Answer:
θ = 16.2 rad
Explanation:
First we find the angular acceleration by using first equation of motion in angular form:
ωf = ωi + αt
where,
ωf =final angular speed = (110 rev/min)(2π rad/1 rev)(1 min/60 s) = 11.5 rad/s
ωi =initial angular speed = (45 rev/min)(2π rad/1 rev)(1 min/60 s) = 4.7 rad/s
α = angular acceleration = ?
t = time = 2 s
Therefore,
11.5 rad/s = 4.7 rad/s + α(2 s)
α = (6.8 rad/s)/(2 s)
α = 3.4 rad/s²
Now, we use 2nd equation of motion:
θ = ωi t + (1/2)αt²
where,
θ = rotation = ?
Therefore,
θ = (4.7 rad/s)(2 s) + (1/2)(3.4 rad/s²)(2 s)²
θ = 9.4 rad + 6.8 rad
θ = 16.2 rad
Which of the following statements is true vibrations ?
Answer:
C. Neither ultrasonic nor infrasonic vibrations can be heard by humans.
Explanation:
The complete question is
Which of the following statements is true of vibrations? A. The frequency of infrasonic vibrations is much too high to be heard by humans. B. Ultrasonic vibrations have a frequency lower than the range for normal hearing. C. Neither ultrasonic nor infrasonic vibrations can be heard by humans. D. Infrasonic vibrations are used in sonar equipment and to detect flaws in steel castings.
Ultrasonic vibrations have frequencies higher than our range of hearing, while infrasonic vibrations have frequencies lower than our range of hearing. Ultrasonic vibrations or sound is used in sonar equipment, and is used for detecting hidden flaws in steel castings and structures. Both infrasonic and ultrasonic fall below and above our normal hearing range respectively, and are only audible to dogs, cats, and some other mammals.
Answer:
Answer is " Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate."
Explanation:
My options were:
A) Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body.
B) Resonance occurs as a result of sympathetic vibrations.
C) A non-vibrating object can begin to vibrate as a result of forced vibrations.
D) Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate.
A is correct
.
A 2-slit arrangement with 60.3 μm separation between the slits is illuminated with 482.0 nm light. Assuming that a viewing screen is located 2.14 m from the slits, find the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side. A. 24.1 mm B. 34.2 mm C. 68.4 mm D. 51.3 mm
Answer:
The distance is [tex]y = 0.03425 \ m[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 60.3 \mu m= 60.3 *10^{-6}\ m[/tex]
The wavelength is [tex]\lambda = 482.0 \ nm = 482.0 *10^{-9} \ m[/tex]
The distance of the screen is [tex]D = 2.14 \ m[/tex]
Generally the distance of a fringe from the central maxima is mathematically represented as
[tex]y = [m + \frac{1}{2} ] * \frac{\lambda * D}{d}[/tex]
For the first dark fringe m = 0
[tex]y_1 = [0 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]
[tex]y_1 = 0.00855 \ m[/tex]
For the second dark fringe m = 1
[tex]y_2 = [1 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]
[tex]y_2 = 0.0257 \ m[/tex]
So the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side is
[tex]y = y_1 + y_2[/tex]
[tex]y = 0.00855 + 0.0257[/tex]
[tex]y = 0.03425 \ m[/tex]
How much time will elapse if a radioisotope with a half-life of 88 seconds decays to one-sixteenth of its original mass?
Answer:
352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.
Explanation:
The decay of radioisotopes are represented by the following ordinary differential equation:
[tex]\frac{dm}{dt} = -\frac{t}{\tau}[/tex]
Where:
[tex]t[/tex] - Time, measured in seconds.
[tex]\tau[/tex] - Time constant, measured in seconds.
[tex]m[/tex] - Mass of the radioisotope, measured in grams.
The solution of this expression is:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]
Where [tex]m_{o}[/tex] is the initial mass of the radioisotope, measured in kilograms.
The ratio of current mass to initial mass is:
[tex]\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]
The time constant is now calculated in terms of half-life:
[tex]\tau = \frac{t_{1/2}}{\ln2}[/tex]
Where [tex]t_{1/2}[/tex] is the half-life of the radioisotope, measured in seconds.
Given that [tex]t_{1/2} = 88\,s[/tex], the time constant of the radioisotope is:
[tex]\tau = \frac{88\,s}{\ln 2}[/tex]
[tex]\tau \approx 126.957\,s[/tex]
Now, if [tex]\frac{m(t)}{m_{o}(t)} = \frac{1}{16}[/tex] and [tex]\tau \approx 126.957\,s[/tex], the time is:
[tex]t = -\tau \cdot \ln\frac{m(t)}{m_{o}}[/tex]
[tex]t = -(126.957\,s)\cdot \ln \frac{1}{16}[/tex]
[tex]t \approx 352\,s[/tex]
352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.
A swimmer is treading water with their head above the surface of a pool and sees a penny at the bottom of the pool 5.0 mm below. How deep does the coin appear to be? (Index of refraction of water = 1.33) [Conceptual note: Does the coin appear to be shallower or deeper?]
Answer:
The apparent depth is [tex]D' = 0.00376 \ m[/tex]
Explanation:
From the question we are told that
The depth of the water is [tex]D = 5.0 \ mm = 5.0 *10^{-3} \ m[/tex]
The refractive index of water is [tex]n = 1.33[/tex]
Generally the apparent depth of the coin is mathematically represented as
[tex]D' = D * [\frac{ n_a}{n} ][/tex]
Here [tex]n_a[/tex] is the refractive index of air the value is [tex]n_a = 1[/tex]
So
[tex]D' = 5.0 *10^{-3} * [\frac{1}{1.33} ][/tex]
[tex]D' = 0.00376 \ m[/tex]
The apparent depth will be 0.00376 m.
What is an index of refraction?
The index of refraction of a substance also known as the refraction index is a dimensionless quantity that specifies how quickly light passes through it in optics.
d is the depth of the water =5.0 mm =5.0 ×10⁻³
n is the refractive index of water =1.33
[tex]\rm n_a[/tex] is the refractive index of wire=1
The apparent depth of the coin is given as;
[tex]\rm D'=D \times \frac{n_a}{n} \\\\ \rm D'=5.0 \times 10^{-3} \times \frac{1}{1.33} \\\\ \rm D'=0.00376 \ m[/tex]
Hence the apparent depth will be 0.00376 m.
To learn more about the index of refraction refer to the link;
https://brainly.com/question/23750645
¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una
resistencia combinada de 15 Ω?
Answer:
60 Ω
Explanation:
R(com) = 15 Ω
1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn
1/15 = 1/20 + 1/R2
1/R2 = 1/15 - 1/20
1/R2 = (4 - 3) / 60
1/R2 = 1/60
R2 = 60 Ω
así, la combinada de resistencia necesaria es 60 Ω
A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height from which it was shot.
A. What is the magnitude of the inital velocity?B. What was the maximum height reached by the cannonball?C. How far, horizontally, did it travel?
Answer:
Explanation:
According to Equations of Projectile motion :
[tex]Time\ of\ Flight = \frac{2vsin(x)}{g}[/tex]
vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec
(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec
[tex]Maximum Height = \frac{(vsinx)^{2} }{2g}[/tex]
(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m
[tex]Horizontal Range = vcosx * t[/tex]
(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m
An electron moving in the direction of the +x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the:______
Answer:
-z axis
Explanation:
According to the left hand rule for an electron in a magnetic field, hold the thumb of the left hand at a right angle to the rest of the fingers, and the rest of the fingers parallel to one another. If the thumb represents the motion of the electron, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the electron. In this case, the left hand will be held out with the thumb pointing to the right (+x axis), and the palm facing your body (-y axis). The magnetic field indicated by the other fingers will point down in the the -z axis.
Changing the speed of a synchronous generator changes A) the frequency and amplitude of the output voltage. B) only the frequency of the output voltage. C) only the amplitude of the output voltage. D) only the phase of the output voltage.
Answer:
A) the frequency and amplitude of the output voltag
Explanation:
Changing the speed of a synchronous generator changes both the output voltage (amplitude of the wave) and frequency as they tend to increase.
Changing the speed regulator will change the engine throttle setting to maintain the speed.
While the power, torque, current, fuel flow rate and torque angle will have decreased.
The main purpose of a written report may be to _____. A. revise a hypothesis B. summarize other scientists' results C. design a procedure for an experiment D. analyze data without drawing conclusions
PLZZZ HURRY TIMED MARK BRAINLIEST
Answer:
analyze data without drawing conclusions
Explanation:
Research reports are written in order to communicate clearly, information obtained primarily from research and analysis of data.
Typical reports of scientific research endeavours are written in such a way that they convey the research process succinctly without excessive extraneous information. A report is typically made up of; summary of the contents, introduction/ background, methods, results, discussion, conclusion and recommendations.
Hence a report does not really make inferences from the research findings.
As the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor:_______.
a. approaches zero.
b. approaches infinity.
c. approaches unity.
d. none of the above.
Answer:
b. approaches infinity
Explanation:
Because Capacitive reactance is given as Xc = 1/ωC
So we can see that the value of capacitive reactance and therefore its overall impedance (in Ohms) decreases to zero as the frequency increases acting like a short circuit.
Same as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC
A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 950 kg. Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift. Express your answer to two significant figures and include the appropriate units.
volume of balloon
= 4/3 T R3
= 4/3 x 3.14 x 6.953
= 1405.47 m3
uplift force
= volume of balloon x density of air x 9.8
= = 1405.47 x 1.29 x 9.8
= 1813.05 x 9.8 N
weight of helium gas
= volume of balloon x density of helium x
9.8
= 1405.47 x .179 x 9.8
= 251.58 x 9.8 N
Weight of other mass = 930 x 9.8 N Total weight acting downwards
= 251.58 x 9.8 +930 x 9.8
= 1181.58 x 9.8 N
If W be extra weight the uplift can balance
1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8
1181.58+W=1813.05
W= 631.47 kg
Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2. Compared to the incident ray, what happens to the refracted ray?
Answer:
It bends away from the normal
Explanation:
From Snell's law of Refraction, when a ray passes from a medium of lower Refractive index to a medium with higher Refractive index, the Refractive ray will bend towards the normal. However, when the ray passes from a medium of higher Refractive index to a medium of lower Refractive index, the Refractive ray will bend away from the normal.
Now, from the question we are told that Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2.
This means from a higher Refractive index to a lower one and from Snell's law as earlier said, the refracted ray will bend away from the normal
The refracted ray is seen to bend away from the normal.
Let us recall that an optically denser medium will have a higher refractive index. This means that the medium with a refractive index of 1.3 is the denser medium and the medium with a refractive index of 1.2 is the less dense medium.
From the statement in the question, we can boldly say that light is travelling from a denser to less dense medium given the values of the refractive index given. When light is travelling from a denser to a less dense medium, the refracted ray bends away from the normal.
Learn more: https://brainly.com/question/1445383
A pulley 326 mm in diameter and rotating initially at 4.00 revolutions per second receives a constant angular acceleration of 2.25 radians per second squared by a drive belt. What is the linear velocity of the belt after 5.00 seconds
Answer:
The linear velocity, v = 5.93 m/s
Explanation:
To find the linear velocity after 5 seconds, we find its angular velocity after 5 seconds using
ω' = ω + αt where ω = initial angular speed = 4.00 rev/s = 4.00 × 2π rad/s = 25.13 rad/s, ω' = = final angular speed, α = angular acceleration = 2.25 rad/s² and t = time = 5.00 s
ω' = ω + αt
= 25.13 rad/s + 2.25 rad/s² × 5.00 s
= 25.13 rad/s + 11.25 rad/s
= 36.38 rad/s
The linear velocity v is gotten from v = rω' where r = radius of pulley = 326 mm/2 = 163 mm = 0.163 m
v = rω'
= 0.163 m × 36.38 rad/s
= 5.93 m/s
So, the linear velocity v = 5.93 m/s
A uniform narrow tube 1.90 m long is open at both ends. It resonates at two successive harmonics of frequencies 280 Hz and 294 Hz.(a) What is the fundamental frequency?_____Hz(b) What is the speed of sound in the gas in the tube?________ m/s
Answer:
a)14Hz
b)26.6m/s
Explanation:
a)we were given
the first harmonics frequencies as 280 Hz
The second harmonic frequency as 294 Hz.
The fundamental frequency is equal to the gap which means the distance that exist between the harmonics, then
the fundamental frequency=(294 - 280 = 10 Hz)
= 14Hz
b) We know the frequency and the wavelength of the sound wave (
We were told that the wavelength must be twice the length of the tube then, velocity can be calculated as
And fundamental frequency= 14Hz, and distance of 1.90 m then
v = f*2L = (14Hz)*2*(1.90 m) = 26.6m/s
Therefore, the speed of sound in the gas in the tubes is 26.6m/s
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
Answer:
A) the moment of inertia of the system decreases and the angular speed increases.
Explanation:
The complete question is
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, It is true to say that
A) the moment of inertia of the system decreases and the angular speed increases.
B) the moment of inertia of the system decreases and the angular speed decreases.
C) the moment of inertia of the system decreases and the angular speed remains the same.
D) the moment of inertia of the system increases and the angular speed increases.
E) the moment of inertia of the system increases and the angular speed decreases
In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as
[tex]I_{1} w_{1} = I_{2} w_{2}[/tex] ....1
where [tex]I_{1}[/tex] and [tex]I_{2}[/tex] are the initial and final moment of inertia respectively.
and [tex]w_{1}[/tex] and [tex]w_{2}[/tex] are the initial and final angular speed respectively.
Also, we know that the moment of inertia of a rotating body is given as
[tex]I = mr^{2}[/tex] ....2
where [tex]m[/tex] is the mass of the rotating body,
and [tex]r[/tex] is the radius of the rotating body from its center.
We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.
From equation 1, we see that in order for the angular momentum to be conserved, the decrease from [tex]I_{1}[/tex] to [tex]I_{2}[/tex] will cause the angular speed of the system to increase from [tex]w_{1}[/tex] to [tex]w_{2}[/tex] .
From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.
Two hoops, staring from rest, roll down identical incline planes. The work done by nonconservative forces is zero. The hoops have the same mass, but the larger hoop has twice the radius. Which hoop will have the greater total kinetic energy at the bottom
Answer:
They both have the same total K.E at the bottom
Explanation:
This Is because If assuming no work is done by non conservative forces, total mechanical energy must be conserved
So
K1 + U1 = K2 + U2
But If both hoops start from rest, and and at the bottom of the incline the level for gravitational potential energy is zero for reference
thus
K1 = 0 , U2 = 0
ΔK = ΔU = m g. h
But if the two inclines have the same height, and both hoops have the same mass m,
So difference in kinetic energy, must be the same for both hoops.
A solenoid is designed to produce a magnetic field of 3.50×10^−2 T at its center. It has a radius of 1.80 cm and a length of 46.0 cm , and the wire can carry a maximum current of 13.0 A.
Required:
a. What minimum number of turns per unit length must the solenoid have?
b. What total length of wire is required?
Answer:
a. 2143 turns/m
b. 111.5 m
Explanation:
a. The minimum number of turns per unit length (N/L) can be found using the following equation:
[tex] B = \frac{\mu_{0}NI}{L} [/tex]
[tex] \frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m [/tex]
Hence, the minimum number of turns per unit length is 2143 turns/m.
b. The total length of wire is the following:
[tex] N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns [/tex]
Since each turn has length 2πr of wire, the total length is:
[tex] L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m [/tex]
Therefore, the total length of wire required is 111.5 m.
I hope it helps you!
A girl is sitting on the edge of a pier with her legs dangling over the water. Her soles are 80.0 cm above the surface of the water. A boy in the water looks up at her feet and wants to touch them with a reed. (nwater =1.333). He will see her soles as being:____
a. right at the water surface.
b. 53.3 cm above the water surface.
c. exactly 80.0 cm above the water surface.
d. 107 cm above the water surface.
e. an infinite distance above the water surface.
Answer:
d. 107 cm above the water surface.
Explanation:
The refractive index of water and air = 1.333
The real height of the girl's sole above water = 80.0 cm
From the water, the apparent height of the girl's sole will be higher than it really is in reality by a factor that is the refractive index.
The boy in the water will therefore see her feet as being
80.0 cm x 1.333 = 106.64 cm above the water
That is approximately 107 cm above the water
PLEASE HELP FAST Five-gram samples of brick and glass are at room temperature. Both samples receive equal amounts of energy due to heat flow. The specific heat capacity of brick is 0.22 cal/g°C and the specific heat capacity of glass is 0.22 cal/g°C. Which of the following statements is true? 1.The temperature of each sample will increase by the same amount. 2.The temperature of each sample will decrease by the same amount. 3.The brick will get hotter than the glass. 4.The glass will get hotter than the brick.
Answer:
1.The temperature of each sample will increase by the same amount
Explanation:
This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.
This is shown by the calculation below
Q = mcΔT
ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.
Since Q, m and c are the same for both substances, thus ΔT will be the same.
So, the temperature of each sample will increase by the same amount
At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.10 ✕ 10−5 T
Answer:
The speed of the proton is 4059.39 m/s
Explanation:
The centripetal force on the particle is given by;
[tex]F = \frac{mv^2}{r}[/tex]
The magnetic force on the particle is given by;
[tex]F = qvB[/tex]
The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.
[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]
where;
r is the radius of the circular path moved by both electron and proton;
⇒For electron;
[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]
⇒For proton
The speed of the proton is given by;
[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]
Therefore, the speed of the proton is 4059.39 m/s
A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for it three wet weather on 3 separate days? Also find its Standard Deviation
Answer:
so the probability will be = 0.062
Standard deviation = 0.8925
Explanation:
The probability of rain = 15% = 15/100= 0.15
and the probability of no rain=q = 1-p= 1-0.15= 0.85
The number of trials = 7
so the probability will be
7C3 * ( 0.15)^3 (0.85)^4= 35* 0.003375 * 0.52200 =0.06166= 0.062
Taking this as binomial as the p and q are constant and also the trials are independent .
For a binomial distribution
Standard deviation = npq= 0.15 *0.85 *7= 0.8925
A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 cm carries a current of 0.25 A. The relative permeability of the core is 80.
(a) What is the magnetic field in the core?
(b) What part of the magnetic field is due to atomic currents?
Answer:
A) 0.0267 T
B) 0.0263 T
Explanation:
Given that
The number of turns, N = 400
Radius of the wire, r = 6 cm = 0.06 m
Current in the wire, I = 0.25 A
Relative permeability, K(m) = 80
See the attached picture for the calculation
A small helium-neon laser emits red visible light with a power of 5.40 mW in a beam of diameter 2.30 mm.
Required:
a. What is the amplitude of the electric field of the light? Express your answer with the appropriate units.
b. What is the amplitude of the magnetic field of the light?
c. What is the average energy density associated with the electric field? Express your answer with the appropriate units.
d. What is the average energy density associated with the magnetic field? Express your answer with the appropriate units.
E) What is the total energy contained in a 1.00-m length of the beam? Express your answer with the appropriate units.
Answer:
A. 990v/m
B.330x10^-8T
C.2.19x10^-6J/m³
D.1.45x10^-11J
Explanation:
See attached file
The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various atmosphericprocesses, including lightning.
What is the excess charge on the surface of the earth? inC
Answer:
[tex]q = -461532.5 \ C[/tex]
Explanation:
From the question we are told that
The electric filed is [tex]E = 102 \ N/C[/tex]
Generally according to Gauss law
=> [tex]E A = \frac{q}{\epsilon_o }[/tex]
Given that the electric field is pointing downward , the equation become
[tex]- E A = \frac{q}{\epsilon_o }[/tex]
Here [tex]q[/tex] is the excess charge on the surface of the earth
[tex]A[/tex] is the surface area of the of the earth which is mathematically represented as
[tex]A = 4\pi r^2[/tex]
Where r is the radius of the earth which has a value [tex]r = 6.3781*10^6 m[/tex]
substituting values
[tex]A = 4 * 3.142 * (6.3781*10^6 \ m)^2[/tex]
[tex]A =5.1128 *10^{14} \ m^2[/tex]
So
[tex]q = -E * A * \epsilon _o[/tex]
Here [tex]\epsilon_o[/tex] s the permitivity of free space with value
[tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}[/tex]
[tex]q = -461532.5 \ C[/tex]
A fan rotating with an initial angular velocity of 1500 rev/min is switched off. In 2.5 seconds, the angular velocity decreases to 400 rev/min. Assuming the angular acceleration is constant, answer the following questions.
How many revolutions does the blade undergo during this time?
A) 10
B) 20
C) 100
D) 125
E) 1200
Answer:
The blade undergoes 40 revolutions, so neither of the given options is correct!
Explanation:
The revolutions can be found using the following equation:
[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
Where:
α is the angular acceleration
t is the time = 2.5 s
[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min
First, we need to find the angular acceleration:
[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]
Now, the revolutions that the blade undergo are:
[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]
Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!
I hope it helps you!
Two wires carry current I1 = 73 A and I2 = 31 A in the opposite directions parallel to the x-axis at y1 = 3 cm and y2 = 13 cm. Where on the y-axis (in cm) is the magnetic field zero?
Answer:
The position on the y-axis where the magnetic field is zero is at y = 10 cm
Explanation:
The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by
B = μ₀i/2πR
Now, let y be the point where the magnetic fields of both wires are equal.
So, the magnetic field due to wire 1 carrying current i₁ = 73 A is
B₁ = μ₀i₁/2π(y - 3) and
the magnetic field due to wire 2 carrying current i₂ = 31 A is
B₂ = μ₀i₂/2π(13 - y)
At the point where the magnetic field is zero, B₁ = B₂. So,
μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)
cancelling out μ₀ and 2π, we have
i₁/(x - y) = i₂/(13 - y)
cross-multiplying, we have
(13 - y)i₁ = (y - 3)i₂
Substituting the values of i₁ and i₂, we have
(13 - y)73 = (y - 3)31
949 - 73y = 31y - 93
Collecting like terms, we have
949 + 93 = 73y + 31y
1042 = 104y
dividing through by 104, we have
y = 1042/104
y = 10.02 cm
y ≅ 10 cm
So, the position on the y-axis where the magnetic field is zero is at y = 10 cm
A satellite of mass m circles a planet of mass M and radius R in an orbit at a height 2R above the surface of the planet. What minimum energy is required to change the orbit to one for which the height of the satellite is 3R above the surface of the planet
Answer:
ΔE = GMm/24R
Explanation:
centripetal acceleration a = V^2 / R = 2T/mr
T= kinetic energy
m= mass of satellite, r= radius of earth
= gravitational acceleration = GM / r^2
Now, solving for the kinetic energy:
T = GMm / 2r = -1/2 U,
where U is the potential energy
So the total energy is:
E = T+U = -GMm / 2r
Now we want to find the energy difference as r goes from one orbital radius to another:
ΔE = GMm/2 (1/R_1 - 1/R_2)
So in this case, R_1 is 3R (planet's radius + orbital altitude) and R_2 is 4R
ΔE = GMm/2R (1/3 - 1/4)
ΔE = GMm/24R
On a separate sheet of paper, tell why scientists in different countries can easily compare the amount of matter in similar objects in their countries
Answer: no u
Explanation: no u
Charge of uniform density (0.30 nC/m2) is distributed over the xy plane, and charge of uniform density (−0.40 nC/m2) is distributed over the yz plane. What is the magnitude of the resulting electric field at any point not in either of the two charged planes?
Answer: E = 39.54 N/C
Explanation: Electric field can be determined using surface charge density:
[tex]E = \frac{\sigma}{2\epsilon_{0}}[/tex]
where:
σ is surface charge density
[tex]\epsilon_{0}[/tex] is permitivitty of free space ([tex]\epsilon_{0} = 8.85.10^{-12}[/tex][tex]C^{2}/N.m^{2}[/tex])
Calculating resulting electric field:
[tex]E=E_{1} - E_{2}[/tex]
[tex]E = \frac{\sigma_{1}-\sigma_{2}}{2\epsilon_{0}}[/tex]
[tex]E = \frac{[0.3-(-0.4)].10^{-9}}{2.8.85.10^{-12}}[/tex]
[tex]E=0.03954.10^{3}[/tex]
E = 39.54
The resulting Electric Field at any point is 39.54N/C.
The magnitude of the resulting electric field at any point should be 28.2 N/C.
Calculation of the magnitude:Since the Charge of uniform density (0.30 nC/m2) should be allocated over the xy plane, and charge of uniform density (−0.40 nC/m2)should be allocated over the yz plane.
So,
E1
= σ1/2ε0
= 0.30e-9/(2*8.85e-12)
= 16.949 N/C
So, direction of E1 is +z
Now
E2 = σ2/2ε0
= 0.40e-9/(2*8.85e-12)
= 22.6 N/C
So, direction of E2 is -x
Now
E = √(E1*E1+E2*E2)
= √(16.949*16.949+22.6*22.6)
= 28.2 N/C
Learn more about magnitude here: https://brainly.com/question/14576767
a car moves for 10 minutes and travels 5,280 meters .What is the average speed of the car?
Answer:use the formular distance over time i.e distance/time. Make sure to convert the distance from metres to kilometers and time from minutes to hours .
Explanation:
The average speed of the car is 31,680 meters per hour.
To calculate the average speed of the car, you need to divide the total distance traveled by the time it took to travel that distance.
Given:
Time taken (t) = 10 minutes = 10 minutes × (1 hour / 60 minutes) = 10/60 hours = 1/6 hours
Distance traveled (d) = 5,280 meters
Average Speed (v) = Distance (d) / Time (t)
Average Speed (v) = 5280 meters / (1/6) hours
To simplify, when you divide by a fraction, it's equivalent to multiplying by its reciprocal:
Average Speed (v) = 5280 meters × (6/1) hours
Average Speed (v) = 31,680 meters per hour
Hence, the average speed of the car is 31,680 meters per hour.
To know more about average speed here
https://brainly.com/question/17661499
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