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A uniform line charge with 16.5 nC total charge creates an electric field that weakens with distance from the line. The exact electric field strength can be calculated using the formula for an infinitely long line charge, while approximating the line as a point charge underestimates the true field strength.
Solving for the Electric Field of a Uniform Line Charge
(a) Total Charge:
The total charge (Q) of a line charge can be found by integrating the linear charge density (λ) over the length (L) of the charge distribution:
Q = ∫λ dx (from x = 0 to x = 5.0 m)
Here, λ = 3.3 nC/m and L = 5.0 m.
Q = (3.3 nC/m) * (5.0 m) = 16.5 nC
Therefore, the total charge is 16.5 nC.
(b) Electric Field at x = 6 m (Exact):
Because the line charge is infinitely long, we can apply the electric field formula:
E = (λ / (2πε₀)) * ln(2a / b)
Plugging in the values:
E = (3.3 x 10⁻⁹ C/m) / (2π * 8.854 x 10⁻¹² C²/N∙m²) * ln(2 * 6 m / 0)
E ≈ 47.7 N/C (rounded to two significant figures)
(c) Electric Field at x = 10.0 m (Exact):
Following the same method as part (b):
E ≈ (3.3 x 10⁻⁹ C/m) / (2π * 8.854 x 10⁻¹² C²/N∙m²) * ln(2 * 10.0 m / 0)
E ≈ 33.1 N/C (rounded to two significant figures)
(d) Electric Field at x = 270 m (Exact):
Using the same formula:
E ≈ (3.3 x 10⁻⁹ C/m) / (2π * 8.854 x 10⁻¹² C²/N∙m²) * ln(2 * 270 m / 0)
E ≈ 0.012 N/C (rounded to three significant figures)
(e) Electric Field at x = 270 m (Approximation):
Assuming the charge is a point charge at x = 2.5 m (center of the line charge):
E ≈ k * Q / (x - 2.5 m)²
(f) Ratio of Approximation to Exact Result:
Ratio = (Approximate Electric Field) / (Exact Electric Field)
Ratio ≈ (2.02 x 10⁻⁴ N/C) / (0.012 N/C) ≈ 0.0017
(g) Comparison of Results:
Since the ratio is less than 1, the approximate result (2.02 x 10⁻⁴ N/C) is smaller than the exact result (0.012 N/C). This is reasonable because approximating the finite line charge as a point charge weakens the effect of the charge, leading to a lower electric field value.
A sailor drops a wrench from the top of a sailboat's vertical mast while the boat is moving rapidly and steadily straight forward. Where will the wrench hit the deck?
(A) ahead of the base of the mast
(B) at the base of the mast
(C)behind the base of the mast
(D)on the windward side of the base of the mast
(E)None of the above choices
Answer:
B
Explanation:
The sailor, the boat and the wrench are all moving at he same constant rate, so the wrench will appear to fall straight down. This due to that fact there is no relative motion among them and all are at rest w.r.t to one another. Hence the correct answer would be B.
The correct option is (B). The wrench will fall straight down relative to the moving boat and will land directly at the base of the mast.
To determine where the wrench will hit the deck, let's analyze the motion of the wrench.
1. Horizontal Motion:
When the wrench is dropped from the top of the mast, it has the same horizontal velocity as the sailboat. This is because, in the absence of air resistance and assuming no external horizontal forces act on the wrench, it retains the horizontal component of its velocity that it had while it was still in the sailor's hand. Therefore, as the wrench falls, it will continue to move forward with the same horizontal velocity as the boat.2. Vertical Motion*: The wrench will accelerate downwards due to gravity.
Since the wrench maintains its horizontal velocity and is only influenced vertically by gravity, it will fall straight down from the perspective of someone moving with the boat. From the perspective of an observer on the boat, the wrench will fall directly downwards.
Calculate the average linear momentum of a particle described by the following wavefunctions: (a) eikx, (b) cos kx, (c) e−ax2 , where in each one x ranges from −[infinity] to +[infinity].
Answer:
a) p=0, b) p=0, c) p= ∞
Explanation:
In quantum mechanics the moment operator is given by
p = - i h’ d φ / dx
h’= h / 2π
We apply this equation to the given wave functions
a) φ = [tex]e^{ikx}[/tex]
.d φ dx = i k [tex]e^{ikx}[/tex]
We replace
p = h’ k [tex]e^{ikx}[/tex]
i i = -1
The exponential is a sine and cosine function, so its measured value is zero, so the average moment is zero
p = 0
b) φ = cos kx
p = h’ k sen kx
The average sine function is zero,
p = 0
c) φ = [tex]e^{-ax^{2} }[/tex]
d φ / dx = -a 2x [tex]e^{-ax^{2} }[/tex]
.p = i a g ’2x [tex]e^{-ax^{2} }[/tex]
The average moment is
p = (p₂ + p₁) / 2
p = i a h ’(-∞ + ∞)
p = ∞
How many times a minute does a boat bob up and down on ocean waves that have a wavelength of 40.0 m and a propagation speed of 5.00 m/s?
Answer:
n = 7.5 times/minute
Explanation:
Given that,
Wavelength of the ocean wave, [tex]\lambda=40\ m[/tex]
The speed of the ocean wave, v = 5 m/s
To find,
Number of times a minute does a boat bob up and down on ocean waves.
Solution,
The relation between the speed of wave, wavelength and frequency is given by :
[tex]v=f\times \lambda[/tex]
[tex]f=\dfrac{v}{\lambda}[/tex]
[tex]f=\dfrac{5\ m/s}{40\ m}[/tex]
f = 0.125 Hz
The number of times per minute the bob moves up and down is given by :
[tex]n=f\times t[/tex]
[tex]n=0.125\times 60[/tex]
n = 7.5 times/minute
So, its will move up and down in 7.5 times/minute. Therefore, this is the required solution.
Taking into account the definition of wavelength, frecuency and propagation speed, the number of times per minute the bob moves up and down is 7.5 times per minute.
WavelengthWavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).
FrequencyFrequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
Propagation speedFinally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.
The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:
v = f× λ
Amount of times in a minute that a boat bob up and down on ocean wavesIn this case, you know:
v= 5 [tex]\frac{m}{s^{2} }[/tex]f= ?λ= 40 mReplacing in the definition of propagation speed:
5 [tex]\frac{m}{s^{2} }[/tex]= f× 40 m
Solving:
f= 5 [tex]\frac{m}{s^{2} }[/tex]÷ 40 m
f= 0.125 Hz= 0.125 [tex]\frac{1}{seconds}[/tex]
Then, a boat bob up and down on ocean waves 0.125 times in a second.
So, the number of times per minute the bob moves up and down is given by:
n= f× time
n= 0.125 Hz× 60 minutes in 1 second
n=7.5 times per minute
Finally, the number of times per minute the bob moves up and down is 7.5 times per minute.
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Starting fom rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far does it go in this time?
Answer:
(A) [tex]a=2.0.37m/sec^2[/tex]
(B) s = 146.664 m
Explanation:
We have given car starts from the rest so initial velocity u = 0 m /sec
Final velocity v = 88 km/hr
We know that 1 km = 1000 m
And 1 hour = 3600 sec
So [tex]88km/hr=88\times \frac{1000}{3600}=24.444m/sec[/tex]
Time is given t = 12 sec
(A) From first equation of motion v = u+at
So [tex]24.444=0+a\times 12[/tex]
[tex]a=2.0.37m/sec^2[/tex]
So acceleration of the car will be [tex]a=2.0.37m/sec^2[/tex]
(b) From third equation of motion [tex]v^2=u^2+2as[/tex]
So [tex]24.444^2=0^2+2\times 2.037\times s[/tex]
s = 146.664 m
Distance traveled by the car in this interval will be 146.664 m
A magnetic field of magnitude 1.30x10-3 T is measured a distance of 0.03 m from a long straight wire. What is the current through the wire?
Final answer:
The current through the wire is 1.95 A.
Explanation:
To find the current through the wire, we can use Ampere's law. Ampere's law states that the magnetic field around a long straight wire is directly proportional to the current through the wire and inversely proportional to the distance from the wire.
So, we can use the equation B = μ0 * I / (2π * r), where B is the magnetic field, μ0 is the magnetic constant, I is the current, and r is the distance from the wire.
Plugging in the given values, we have 1.30x10-3 T = (4πx10-7 T*m/A) * I / (2π * 0.03 m). Solving for I, we get I = 1.30x10-3 * (2*0.03)/(4x10-7) = 1.95 A.
Why isn't Coulomb's law valid for dielectric objects, even if they are spherically symmetrical?
Answer:
Explanation:
The "traditional" form of Coulomb's law, explicitly the force between two point charges. To establish a similar relationship, you can use the integral form for a continuous charge distribution and calculate the field strength at a given point.
In the case of moving charges, we are in presence of a current, which generates magnetic effects that in turn exert force on moving charges, therefore, no longer can consider only the electrostatic force.
Raindrops acquire an electric charge as they fall. Suppose a 2.4-mm-diameter drop has a charge of +18 pC, fairly typical values.
What is the potential at the surface of the raindrop?
To solve this problem we will apply the concepts related to the potential, defined from the Coulomb laws for which it is defined as the product between the Coulomb constant and the load, over the distance that separates the two objects. Mathematically this is
[tex]V = \frac{kq}{r}[/tex]
k = Coulomb's constant
q = Charge
r = Distance between them
[tex]q = 18 pC \rightarrow q = 1.8*10^-11 C[/tex]
[tex]d = 2.4mm \rightarrow r = 1.2 mm = 1.2*10^-3 m[/tex]
Replacing,
[tex]V = \frac{kq}{r}[/tex]
[tex]V = \frac{ (9*10^9)*(1.8*10^{-11})}{(1.2*10^{-3})}[/tex]
[tex]V = 135 V[/tex]
Therefore the potential at the surface of the raindrop is 135 V
A juggler throws a bowling pin straight up in the air. After the pin leaves his hand and while it is in the air, which statement is true?(a) The velocity of the pin is always in the same direction as its acceleration.(b) The velocity of the pin is never in the same direction as its acceleration.(c) The acceleration of the pin is zero.(d) The velocity of the pin is opposite its acceleration on the way up. (e) The velocity of the pin is in the same direction as its acceleration on the way up.
Answer:
The velocity of the pin is opposite its acceleration on the way up.
(d) option is correct.
Explanation:
when the juggler throws a bowling pin straight in the air, the acceleration working on the pin is in the downward direction due to the gravitational force of the earth.
According to Newton's Universal Law of Gravitation
''The gravitational force is a force that attracts any objects with mass''
Final answer:
The acceleration due to gravity is always downward, so when the pin is thrown up, its velocity is opposite to its acceleration. At the peak, the velocity is zero and then aligns with the direction of gravity on the descent.
Explanation:
When a juggler throws a bowling pin straight up in the air, the acceleration due to gravity is always directed towards the ground, which means it is downwards. As the pin moves upwards, its velocity is in the opposite direction of the acceleration. At the peak of its motion, the velocity of the pin is zero, after which it starts to fall back down, and its velocity is then in the same direction as acceleration. Thus, the correct statement is (d) The velocity of the pin is opposite its acceleration on the way up.
A circular ring of charge with radius b has total charge q uniformly distributed around it.
What is the magnitude of the electric field at the center of the ring?
a) 0
b) kₑq/b²
c) kₑq²/b²
d) kₑq²/b
e) None of these
Answer:
Option A is correct (0) ( The electric field at the center of circular charged ring is zero)
Explanation:
Option A is correct (0) ( The electric field at the center of circular charged ring is zero)
The reason why electric field at the center of circular charged ring is zero because the fields at the center of the circular ring due to any point are cancelled by electric fields of another point which is at 180 degree to that point i.e opposite to that charge.
Answer: a) 0
The electric field from opposite directions at the centre of the circular ring cancel each other out and give a resultant of zero
Explanation:
Given that the ring is perfectly circular with radius b which has a uniform distributed charge q around it.
The electric field experienced at the centre of the ring from opposite directions are given as
E1 = kₑq/b²
E2 = -kₑq/b²
It experience the two electric field E1 and E2 from opposite directions at the centre. So the resultant electric field is given by:
E = E1 + E2
E = kₑq/b² - kₑq/b²
E = 0
The electric field from opposite directions at the centre of the circular ring cancel each other out and give a resultant of zero
An astronaut is on a 100-m lifeline outside a spaceship, circling the ship with an angular speed of
0.100 rad/s. How far inward can she be pulled before the centripetal acceleration reaches 5g = 49 m/s2?
Answer:
D = 72.68 m
Explanation:
given,
R = 100 m
angular speed = 0.1 rad/s
distance she can be pulled before the centripetal acceleration reaches 5g = 49 m/s².
using conservation of Angular momentum
[tex]I_i\omega_i= I_f\omega_f[/tex]
[tex]mr_i^2\omega_i=m r_f^2\omega_f[/tex]
[tex]\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i[/tex]
[tex]\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i[/tex]
we know,
centripetal acceleration
[tex]a = \dfrac{v^2}{r}[/tex]
v = r ω
[tex]a =\omega_f^2 r_f [/tex]
[tex]a =(\dfrac{r_i^2}{r_f^2}\times \omega_i)^2 r_f [/tex]
[tex]a =\dfrac{r_i^4\times \omega_i^2}{r_f^3}[/tex]
[tex]r_f^3=\dfrac{100^4\times 0.1^2}{5\times 9.8}[/tex]
[tex]r_f^3=20408.1632[/tex]
[tex]r_f = 27.32\ m[/tex]
distance she has reached inward is equal to
D = 100 - 27.23
D = 72.68 m
According to one set of measurements, the tensile strength of hair is 196 MPa , which produces a maximum strain of 0.380 in the hair. The thickness of hair varies considerably, but let's use a diameter of 50.0 μm
Part A
What is the magnitude of the force giving this tensile stress?
F = ? N
Part B
If the length of a strand of the hair is 12.0 cm at its breaking point, what was its unstressed length? original length = ?cm
Answer:
(a). The magnitude of the force is 0.38416 N.
(b). The original length is 0.0869 m.
Explanation:
Given that,
Tensile strength = 196 MPa
Maximum strain = 0.380
Diameter = 50.0 μm
Length = 12.0 cm
We need to calculate the area
Using formula of area
[tex]A=\dfrac{\pi}{4}\times d^2[/tex]
Put the value into the formula
[tex]A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2[/tex]
[tex]A=1.96\times10^{-9}\ m^2[/tex]
We need to calculate the magnitude of the force
Using formula of force
[tex]F=\sigma A[/tex]
Put the value into the formula
[tex]F=196\times10^{6}\times1.96\times10^{-9}[/tex]
[tex]F=0.38416\ N[/tex]
(b). If the length of a strand of the hair is 12.0 cm at its breaking point
We need to calculate the unstressed length
Using formula of strain
[tex]strain=\dfrac{\Delta l}{l_{0}}[/tex]
[tex]\Delta l=strain\times l_{0}[/tex]
Put the value into the formula
[tex]\Delta l=0.380\times l_{0}[/tex]
Length after expansion is 12 cm
We need to calculate the original length
Using formula of length
[tex]l=l_{0}+\Delta l[/tex]
Put the value into the formula
[tex]I=l_{0}+0.380\times l_{0}[/tex]
[tex]l=1.38l_{0}[/tex]
[tex]l_{0}=\dfrac{l}{1.38}[/tex]
[tex]l_{0}=\dfrac{12\times10^{-2}}{1.38}[/tex]
[tex]l_{0}=0.0869\ m[/tex]
Hence, (a). The magnitude of the force is 0.38416 N.
(b). The original length is 0.0869 m.
Maximum current problem. If the current on your power supply exceeds 500 mA it can damage the supply. Suppose the supply is set for 37 V. What is the smallest resistance you can measure?
To solve this problem we will apply Ohm's law. The law establishes that the potential difference V that we apply between the ends of a given conductor is proportional to the intensity of the current I flowing through the said conductor. Ohm completed the law by introducing the notion of electrical resistance R. Mathematically it can be described as
[tex]V = IR \rightarrow R = \frac{V}{I}[/tex]
Our values are
[tex]I = 500mA = 0.5A[/tex]
[tex]V = 37V[/tex]
Replacing,
[tex]R = \frac{V}{I}[/tex]
[tex]R = \frac{37}{0.5}[/tex]
[tex]R = 74 \Omega[/tex]
Therefore the smallest resistance you can measure is [tex]74 \Omega[/tex]
For the merry-go-round problem, do the magnitudes of the position, velocity, and acceleration vectors change with time?
Answer:
No
Explanation:
Although the direction of position, velocity or acceleration of an object in marry-go-round problem changes continuously,however the magnitude of the position, velocity and acceleration do remains the same. Marry-go-round is nothing but a machine found in fairs that turn round in circular motion. So, the laws of circular motion are applicable in it.
In his famous 1909 experiment that demonstrated quantization of electric charge, R. A. Millikan suspended small oil drops in an electric field. With a field strength 0f 20 MN/C, what mass drop can be suspended when the drop carries a net charge of 10 elementary charges?
Answer:
[tex]3.26198\times 10^{-12}\ kg[/tex]
Explanation:
E = Electric field = 20 MN/C
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
g = Acceleration due to gravity = 9.81 m/s²
m = Mass of drop
The electrical force will balance the weight
[tex]Eq=mg\\\Rightarrow 20\times 10^{6}\times 10\times 1.6\times 10^{-19}=m\times 9.81\\\Rightarrow m=\dfrac{20\times 10^{6}\times 10\times 1.6\times 10^{-19}}{9.81}\\\Rightarrow m=3.26198\times 10^{-12}\ kg[/tex]
The mass that can be suspended is [tex]3.26198\times 10^{-12}\ kg[/tex]
From charge to mass ratio, the mass of the charges is 3.2 × 10^-12 Kg.
The charge to mass ratio experiment was used by R. A. Millikan to accurately determine the charge to mass ratio of the electron. We have the following information from the question;
Field strength = 20 MN/C
Number of charges = 10
Now;
The magnitude of electric field strength is obtained from;
E = F/q
F = Eq
Where;
F = electric force
E = electric field intensity
q = magnitude of charge
F = 10 × 20 × 10^6 × 1.6 × 10^-19 = 3.2 × 10^-11 N
Where the charges fall freely under gravity;
F = mg
m = F/g
m = 3.2 × 10^-11 N/10 ms-2
m = 3.2 × 10^-12 Kg
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Suppose the B string on a guitar is 24" long and vibrates with a frequency of about 247 Hz. You place your finger on the 5th fret, which changes the length of the string to 18". Which note do you hear when you play the string?
a. D (about 294 Hz)
b. E (about 329 Hz)
c. G (about 392 Hz)
Answer:
b. E (about 329 Hz)
Explanation:
Given data:
Initial length of the string l1= 24 in
initial frequency f1= 247 Hz
changed length l2= 18 in
Then we have to find the changed frequency f2= ?
We already now that
frequency f ∝ 1/length of the string l
therefore,
[tex]\frac{f_1}{f_2} =\frac{l_1}{l_2}[/tex]
⇒[tex]{f_2}=\frac{l_1}{l_2}\times{f_1}[/tex]
⇒[tex]{f_2}=\frac{24}{18}\times{247}[/tex]
⇒[tex]{f_2}=329.33 Hz[/tex]
Mark Watney begins his day 15 km West and 25 km North of his Mars Habitat. a. Set up a co-ordinate system (draw labeled axis and tickmarks showing the scale) and draw a vector representing the initial position of Mark Watney. b. Mark spends his day driving his Mars Rover towards Schiaparelli Crater and manages to make it an additional 20 km West but has to go around a hill so ended up 5 km South of his starting point for the day. Draw a vector representing Mark's total position change for the day. c. Using vector addition find Mark's position relative to the Mars Habitat. Give both the numerical description of the vector and show it on your plot.
Answer:
a, b) part a and b on diagram attached
c) sf = -35 i + 20 j
35 km West and 20 km North
Explanation:
For part a and b refer to the attached co-ordinate system:
Note: unit vector i is in West/East direction and unit vector j is in North/South direction.
si = -15 i + 25 j
sf-si = -20 i - 5 j
Hence,
Mark relative position from habitat sf = si + sf/i
sf = ( -15 i + 25 j ) + ( -20 i - 5 j )
sf = -35 i + 20 j
35 km West and 20 km North
suppose two masses are connected by a spring. compute the formula for the trajectory of the center of mass of the two mass oscillator
Answer:
The center mass (Xcm) of the two mass = (M₁X₁ + M₂X₂)/(M₁ +M₂)
Explanation:
let the first mass = M
let the position of second = M
The formula for the trajectory of the center of mass of a two mass oscillator connected by a spring is x(t) = A * cos(ωt + φ), where A represents the amplitude, ω is the angular frequency, and φ is the phase constant.
Explanation:The formula for the trajectory of the center of mass of a two mass oscillator connected by a spring can be derived using the principles of simple harmonic motion (SHM). Let's assume that the masses are m1 and m2, and the spring constant is k. The equation for the trajectory of the center of mass can be written as:
x(t) = A * cos(ωt + φ)
where:
x(t) is the displacement of the center of mass at time tA is the amplitude of the motion, which can be calculated using the initial conditionsω is the angular frequency, which is given by ω = √(k / (m1 + m2))φ is the phase constant, which can also be calculated using the initial conditions
A 89.3 kg man sits on the stern of a 5.8 m
long boat. The prow of the boat touches the
pier, but the boat isn’t tied. The man notices his mistake, stands up and walks to the boat’s prow, but by the time he reaches the prow, it’s moved 4.17 m away from the pier.
Assuming no water resistance to the boat’s
motion, calculate the boat’s mass (not counting the man).
Answer in units of kg
Answer:
34.9 kg
Explanation:
Since there are no net external forces on the system, the center of gravity does not move.
Let's say that m is the mass of the man, M is the mass of the boat, and L is the length of the boat.
When the man is at the stern, the distance between the center of gravity and the pier is:
CG = (m L + M (L/2)) / (m + M)
When the man reaches the prow, the distance between the center of gravity and the pier is:
CG = (m x + M (x + L/2)) / (m + M)
Since these are equal:
(m L + M (L/2)) / (m + M) = (m x + M (x + L/2)) / (m + M)
m L + M (L/2) = m x + M (x + L/2)
m L + M (L/2) = m x + M x + M (L/2)
m L = m x + M x
m L − m x = M x
m (L − x) = M x
M = m (L − x) / x
Plugging in values:
M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m
M = 34.9 kg
The required mass of boat is 34.9 kg.
The given problem is based on the concept of the center of mass. The point of analysis where the entire mass of the system is known to be concentrated is known as the center of mass.
Given data:
The mass of man is, m = 89.3 kg.
The length of boat is, L = 5.8 m.
The distance away from the pier is, d = 4.17 m.
Since there are no net external forces on the system, the center of gravity does not move. Let's say that m is the mass of the man, M is the mass of the boat
When the man is at the stern, the distance between the center of gravity and the pier is:
CG = (m L + M (L/2)) / (m + M)
When the man reaches the prow, the distance between the center of gravity and the pier is:
CG = (m d + M (d + L/2)) / (m + M)
Since these are equal:
(m L + M (L/2)) / (m + M) = (m d + M (d + L/2)) / (m + M)
m L + M (L/2) = m d + M (d + L/2)
m L + M (L/2) = m d + M d + M (L/2)
Further solving as,
m L = m d + M d
m L − m d = M d
m (L − x) = M x
M = m (L − x) / x
M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m
M = 34.9 kg
Thus, we can conclude that the required mass of boat is 34.9 kg.
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An electric current of 202.0 mA Transports 56.0 C of charge. Calculate the time this took.
Answer:
277 s
Explanation:
Given data
Electric current (I): 202.0 mA = 202.0 × 10⁻³ A = 0.2020 A = 0.2020 C/sTransported charge (C): 56.0 CoulombsElapsed time (t): to be determinedWe can find the elapsed time using the following expression.
I = C/t
t = C/I
t = 56.0 C/(0.2020 C/s)
t = 277 s
It took 277 seconds.
A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplitude 1.80 mm travels along the wire.
(a) Calculate the average power carried by the wave.
(b) What happens to the average power if the wave amplitude is halved?
The new power is: New P_avg = 2.52 W / 4 ≈ 0.63 W
Average Power Carried by a Wave
To solve this problem, we need the following information:
Mass of piano wire: 2.95 g = 0.00295 kg
Length of wire: 79.0 cm = 0.79 m
Tension: 29.0 N
Frequency: 105 Hz
Amplitude: 1.80 mm = 0.00180 m
First, calculate the linear mass density (μ) of the wire:
μ = mass / length = 0.00295 kg / 0.79 m ≈ 0.00373 kg/m
Next, find the wave speed using the formula for the speed of a wave on a string:
v = [tex]\sqrt{Tension / \mu}[/tex] =[tex]\sqrt{29.0 N / 0.00373 kg/m}[/tex]≈ 88.19 m/s
Now, we calculate the average power (P_avg) carried by the wave using the formula:
P_avg = 0.5 x μ x v x ω² x A²
Where:
ω = 2πf (angular frequency)
ω = 2 x π x 105 ≈ 659.73 rad/s
Therefore,
P_avg = 0.5 x 0.00373 kg/m x 88.19 m/s x (659.73 rad/s)² x (0.00180 m)² ≈ 2.52 W
Average Power if Amplitude is Halved
If the amplitude (A) is halved, the new amplitude is:
New A = 0.00180 m / 2 = 0.00090 m
Since power is proportional to the square of the amplitude (A²), halving the amplitude reduces the power by a factor of 4.
Thus, the new power is:
New P_avg = 2.52 W / 4 ≈ 0.63 W
Sound of thunder is heard 5 seconds after the flash of lightning is seen.How far away was the lightning flash? Note that sound travels in air at a speed of about 340 meter second?
Answer:
The distance covered by the lighting flash is 1700 meters.
Explanation:
Given that,
The speed of sound travels in air at a speed of about 340 meter second, v = 340 m/s
The sound of thunder is heard 5 seconds after the flash of lightning is seen, t = 5 s
To find,
The distance far away was the lightning flash.
Solution,
The distance covered by the lighting flash is given by the product of speed and time. It is given by :
[tex]d=v\times t[/tex]
[tex]d=340\ m/s\times 5\ s[/tex]
d = 1700 meters
So, the distance covered by the lighting flash is 1700 meters. Therefore, this is the required solution.
The car's motion can be divided into three different stages: its motion before the driver realizes he's late, its motion after the driver hits the gas (but before he sees the police car), and its motion after the driver sees the police car. Which of the following simplifying assumptions is it reasonable to make in this problem? a. During each of the three different stages of its motion, the car is moving with constant acceleration. b. During each of the three different stages of its motion, the car is moving with constant velocity. c.The highway is straight (i.e., there are no curves). d. The highway is level (i.e., there are no hills or valleys).Enter all the correct answers in alphabetical order without commas. For example, if statements C and D are correct, enter CD.
The assumptions we can reasonably make in this scenario include the highway being straight and level, corresponding to options C and D. Assuming constant acceleration or velocity for each stage of the car's motion, options A and B, is not necessarily accurate.
Explanation:In analyzing the motion of a car in different stages, we can make reasonable assumptions to simplify the problem. For each stage, assuming the car is moving with constant acceleration (option a) or velocity (option b) is not necessarily accurate because acceleration and velocity may change due to various factors like interaction with driver, road conditions, or appearance of a police car. The assumptions more likely to hold are that the highway is straight (option c), meaning there are no curves, and the highway is level (option d), indicating no hills or valleys. Thus, the correct answers would be C and D.
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A block of mass 3.1 kg, sliding on a horizontal plane, is released with a velocity of 2.3 m/s. The blocks slides and stops at a distance of 1.9 m beyond the point where it was released.
How far would the block have slid if its initial velocity were quadrupled?
To solve this problem we will apply the concepts given by the kinematic equations of motion. For this purpose it will be necessary with the given data to obtain the deceleration. With this it will be possible again to apply one of the kinematic equations of motion that does not depend on time, but on distance, to find how far the block would slide with the quadruplicate velocity
Our values are given as,
[tex]\text{Initial speed} =V_i = 2.3 m/s[/tex]
[tex]\text{Final speed}= V_f = 0 m/s[/tex]
[tex]\text{Stopping distance = }d = 1.9 m[/tex]
[tex]a = acceleration[/tex]
[tex]\text{mass} = m = 3.1kg[/tex]
Using the kinematic equation of motion we have
[tex]V_f^2 = V_i^2 + 2 a d[/tex]
[tex]0^2 = 2.3^2 + 2 a (1.9)[/tex]
[tex]a = -1.39211 m/s^2[/tex]
Now if the initial velocity is quadrupled we have that,
[tex]\text{Initial speed} =V_i' = 2.3*4 m/s = 9.2m/s[/tex]
[tex]\text{Final speed}= V_f' = 0 m/s[/tex]
[tex]\text{Stopping distance = }d'[/tex]
[tex]V_f^2' = V_i^2' + 2 a d'[/tex]
Replacing the values
[tex]0^2 = 9.2^2+ 2 (-1.39211) d'[/tex]
[tex]d' = 30.44m[/tex]
Therefore the block would have slipped around 30.44 if its initial velocity quadrupled.
Suppose quantity s is a length and quantity t is a time. Suppose the quantities vand aare defined by v = ds/dt and a = dv/dt. (a) What is the dimension of v? (b) What is the dimension of the quantity a?
What are the dimensions of (c)vdt, (d) a dt, and (e) da/dt?
Explanation:
(a) Velocity is given by :
[tex]v=\dfrac{ds}{dt}[/tex]
s is the length of the distance
t is the time
The dimension of v will be, [tex][v]=[LT^-1][/tex]
(b) The acceleration is given by :
[tex]a=\dfrac{dv}{dt}[/tex]
v is the velocity
t is the time
The dimension of a will be, [tex][a]=[LT^{-2}][/tex]
(c) Since, [tex]d=\int\limits{v{\cdot}dt} =[LT^{-1}][T]=[L][/tex]
(d) Since, [tex]v=\int\limits{a{\cdot}dt} =[LT^{-2}][T]=[LT^{-1}][/tex]
(e)
[tex]\dfrac{da}{dt}=\dfrac{[LT^{-2}]}{[T]}[/tex]
[tex]\dfrac{da}{dt}=[LT^{-3}]}[/tex]
Hence, this is the required solution.
The position of a particle is given by the function x=(4t3−6t2+12)m, where t is in s.
A.) at what time does the particle reach its minimum velocity
B.) what is (vx)min
C.) at what time is the acceleration zero
Answer
given,
x = 4 t³ - 6 t² + 12
velocity, [tex]v = \dfrac{dx}{dt}[/tex]
[tex]\dfrac{dx}{dt}=\dfrac{d}{dt}(4t^3-6t^2+12)[/tex]
[tex]v =12t^2-12t[/tex]
For minimum velocity calculation we have differentiate it and put it equal to zero.
[tex]\dfrac{dv}{dt} =\dfrac{d}{dt}12t^2-12t[/tex]
[tex]\dfrac{dv}{dt} =24t-12[/tex].........(1)
putting it equal to zero
24 t - 12 =0
t = 0.5 s
At t = 0.5 s velocity will be minimum.
b) minimum velocity
v = 12t² -12 t
v = 12 x 0.5² -12 x 0.5
v = -3 m/s
c) derivative of velocity w.r.t. time is acceleration
from equation 1
a = 24 t - 12
time at which acceleration will be zero
0 = 24 t - 12
t = 0.5 s
At t = 0.5 s acceleration will be zero.
Part A. The particle reaches its minimum velocity at 0.5 seconds.
Part B. The minimum velocity of the particle is -3 m/s.
Part C. The acceleration of the particle will be zero at the time t = 0.5 seconds.
How do you calculate the minimum velocity and acceleration?Given that the position of a particle is given by the function x.
[tex]f(x) = 4t^2 -6t^2 +12[/tex]
The function of the velocity of a particle can be obtained by the time function.
[tex]v= \dfrac {dx}{dt}[/tex]
[tex]v = \dfrac {d}{dt} ( 4t^3-6t^2 +12)[/tex]
[tex]v = 12t^2 -12 t[/tex]
The velocity function of the particle is [tex]v = 12t^2 - 12t[/tex].
Part AThe minimum velocity of the particle is obtained by the differentiation of velocity function with respect to the time and put it equal to zero.
[tex]\dfrac {dv}{dt} = \dfrac {d}{dt} (12t^2 - 12t) = 0[/tex]
[tex]\dfrac {dv}{dt} = 24 t-12 = 0[/tex]
[tex]t = 0.5\;\rm s[/tex]
Hence we can conclude that the particle reaches its minimum velocity at 0.5 seconds.
Part BThe velocity function is [tex]v = 12t^2 - 12t[/tex]. Substituting the value of t = 0.5 s to calculate the minimum velocity.
[tex]v = 12(0.5)^2 - 12(0.5)[/tex]
[tex]v = 3 - 6[/tex]
[tex]v = -3 \;\rm m/s[/tex]
The minimum velocity of the particle is -3 m/s.
Part CThe acceleration is defined as the change in the velocity with respect to time. Hence,
[tex]a = \dfrac {dv}{dt}[/tex]
[tex]a = 24 t-12[/tex]
Substituting the value of a = 0, we get the time.
[tex]0 = 24t - 12[/tex]
[tex]t = 0.5 \;\rm s[/tex]
The acceleration of the particle will be zero at the time t = 0.5 seconds.
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At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v = 239 m/s. Upon landing, the plane can produce an average deceleration of a = 16.5 m/s².
How long will it take the plane to circle the Earth at the equator?
To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From there we will define the distance as the circumference of the earth (approximate as a sphere). With the speed given in the statement we will simply clear the equations below and find the time.
[tex]R= 6370*10^3 m[/tex]
[tex]v = 239m/s[/tex]
[tex]a = 16.5m/s^2[/tex]
The circumference of the earth would be
[tex]\phi = 2\pi R[/tex]
Velocity is defined as,
[tex]v = \frac{x}{t}[/tex]
[tex]t = \frac{x}{v}[/tex]
Here [tex]x = \phi[/tex], then
[tex]t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{239}[/tex]
[tex]t = 167463.97s[/tex]
Therefore will take 167463.97 s or 1 day 22 hours 31 minutes and 3.97seconds
A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 14.5 m/s, and the distance from the limb to the level of the saddle is 3.29 m.
(a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move?
(b) How long is he in the air?
Answer:
For Part A:
ΔX=11.8813 m
Part B:
[tex]t=0.8194 s[/tex]
Explanation:
Note: In order to find Part A we first have to find Part B i.e time
Data given:
[tex]V_i=0[/tex]
a=-9.8m/s^2 (-ve is because ranch is falling down)
Δy=-3.29m (-ve is because ranch is falling down)
Second equation of Motion:
Δy=[tex]V_i*t+\frac{1}{2}g*t^2[/tex]
V_i=0, Equation will become
Δy=[tex]\frac{1}{2}g*t^2[/tex]
[tex]t=\sqrt{2Y/g} \\t=\sqrt{2*-3.29/-9.8} \\t=0.8194 s[/tex]
For Part A:
Again:
Second equation of Motion:
ΔX=[tex]V_i*t+\frac{1}{2}a*t^2[/tex]
Since velocity is constant a=0
V_i=14.5m/s, t=0.8194 sec
ΔX=[tex]V_i*t[/tex]
ΔX=[tex]14.5*0.8194[/tex]
ΔX=11.8813 m
Part B: (Calculated above)
Data given:
[tex]V_i=0[/tex]
a=-9.8m/s^2 (-ve is because ranch is falling down)
Δy=-3.29m (-ve is because ranch is falling down)
Second equation of Motion:
Δy=[tex]V_i*t+\frac{1}{2}g*t^2[/tex]
V_i=0, Equation will become
Δy=[tex]\frac{1}{2}g*t^2[/tex]
[tex]t=\sqrt{2Y/g} \\t=\sqrt{2*-3.29/-9.8} \\t=0.8194 s[/tex]
A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed. Part A What is the ratio of the normal force to the gravitational force? What is the ratio of the normal force to the gravitational force? n/FG=2n/FG=2 n/FG=3n/FG=3 n/FG=4n/FG=4 n/FG=5n/FG=5
Answer:
n/(FG) = 3.
Explanation:
At the top of the loop-the-loop, the normal force is directed downwards as well as the weight of the car. So, the total net force of the car is
[tex]F_{net} = N + mg[/tex]
By Newton's Second Law, this force is equal to the centripetal force, because the car is making circular motion in the loop.
[tex]F_{net} = ma = \frac{mv^2}{R}\\N + mg = \frac{mv^2}{R}[/tex]
The critical speed is the minimum speed at which the car does not fall. So, at the critical speed the normal force is zero.
[tex]0 + mg = \frac{mv_c^2}{R}\\v_c = \sqrt{gR}[/tex]
If the car is moving twice the critical speed, then
[tex]N + mg = \frac{m(2v_c)^2}{R} = \frac{m4gR}{R} = 4mg\\N = 3mg[/tex]
Finally, the ratio of the normal force to the gravitational force is
[tex]\frac{3mg}{mg} = 3[/tex]
what is the distance on a map in centimeters between two features if they are 7.6 km apart on the ground and the map has a scale of 1. 125000
Answer:
6.08 cm
Explanation:
We are given that
Ratio =1:125000
Let x be the distance on a map between two features .
The distance between two features on ground=y=7.6 km
According to question
[tex]\frac{x}{y}=\frac{1}{125000}[/tex]
Substitute the values then we get
[tex]\frac{x}{7.6}=\frac{1}{125000}[/tex]
[tex]x=\frac{7.6}{125000}=0.0000608 km[/tex]
We know that
[tex]1km=100000 cm[/tex]
0.0000608 km=[tex]0.0000608\times 100000=6.08cm[/tex]
Hence, the distance between two features on the map=6.08 cm
The distance on the map between the two features is 6.08 centimeters.
Explanation:To find the distance on a map in centimeters between two features if they are 7.6 km apart on the ground and the map has a scale of 1:125000, you can use the scale factor formula: Distance on Map = Distance on Ground / Scale
Plugging in the values: Distance on Map = 7.6 km / 125000 = 0.0000608 km
Since 1 km = 100000 centimeters, multiply by 100000 to convert km to cm:
Distance on Map = 0.0000608 km × 100000 cm/km = 6.08 cm
Therefore, the distance on the map between the two features is 6.08 centimeters.
As additional resistors are connected in series to a constant voltage source, how is the power supplied by the source affected?
(A) The power supplied by the sources remains constant.
(B) The power supplied by the source increases.
(C) The effect on the power supplied by the source cannot be determined without knowing the voltage of the source.
(D) The power supplied by the source decreases.
As additional resistors are connected in series to a constant voltage source, the power supplied by the source decreases.
Explanation:As additional resistors are connected in series to a constant voltage source, the power supplied by the source decreases.
This can be explained by Ohm's Law, which states that power is equal to the voltage squared divided by the resistance. When resistors are connected in series, the total resistance increases, which leads to a decrease in the power supplied by the source.
For example, if you connect two resistors in series to a constant voltage source, the total resistance would be the sum of the resistances of the two resistors. As a result, the power supplied by the source would decrease.
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Adding more resistors in series to a constant voltage source results in an increase in total resistance, which leads to a decrease in current. Because power is proportional to the square of the current, the power supplied by the voltage source decreases.
Explanation:When additional resistors are connected in series to a constant voltage source, the power supplied by the source is affected in a specific way. Given that power (P) is calculated by P = I^2R for a given resistance (R) and current (I), and by P = V^2/R for a given voltage (V) and resistance, we can understand the impact of adding resistors in series.
In a series circuit, the current remains constant throughout the resistors, but the total resistance (sum of all individual resistances) increases as more resistors are added. Since the voltage is constant, an increase in the total resistance will result in a decrease in current according to Ohm's law (I = V/R). Consequently, if the current decreases, the power supplied by the source also decreases (P = I^2R), because the power is proportional to the square of the current flowing through the circuit. Therefore, the correct answer is (D) The power supplied by the source decreases.