To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d = 4.00 m apart. A listener observes maximum constructive interference while standing in front of the loudspeakers, equidistant from both of them. The distance from the listener to the point halfway between the speakers is l = 5.00 m . One of the loudspeakers is then moved directly away from the other. Once the speaker is moved a distance r = 60.0 cm from its original position, the listener, who is not moving, observes destructive interference for the first time. Find the speed of sound v in the air if both speakers emit a tone of frequency 700 Hz .
Answers
Complete Question
The compete question is shown on the first uploaded question
Answer:
The speed is [tex] v = 350 \ m/s [/tex]
Explanation:
From the question we are told that
The distance of separation is d = 4.00 m
The distance of the listener to the center between the speakers is I = 5.00 m
The change in the distance of the speaker is by [tex]k = 60 cm = 0.6 \ m[/tex]
The frequency of both speakers is [tex]f = 700 \ Hz[/tex]
Generally the distance of the listener to the first speaker is mathematically represented as
[tex]L_1 = \sqrt{l^2 + [\frac{d}{2} ]^2}[/tex]
[tex]L_1 = \sqrt{5^2 + [\frac{4}{2} ]^2}[/tex]
[tex]L_1 = 5.39 \ m [/tex]
Generally the distance of the listener to second speaker at its new position is
[tex]L_2 = \sqrt{l^2 + [\frac{d}{2} ]^2 + k}[/tex]
[tex]L_2 = \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}[/tex]
[tex]L_2 = 5.64 \ m [/tex]
Generally the path difference between the speakers is mathematically represented as
[tex]pD = L_2 - L_1 = \frac{n * \lambda}{2}[/tex]
Here [tex]\lambda[/tex] is the wavelength which is mathematically represented as
[tex]\lambda = \frac{v}{f}[/tex]
=> [tex] L_2 - L_1 = \frac{n * \frac{v}{f}}{2}[/tex]
=> [tex] L_2 - L_1 = \frac{n * v}{2f}[/tex]
=> [tex] L_2 - L_1 = \frac{n * v}{2f}[/tex]
Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves
=> [tex] 5.64 - 5.39 = \frac{1 * v}{2*700}[/tex]
=> [tex] v = 350 \ m/s [/tex]
The speed of sound in air is 350 m/s
Since the distance between both speakers is 4 m, and the listener is standing 5 m away from halfway between them, the distance L from each loudspeaker to the listener, since the arrangement forms a right-angled triangle, using Pythagoras' theorem,
L = √[(5 m)² + (4/2 m)²]
= √[25 m² + (2 m)²]
= √[25 m² + 4 m²]
= √29 m² = 5.39 m.
Now, when one speaker is moved 60 cm = 0.6 m away from its original position, its distance from the listener is now
L' = √[(5 m)² + (4/2 + 0.6 m)²]
= √[25 m² + (2 m + 0.6 m)²]
= √[25 m² + (2.6 m)²]
= √[25 m² + 6.76 m²]
= √31.76 m²
= 5.64 m.
Now, the path difference when we first have destructive interference is
ΔL = L' - L
= 5.64 - 5.39
= 0.25
Since we have destructive interference for the first time when the speaker is moved, the path difference, ΔL = (n + 1/2)λ where λ = wavelength = v/f where v = speed of sound in air and f = frequency = 700 Hz.
Now, since we have destructive interference for the first time, n = 0.
So, ΔL = (n + 1/2)λ
ΔL = (0 + 1/2)v/f
ΔL = v/2f
Making v subject of the formula, we have
v = 2fΔL
Substituting the values of the variables into the equation, we have
v = 2fΔL
v = 2 × 700 Hz × 0.25 m
v = 350 m/s
So, the speed of sound in air is 350 m/s
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Related Questions
If you weigh 660 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 20.0 km? Take the mass of the sun to be 1.99×10^30, the gravitational constant to be G = 6.67×10^−11Nm^2/kg^2, and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2.p
Answers
Answer:
8.93*10^13 N.
Explanation:
Assuming that in this case, the weight is just the the force exerted on you by the mass of the star, due to gravity, we can apply the Universal Law of Gravitation:[tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]
where, m1 = mass of the man = 660 N / 9.81 m/s^2 = 67.3 kg, ms = mass of the star = 1.99*10^30 kg, G= Universal Constant of Gravitation, and rs= radius of the star = 10.0 km. = 10^4 m.Replacing by the values, we get:[tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]
Fg = 8.93*10^13 N.Why does the brightness of a bulb not change noticeably when you use longer copper wires to connect it to the battery?
a. All the current is used up in the bulb, so the connecting wires don't matter.
b. Very little energy is dissipated in the thick connecting wires.
c. Electric field in the connecting wires is zero, so emf = E_bulb * L_bulb.
d. Current in the connecting wires is smaller than current in the bulb.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
Answers
Answer:
Options B & E are correct
Explanation:
Looking at all the options, B & E are the correct ones.
Option B is correct because the thicker the wire per unit length, the lesser resistance it will posses and the lesser the energy that will be dissipated by the wire and in return more energy will be dissipated by the bulb.
Option E is also correct because the resistance of the copper wires is low enough to ensure that there's not much drop in voltage across the copper wires. Thus, there will not be any noticeable differences in the voltage across the bulb.
Option A is not correct because the current is not used up and thus the charge is conserved, and it will circulate just through the circuit.
Option C is not correct because although the Electric field along the wire is not zero, it is very small.
Option D is not correct because the wires and the light bulb are connected in series and as such, the current in both the wires and the light bulb will be identical.
The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is :
b. Very little energy is dissipated in the thick connecting wires.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
"Energy"The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is very little energy is dissipated in the thick connecting wires and the electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
The thicker the wire per unit length, the lesser resistance it'll posses and the lesser the vitality that will be scattered by the wire and in return more vitality will be disseminated by the bulb.
The resistance of the copper wires is low sufficient to guarantee that there's not much drop in voltage over the copper wires. Hence, there will not be any noticeable contrasts within the voltage over the bulb.
Thus, the correct answer is B and E.
Learn more about "Circuit":
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If it takes you 5 minutes to dry your hair using a 1200-W hairdryer plugged into a 120-V power outlet, how many Coulombs of charge pass through your hair dryer
Answers
Answer:
The charge pass through your hair dryer is 3000 C.
Explanation:
Given that,
Power = 1200 W
Voltage = 120 V
Flow time = 5 min
We need to calculate the current
Using formula of power
[tex]P=VI[/tex]
[tex]I=\dfrac{P}{V}[/tex]
Put the value into the formula
[tex]I=\dfrac{1200}{120}[/tex]
[tex]I=10\ A[/tex]
We need to calculate the charge pass through your hair dryer
Using formula of current
[tex]I=\dfrac{Q}{t}[/tex]
[tex]Q=It[/tex]
Put the value into the formula
[tex]Q=10\times5\times60[/tex]
[tex]Q=3000\ C[/tex]
Hence, The charge pass through your hair dryer is 3000 C.
While making some observations at the top of the 66 m tall Astronomy tower, Ron
accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?
Answers
Analysing the question:
Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s
We are given:
height of the tower (h) = 66 m
mass of the stone (m) = 0.5 kg
initial velocity of the stone (u) = 0 m/s
time taken by the stone to reach the ground (t) = t seconds
acceleration due to gravity = 10 m/s²
** Neglecting air resistance**
Finding the time taken by the stone to reach the ground:
from the second equation of motion
h = ut + 1/2at²
replacing the variables
66 = (0)(t) + 1/2 (10)(t)²
66 = 5t²
t² = 13.2
t = 3.6 seconds
I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds
but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved
7. A 1,500-N force is applied to a 1,000-kg car. What is the car's acceleration?
Answers
Answer:
1.5m/s^2
Explanation:
Answer:
1.5 m/s2. accerelation =force ÷mass
A soccer player kicking a ball; the ball soaring through the air and landing on the ground
Answers
Jumping on a trampoline cause you to fly up in the air. What type of newton’s law is it ?
Answers
Answer:
The Third law
Explanation:
For every action there is an equal and opposite reaction.
Answer:
First Law
Explanation:
An object at rest (not moving) will stay at rest unless an unbalanced force acts on it.
An object in motion will stay in motion (in a straight line and at a constant speed) unless an unbalanced force acts on it.
You jump down on a trampoline and fly up in the air as a result.
I am a cell. I am long and thin. I reach all the way from the brain
to the tip of a finger. I have a special coat of fat that helps me do
my job. My job is to send electrical signals from one part of the
body to another.
Answers
Answer:
Neurons
Explanation:
We humans have a nervous system that coordinates our behavior and transmits signals between different parts of our body.
Now, this nervous system contains a lot of nerve cells which we call Neurons. These Neurons have a cell like body and their job is to transmit signals from one part of our body to another.
Thus, the cell is called Neurons.
A coin rests on a record 0.15 m from its center. The record turns on a turntable that rotates at variable speed. The coefficient of static friction between the coin and the record is 0.30.
Required:
What is the maximum coin speed at which it does not slip?
Answers
Answer:
0.66m/sExplanation:
We are expected to solve for the velocity with no slip condition
we know that the expression that relate coefficient of friction and velocity is given as
μs = v^2/rg
Given
coefficient of friction μs = 0.3
radius r= 0.15
assume g=9.81m/s^2
substituting into the expression we have
0.3= v^2/0.15*9.81
v^2=0.3*0.15*9.81
v^2=0.44145
v=√0.44145
v=0.66
therefore the velocity is 0.66m/s
A small compass is held horizontally, the center of its needle has a distance of 0.270 m directly north
of a long wire that is perpendicular to the Earth's surface. When there is no current in the wire, the
compass needle points due north, which is the direction of the horizontal component of the Earth's
magnetic field at that location. This component is parallel to the Earth's surface. When the current in
the wire is 26.3 A, the needle points 22.9∘ east of north.
(a) Does the current in the wire flow toward or away from the Earth's surface? ( 2 marks)
(b) What is the magnitude of the horizontal component of the Earth's magnetic field at the location of
the compass? (3 marks)
Answers
Answer:
Explanation:
The needle is showing north south direction . when current starts flowing in the wire which is held vertical to the ground , it deflects towards east .
a )
Therefore a magnetic field towards east has been created . It is possible only if current flows towards the surface in the vertical wire .
b )
magnetic field created at the magnetic needle B = 10⁻⁷ x 2I / d where I is current and d is distance .
B = 10⁻⁷ x 2 x 26.3 / .27
= 194.81 x 10⁻⁷ T
angle of deflection of solenoid = 22.9°
Tan 22.9 = B /H
.422 = 194.81 x 10⁻⁷ / H
H = 461.63 x 10⁻⁷ T
= .46 x 10⁻⁴ T .
A) The current in the wire flows towards the Earth's surface
B) The magnitude of the horizontal component of the Earth's magnetic field is : 0.46 x 10⁻⁴ T
A) The compass needle held horizontally points in a North-south direction of the earth and also deflects eastwards when current is allowed to flow through it. The deflection of the needle indicates the presence/generation of a magnetic field on the earth surface. which is facilitated by the flow of the current in the wire towards the Earth's surface
B) Determine The magnitude of the horizontal component of the Earth's magnetic field
B ( magnetic field ) = 10⁻⁷ * 2I / d ---- ( 1 )
where : l = 26.3 A, d = 0.27 m
Back to equation ( 1 )
B = 10⁻⁷ * 2 * 26.3 / 0.27
= 194.81 * 10⁻⁷ T
Final step : Calculate the magnitude of horizontal component ( H )
Tan ∅ = B / H ---- ( 2 )
where : ∅ ( angle of deflection ) = 22.9°
∴ H = B / Tan ( 22.9° )
= ( 194.81 * 10⁻⁷ ) / 0.422
= 0.46 x 10⁻⁴ T
Hence we can conclude that The current in the wire flows towards the Earth's surface and The magnitude of the horizontal component of the Earth's magnetic field is : 0.46 x 10⁻⁴ T
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An electric bulb rated 100 W, 100 V has to be
operated aross 141.4 V, 50 Hz A.C. supply. The
capacitance of the capacitor which has to be
connected in series with bulb so that bulb will
glow with full intensity is [NCERT Pg. 251]
Answers
Answer:
The capacitance of the capacitor is 31.84 μF.
Explanation:
Given;
power rating of the bulb, P = 100 W
voltage rating of the bulb, Vr = 100 V
operating voltage of the bulb, V= 141.4 V
frequency of the AC = 50 Hz
P = IV = 100 W
V = 100 V
I =
Ic = 1 A
The voltage across the capacitor is given by;
[tex]V_c = \sqrt{V^2 - V_R^2} \\\\V_c = \sqrt{141.4^2 - 100^2} \\\\V_c =99.97 \ V[/tex]
[tex]V_c = I_cX_c\\\\V_c = I_C* \frac{1}{2\pi fC}\\\\ 99.97 = 1 * \frac{1}{2\pi *50 *C}\\\\ C=\frac{1}{2\pi *50*99.97}\\\\ C = 31.84*10^{-6} \ F\\\\C = 31.84 \ \mu F[/tex]
Therefore, the capacitance of the capacitor is 31.84 μF.
help me get the answer in Physical Science.
Answers
Answer:
lithium
Explanation:
I took physical science 2 years ago and passed with an A
Based on the information in the table, which elements are most likely in the same periods of the periodic table?
Answers
Answer:
Just to help, periods on the periodic table are those running horizontally from left to right
Answer:
The answer is A.Boron and carbon are likely together in one period because they have very close atomic numbers, while gallium and germanium are likely together in another period because they have very close atomic numbers.
Explanation:
just took test
. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point? What is your displacement vector? What is the direction of your displacement? Assume the +x-axis is to the east.
Answers
Answer:
Explanation:
The total distance is how far you walk from the starting point.
Distance through west = 18.0m
Distance through north = 25.0m
Total distance covered = 18.0+25.0m
Total distance covered = 43.0m
This means that I am 43.0m from the starting point
Displacement is the distance covered in a specified direction. The displacement will be gotten using the Pythagoras theorem as shown:
[tex]d^2 = 25^2 + 18^2\\d^2 = 625+324\\d^2 = 949\\d = \sqrt{949}\\ d = 30.81m[/tex]
The direction of your displacement is 30.81m
Direction is gotten according to the formula;
[tex]\theta = tan ^{-1}{\frac{y}{x} }\\\theta = tan ^{-1}{\frac{25}{-18} }\\\theta = tan ^{-1}-1.3889}\\\theta = -60.27^0\\\theta = 180-60.27\\\theta = 119.7^0[/tex]
Note that the direction to the west is negative, that is why the x is -18.0m
The distance from the starting point is 43 m, the displacement vector is 30.81 m and the direction of the displacement is 119.7 degrees.
Given-
Distance travel through the west is 18 m.
Distance travel through the north is 25 m.
Distance from starting point-
To know the total distance, add both the covered distance. Thus total distance x is,
[tex]x=18+25[/tex]
[tex]x=43[/tex]
Hence, the distance from the starting point is 43 m.
The displacement vector-
Displacement is calculated as the shortest distance between starting and final point. This shortest distance can be calculated using the Pythagoras theorem which states that in a right-angled triangle, the square of the hypotenuse [tex]d[/tex] is equal to the sum of the squares of the other two sides. Therefore,
[tex]d^2=18^2+25^2[/tex]
[tex]d^2=324+625[/tex]
[tex]d^2=949[/tex]
[tex]d=\sqrt{949}[/tex]
[tex]d=30.81[/tex]
The displacement vector is 30.81 m.
The Direction of displacement-The direction of displacement [tex]\theta[/tex] with these two sides can be calculated with the formula,
[tex]\theta=tan^{-1}\dfrac{25}{-18}[/tex]
Here due to the west direction(opposite side), the sign is taken negatively.
[tex]\theta=tan^{-1}(-1.389)[/tex]
[tex]\theta=-60.27^o[/tex]
For the other quarter,
[tex]\theta=180-60.27=119.7^o[/tex]
Hence, the distance from the starting point is 43 m, the displacement vector is 30.81 m and the direction of the displacement is 119.7 degrees.
For more about the displacement, follow the link below-
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Question 1-1: In each case, lifting or pushing, why must you exert a force to keep the object moving at a constant velocity?
Answers
Answer:
We must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
Explanation:
LIFTING:
When an object is lifted, we first need to overcome the force exerted on it by the field of gravity. Due to this force, which is also called the weight of object, we must apply a force on the object to keep it moving at constant speed, otherwise the gravity force will cause the object to slow down and eventually fall back on ground.
PUSHING:
When pushing an object the person must apply the force to first overcome the frictional force. The frictional force acts in opposite direction of motion. Thus, to move the object at constant speed we must apply force to it.
Hence, we must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
the diagram shows a contour map. letter a through k are reference points on the map. which points are located at the same elevation above sea level?
Answers
Answer:
K and I
Explanation:
Contour maps use lines that represent spaces in a map that have the same elevation, this means that all the lines should be continuous and closed, in this case, we are not able to see the full extent of most of the lines, but since the points are located in different lines we can assume that they are at different heights, so since only point K and point I are on the same line, we know that these two points are at the same height.
If the shoe has less mass, it will experience _______________ (more, less, the same) friction as it would with more mass.
Answers
You release a ball from rest at the top of a ramp. 6 s later it is moving at 4.0
m/s. What is the acceleration? (in meters per second squared) *
Your answer
Answers
[tex]a = \frac{vf - vi}{t} [/tex]
here initial velocity vi=0 as ball release from rest
the final velocity is vf=4.0
time is t=6
so putting all these values in above equation
[tex]a = \frac{ 4.0- 0}{6} [/tex]
[tex]a = 0.6667m \s {}^{2} [/tex]
what is the force produced on a body of 30kg mass when a body moving with the velocity of 26km/hr is acceleted to gain the velocity of 54 km/hr in 4 sec
Answers
Answer:
F = 58.35 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that force is equal to the product of mass by acceleration. But first we must use the following equation of kinematics.
We have to convert speeds from kilometers per hour to meters per second
[tex]\frac{26km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=\frac{7.22m}{s} \\\frac{54km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex]
[tex]v_{f}=v_{o}+(a*t) \\[/tex]
where:
Vf = final velocity = 15 [m/s]
Vi = initial velocity = 7.22 [m/s]
a = acceleration [m/s^2]
t = time = 4 [s]
Note: the positive sign of the above equation is because the car increases its speed
15 = 7.22 + (a*4)
a = 1.945 [m/s^2]
Now we can use the Newton's second law:
F = m*a
F = 30*1.945
F = 58.35 [N]
A daring stunt woman sitting on a tree limb
wishes to drop vertically onto a horse gallop-
ing under the tree. The constant speed of the
horse is 6.8 m/s, and the woman is initially
1.91 m above the level of the saddle.
How long is she in the air? The acceleration
of gravity is 9.8 m/s.
Answer in units of s.
Answers
Answer:
she is in the air for approximately 0.62 seconds
Explanation:
We want to find the time for a free fall under the acceleration of gravity, covering a distance of 1.91 m, and considering that the woman doesn't impart initial velocity in the vertical direction. So we use the kinematic equation:
[tex]d=v_i\,t+ \frac{g}{2} \,t^21.91 = 0 +4.9\, t^2\\t^2=1.91/4.9\\t=\sqrt{1.91/4.9} \\t\approx 0.624\,\,sec[/tex]
Then she is in the air for approximately 0.62 seconds
21. Prediction: If you were to measure the current at points A, B and C, how do you think the values would compare? Why? 22. Prediction: If you were to measure the potential differences across these bulbs (what the voltmeter measures) how do you think the values will compare to each other and to the potential difference across the battery pack or the power supply? Why?
Answers
Answer:
hello your question is incomplete attached below is the complete question
21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
Explanation:
The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
An FM radio station, 20 miles away, broadcast at a 93.4 MHz frequency(a) What is the wavelength of the radio wave associated with this signal ?(b) How long does it take for the signal to reach your radio from the station ?
Answers
Answer:
(a) Wavelength = 3.21 m (b) Time = [tex]1.07\times 10^{-4}\ s[/tex]
Explanation:
Given that,
The frequency of FM radio station, f = 93.4 MHz
(a) We need to find the wavelength of the radio wave associated with this signal. The relation between wavelength and frequency is given by :
[tex]c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{93.4\times 10^6}\\\\\lambda=3.21\ m[/tex]
(b) It is given that, an FM radio station, 20 miles away. Let t is time taken for signal to reach your radio from the station. So,
[tex]t=\dfrac{d}{c}\\\\t=\dfrac{20\times 1609.34}{3\times 10^8}\\\\t=1.07\times 10^{-4}\ s[/tex]
Hence, this is the required solution.
A plane travelling at 100 m/s accelerates at 5 m/s² for a distance of 125 m. What is the final velocity of the plane?
Answers
Analyzing the question:
We are given:
initial velocity (u) = 100 m/s
final velocity (v) = v m/s
distance (s) = 125 m
acceleration (a) = 5 m/s²
Solving for Final Velocity (v):
from the third equation of motion:
v² - u² = 2as
v² - (100)² = 2(5)(125)
v² - 10000 = 1250
v² = 1250 + 10000
v² = 11250
v = 106.06 m/s
You have a friend who reports that he falls asleep easily around 11 PM but then awakens for about an hour most nights around two or 3 AM he seems near exhausted what would be the traditional exclamation for his problem how much the information contribute by anthropologists change this view? Give the anthropological view what recommendations would you make your friend?
Answers
It is highly recommended that this friend who is suffering from insomnia visits the doctor and eat foods rich in serotonin.
What is insomnia?Insomnia is a medical condition in which an individual is unable to sleep or has short periods of interrupted sleep.
A friend who falls asleep easily and then has difficulty going to sleep is probably suffering from insomnia.
It is recommended that this friend who is suffering from insomnia visits the doctor and eat foods rich in serotonin.
Learn more about insomnia at: https://brainly.com/question/816019
During a thunderstorm the electric field at a certain point in the earth's atmosphere is 1.07 105 N/C, directed upward. Find the acceleration of a small piece of ice of mass 1.08 10-4 g, carrying a charge of 1.05 10-11 C.
Answers
Answer:
The acceleration of a small piece of ice is 10.40 m/s².
Explanation:
The electric force is given by:
[tex]F = Eq[/tex]
Where:
E is the electric field = 1.07x10⁵ N/C
q is the charge = 1.05x10⁻¹¹ C
The electric force is equal to Newton's second law:
[tex] Eq = ma [/tex]
Where:
m is the mass = 1.08x10⁻⁴ g = 1.08x10⁻⁷ kg
a is the acceleration
Hence, the acceleration is:
[tex] a = \frac{Eq}{m} = \frac{1.07 \cdot 10^{5} N/C*1.05 \cdot 10^{-11} C}{1.08 \cdot 10^{-7} kg} = 10.40 m/s^{2} [/tex]
Therefore, the acceleration of a small piece of ice is 10.40 m/s².
I hope it helps you!
How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential difference, it is connected to the power supply and resistor and onehalf the characteristic time has passed (i.e. t= T(tau)/2)?
Answers
Answer:
The voltage is [tex]V = 0.993V_b[/tex]
Explanation:
From the question we are told that
The time that has passed is [tex]t = \frac{\tau}{2}[/tex]
Here [tex]\tau[/tex] is know as the time constant
The voltage of the power source is [tex]V_b[/tex]
Generally the voltage equation for charging a capacitor is mathematically represented as
[tex]V = V_b [1 - e^{- \frac{t}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\tau}{2\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{1}{2} }][/tex]
=> [tex]V = 0.993V_b[/tex]
the peripheral nervous system is responsible for both sending and receiving signals to and from the brain
Answers
Answer:
its true trust me
Explanation:
Answer: true
Explanation: edge
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A sharpshooter fires a 0.22 caliber rifle horizontally at 100 m/s at a target 75m away. How far does the
bullet drop by the time it reaches the target?
Answers
This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.
The bullet drops "2.76 m" by the time it reaches the target.
First, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here to find the total time to reach the target:
[tex]s = vt\\\\t = \frac{s}{v}[/tex]
where,
s = distance = 75 m
v = velocity = 100 m/s
t = time = ?
Therefore,
[tex]t = \frac{75\ m}{100\ m/s}[/tex]
t = 0.75 s
Now, we will analyze the vertical motion of the bullet. We will use the second equation of motion in the vertical direction to find the height dropped by the bullet.
[tex]h = v_it+\frac{1}{2}gt^2[/tex]
where,
h = height dropped = ?
vi = initial vertical speed = 0 m/s
t = time interval = 0.75 s
g = acceleration due to gravity = 9.81 m/s²
therefore,
[tex]h = (0\ m/s)(0.75\ s)+\frac{1}{2}(9.81\ m/s^2)(0.75\ s)^2[/tex]
h = 2.76 m
Learn more about equations of motion here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion in the horizontal and vertical directions.
It takes 525 J of work to compress a spring 25 cm. What is the force constant of the spring (in kN/m)?
Answers
Answer:
1.680kN/m
Explanation:
Work done by the spring is expressed as shown:
[tex]W = \frac{1}{2}ke^2[/tex] where:
k is the spring constant
e is the extension
Given
W = 525Joules
extension = 25cm = 0.25m
Substitute into the formula:
[tex]525 = \frac{1}{2}k(0.25)^{2} \\525 = \frac{0.0625k}{2}\\ 525 = 0.03125k\\k = \frac{525}{0.3125}\\k = 1680N/m\\k = 1.680kN/m[/tex]
Hence the force constant of the spring is 1.680kN/m
For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same. Describe how they are different.
Answers
Answer:
he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.
Explanation:
In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values where some natural frequency of the system coincides with the excitation frequency.
In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.
From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.