction. The brake pads rub on the turning braking discs or drums when the brakes are applied.
The friction slows the moving vehicle down and eventually causes it to halt by converting its kinetic energy into heat energy. The effectiveness of the braking action will increase with the level of friction between the brake parts. For rapid stops, friction between the tyre and the road is also important.
The vehicle's forward motion is opposed by friction created by the tyres' gripping action on the road surface, which helps in stopping and decelerating. Rapid vehicle stopping would be incredibly difficult and risky without friction.
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1. A drag racer accelerates from rest at 18ft/sec^2. How long does it take to acquire a speed of 60mph? What is required?
2. A contestant ran a 100-m dash in 10.6sec. What was his speed a) In feet per second and b) in miles per hr?
(1) The time taken for the drag racer to accelerate is 4.89 s
(2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.
(1) To calculate the time required to accelerate 18 ft/sec² from rest to a velocity of 60 mph, we use the formula below.
Formula:
t = (v-u)/a........... Equation 1Where:
t = timev = Final velocityu = initial velocitya = acceleration.From the question,
Given:
a = 18 ft/sec² = (18×0.3048) = 5.4864 m/s²v = 60 mph = (60×0.44704) = 26.82 m/su = 0 m/s ( from rest)Substitute these values into equation 1
t = (26.82-0)/5.4864t = 4.89 seconds(2) To calculate the speed of the contestant, we use the formula below
Formula:
s = d/t............ Equation 1Where:
s = speed of the contestantd = distancet = time.From the question,
Given:
d = 100 mt = 10.6 sSubstitute these values into equation 1
s = 100/10.6s = 9.43 m/s(a) In feet = (9.43/0.3048) = 30.94 ft/s
(b) in miles per hr = (9.43×2.24) = 21.12 miles per hr
Hence, (1) The time taken for the drag racer to accelerate is 4.89 s (2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.
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Which angle is the angle of refraction?
Explanation:
[tex]\longmapsto\: \red{ hello \: mate}[/tex]
OPTION (C) 3
is the angle of refraction
Answer:
Option C. 3
Explanation:
Refraction is the change in direction of light passing from one medium to another. Angle 1 and 2 are reflection due to the bouncing of light.
✓ 4.) A certain molecule has an energy-level diagram for its vibrational energy in
which two levels are 0.0141eV apart. Find the wavelength of the emitted
line for the molecule as it falls from one of these levels to the other. [h
6.63 x 10-34 kgm s-?, C = 3 x 108ms-1, lev = 1.602 x 10^-19j
Answer:
Explanation:
You can use [tex]\Delta E=\frac{\hbar c}{\lambda} \rightarrow \lambda =\frac{\hbar c}{E} = \frac{1240 eV.nm}{0.0141 eV} \approx 87943.26[/tex] in nm.
Here the hc = 1240 eV.nm in eV and nanometer unit.
Which statement is true about a wave that is created by causing vibrations in
particles of matter?
A. It can travel through empty space.
B. It is not an electromagnetic wave.
C. It does not need a medium.
D. It moves through a magnetic or electric field.
Answer:
A
Explanation:
A. it can travel through empty space
.. In the reaction 238 92U ---- X + 2 4 He, the particle represented by X is . answer choices . 234 90 Th. 234 92 U. 238 93 Np. 242 94Pu. ... Which conditions are required
My question
• Complete the following reaction:
238 92U = ______ + ________ + 4 2He
…Is this the same question as the answer
Answer:
234 90 +energy
Explanation:
because alpha decrease by 4 2helium
At a typical bowling alley the distance from the line where the ball is released (foul line) to the first pin is 60ft. Assume it takes 5.0s for the ball to reach the pins after you release it, if it rolls without slipping and has a constant translational speed. Also assume the ball weights 12lb and has a diameter of 8.5 inches.
A- Calculate the rotation rate of the ball in rev/s
B- What is the total kinetic energy in pound-feet? Ignore the finger holes and treat the bowling ball as a uniform sphere
A) The rotation rate of the ball in rev/secs is ; 10.79 rev/secs
B) The total kinetic energy of in pound-feet ignoring the finger holes is : 37.6157 pound-feet
Given data :
mass of ball = 12 Ib
diameter of ball = 8.5 inches
Radius = 4.25 inches = 0.10795 m
Time = 5 secs
distance travelled = 60 ft = 18.288 m
A) Determine the rotation rate of the ballFirst step : calculate the velocity of ball
V = distance / time
= 18.288 / 5 = 3.66 m/s
Next step :
angular velocity ( w ) = V / r
= 3.66 / ( 0.10795 ) = 33.90 rad/sec
convert to rev/sec = 33.90 / π
= 10.79 rev/secs
B) Determine the total kinetic energy
given that ball rolls without slipping
Vcon = Rw
Total K.E = 1/2 IW² + 1/2 mv²
= 7/10 MR²W²
Total K.E = 7/10 * 5.44 * ( 0.10795 )² * 33.90²
= 50.99 ≈ 51 J
convert to pound-feet = 37.6157 pound-feet
Hence we can conclude that The rotation rate of the ball in rev/secs is ; 10.79 rev/secs and The total kinetic energy of in pound-feet ignoring the finger holes is : 37.6157 pound-feet.
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Females are theoretically more prone to anxiety disorders due to __________. A. their habit of overreacting to physiological changes B. the high levels of serotonin causing hormone imbalance C. increased levels of norepinephrine causing anxiety sensitivity D. stress levels, life experiences, and friends’ stresses Please select the best answer from the choices provided A B C D
Answer:
fluctuations in the levels of female reproductive hormones
Explanation:
Women are more prone to anxiety due to a variety of biological, psychological, and cultural factors. Although the exact cause is unknown, recent research suggests that fluctuations in the levels of female reproductive hormones and cycles play an important role in women's enhanced vulnerability to anxiety.
Answer: D. stress levels, life experiences, and friends’ stresses
Explanation:
All machines are less than 100% efficient. True False
Answer:
false
Explanation:
because no machines have 100% efficient
Is water Wet or Dry. Explain.
Explanation:
Water is wet, in the sense of being a liquid which flows easily, because its viscosity is low, which is because its molecules are rather loosely joined together.
Answer:
Liquid water is not itself wet, but can make other solid materials wet.
Wetness is the ability of a liquid to adhere to the surface of a solid, so when we say that something is wet, we mean that the liquid is sticking to the surface of a material.
Whether an object is wet or dry depends on a balance between cohesive and adhesive forces. Cohesive forces are attractive forces within the liquid that cause the molecules in the liquid to prefer to stick together. Cohesive forces are also responsible for surface tension. If the cohesive forces are very strong, then the liquid molecules really like to stay close together and they won't spread out on the surface of an object very much. On the contrary, adhesive forces are the attractive forces between the liquid and the surface of the material. If the adhesive forces are strong, then the liquid will try and spread out onto the surface as much as possible. So how wet a surface is depends on the balance between these two forces. If the adhesive forces (liquid-solid) are bigger than the cohesive forces (liquid-liquid), we say the material becomes wet, and the liquid tends to spread out to maximize contact with the surface. On the other hand, if the adhesive forces (liquid-solid) are smaller than the cohesive forces (liquid-liquid), we say the material is dry, and the liquid tends to bead-up into a spherical drop and tries to minimize the contact with the surface.
Water actually has pretty high cohesive forces due to hydrogen bonding, and so is not as good at wetting surfaces as some liquids such as acetone or alcohols. However, water does wet certain surfaces like glass for example. Adding detergents can make water better at wetting by lowering the cohesive forces . Water resistant materials such as Gore-tex fabric is made of material that is hydrophobic (water repellent) and so the cohesive forces within the water (liquid-liquid) are much stronger than the adhesive force (liquid-solid) and water tends to bead-up on the outside of the material and you stay dry.
In the figure below (Figure 1), the upper ball is released from rest, collides with the stationary lower ball, and sticks to it. The strings are both 50.0 cm long. The upper ball has a mass of 2.20 kg and it is initially 10.0 cm higher than the lower ball, which has a mass of 2.70 kg. Find the frequency of the motion after the collision. Find the maximum angular displacement of the motion after the collision.
(a) The frequency of the motion after the collision is 0.71 Hz.
(b) The maximum angular displacement of the motion after the collision is 16.3⁰.
Speed of the 2.2 kg ball when it collides with 2.7 kg ballThe speed of the 2.2 kg ball which was initially 10 cm higher that 2.7 kg ball is calculated as follows;
K.E = P.E
¹/₂mv² = mgh
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 0.1)
v = 1.4 m/s
Final speed of both balls after collisionThe final speed of both balls after the collision is determined from the principle of conservation of linear momentum.
Pi = Pf
m₁v₁ + m₂v₂ = vf(m₁ + m₂)
2.2(1.4) + 2.7(0) = vf(2.2 + 2.7)
3.08 = 4.9vf
vf = 3.08/4.9
vf = 0.63 m/s
Maximum displacement of the balls after the collisionP.E = K.E
[tex]mgh_f = \frac{1}{2} mv_f^2\\\\h_f = \frac{v_f^2}{2g} \\\\h_f = \frac{(0.63)^2}{2(9.8)} \\\\h_f = 0.02 \ m[/tex]
Maximum angular displacementThe maximum angular displacement of the balls after the collision is calculated as follows;
[tex]cos \theta = \frac{L - h_f}{L} \\\\cos\theta = \frac{0.5 - 0.02}{0.5} \\\\cos\theta = 0.96\\\\\theta = cos^{-1}(0.96)\\\\\theta = 16.3 \ ^0[/tex]
Frequency of the motion[tex]f = \frac{1}{2\pi} \sqrt{\frac{g}{L} } \\\\f = \frac{1}{2\pi } \sqrt{\frac{9.8}{0.5} } \\\\f = 0.71 \ Hz[/tex]
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Why is gravitational force always towards the center?
Answer:
i beleave cuz of the Earth is spherical
Explanation:
A 2 kg solid disk with a radius of 0.22 m has a tangential force of 300N applied to it.
a) What is the torque acting on the disk?
b) What is the moment of inertia of the disk?
c) What angular acceleration is produced by the torque?
d) If the disk starts from rest and the acceleration is constant for 3.0s, what is the angular velocity of the disk at the end of 3.0s?
e) Through what angle in radians has the disk rotated during this time?
(a) The torque acting on the disk is 66 Nm.
(b) The moment of inertia of the disk is 0.05 kgm².
(c) The angular acceleration is produced by the torque is 1,320 rad/s².
(d) The final angular velocity of the disk is 3,960 rad/s.
(e) The angle of rotation of the disk is 5,940 rad.
Torque acting on the diskThe torque acting on the disk is calculated as follows;
τ = Fr
τ = 300 x 0.22
τ = 66 Nm
Moment of inertiaThe moment of inertia of a solid disk is calculated as follows;
I = ¹/₂MR²
I = ¹/₂ x 2 x (0.22)²
I = 0.05 kgm²
Angular acceleration of the diskThe angular acceleration of the disk is calculated as follows;
τ = Iα
[tex]\alpha = \frac{\tau }{I} \\\\\alpha = \frac{66}{0.05} \\\\\alpha = 1,320 \ rad/s^2[/tex]
Angular velocity of the disk after 3 sωf = ωi + αt
ωf = 0 + (1320 x 3)
ωf = 3,960 rad/s
Angle of rotation of the diskωf² = ωi²+ 2αθ
(3,960)² = 0 + 2(1320)θ
θ = (3,960²) / (2 x 1320)
θ = 5,940 rad
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A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to the block. The magnitude of the block's acceleration is 1.2 m/s2. If P is increased to 5.0 N, determine the magnitude of the block's
When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
Frictional force between the block and the horizontal surfaceThe frictional force between the block and the horizontal surface is determined by applying Newton's law;
∑F = ma
F - Ff = ma
Ff = F - ma
Ff = 4 - 2(1.2)
Ff = 4 - 2.4
Ff = 1.6 N
When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;
F - Ff = ma
5 - 1.6 = 2a
3.4 = 2a
a = 3.4/2
a = 1.7 m/s²
Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
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• (-/1 Points) DETAILS OSCOLPHYS2016 9.5.WA.040.
MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER
A supertanker uses a windlass (a type of winch) like the one shown in the figure below to hoist its 21,200-kg anchor. Determine the force that must be exerted on the
outside wheel to lift the anchor at constant speed, neglecting friction and assuming the anchor is out of the water.
IN
doorg
-0.45 m
- 1.5m
The force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
Force exerted outside the wheelThe force exerted on the outside of the wheel can be determined by applying the principle of conservation of angular momentum as shown below.
∑τ = 0
Let the distance traveled by the load = 1.5 mLet the radius of the wheel or position of the force = 0.45 m∑τ = R(mg) - r(F)
rF = R(mg)
0.45F = 1.5(21,200 x 9.8)
F = 6.925 x 10⁵ N.
Thus, the force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
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The force between a pair of charges is 900 newtons. The distance between the charges is 0.01 meters. If one of the charges is 2e-10 C what is the strength of the other charge ?
Answer:
[tex] \fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}[/tex]
Explanation:
Given:
Force between pair of charges= 900 newtons
The distance between the charges = 0.01 meters
Strength of Charge first q1 = 2e-10 Coulomb
To find:
Strength of Charge second q2 = ____ Coulomb?
Solution:
We know that,
Force between two charges separate by distance r is given by the equation,
[tex]|F| = K_e \frac{q1 \cdot \: q2}{ {r}^{2} } \\ 900 =K_e \frac{(2e - 10)\cdot \: q2}{ {0.01}^{2} } \\ 900 \times {10}^{ - 4} = K_e {(2e - 10)\cdot \: q2} \\ q2 = \frac{9 \times {10}^{ - 2} }{(2e - 10) K_e} \\ \\ \fbox{We \: know \: that \: e = 2.71 } \\ substituting \: the \: value \: \\ q2 = \frac{9 \times {10}^{ - 2} }{(2 \times 2.71 - 10)K_e} \\ q2 = \frac{0.09}{ - 4.58 K_e} \\ q2 = \frac{-0.0196}{K_e}\: coulomb[/tex]
[tex] \fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}[/tex]
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you move a 25 N object 5.0 meters. how much work did you do?
Answer:
125J
Explanation:
[tex]work \: = force \: \times distance \\ = 25 \times 5 \\ = 125joules[/tex]