Answer:
the water that remain in the tank in one hour will be 200 lb
Explanation:
Initial mass of water in the tank = 2000 lb
hot water is delivered through the first inlet pipe at a rate of = 0.8 lb/s
cold water is delivered through the second inlet pipe at a rate of = 1.2 lb/s
exit pipe flow rate = 2.5 lb/s
amount of water in the tank after one hour = ?
In one hour, there are 60 x 60 seconds = 3600 sec, therefore
the water through the first inlet pipe in one hour = 0.8 x 3600 = 2880 lb
the water through the second inlet pipe in one hour = 1.2 x 3600 = 4320 lb
the water through the exit in one hour = 2.5 x 3600 = 9000 lb
The total amount of water in the tank = 2000 + 2880 + 4320 = 9200 lb
The total amount of water that leaves the tank = 9000 lb
therefore, in one hour, the water that remain in the tank will be
==> 9200 lb - 9000 lb = 200 lb
Red light is bent the least of all colors as it passes through a prism. What does this tell you about red light? It has a short wavelength. It has a long wavelength. It has a high intensity. It has a low intensity.
Answer:
Longest wavelength, lowest intensity
Explanation:
Answer:
It has a long wavelength
Explanation:
GRADPOINT
. A 24-V battery is attached to a 3.0-mF capacitor and a 100-ohm resistor. If the capacitor is initially uncharged, what is the voltage across the capacitor 0.16 seconds after the circuit is connected to the battery
Answer:
The voltage is [tex]V_c = 9.92 \ V[/tex]
Explanation:
From the question we are told that
The voltage of the battery is [tex]V_b = 24 \ V[/tex]
The capacitance of the capacitor is [tex]C = 3.0 mF = 3.0 *10^{-3} \ F[/tex]
The resistance of the resistor is [tex]R = 100\ \Omega[/tex]
The time taken is [tex]t = 0.16 \ s[/tex]
Generally the voltage of a charging charging capacitor after time t is mathematically represented as
[tex]V_c = V_o (1 - e^{- \frac{t}{RC} })[/tex]
Here [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged which in the case of this question is equivalent to the voltage of the battery so
[tex]V_c = 24 (1 - e^{- \frac{0.16}{100 * 3.0 *10^{-1}} })[/tex]
[tex]V_c = 9.92 \ V[/tex]
The rotor of a gas turbine is rotating at a speed of 7000 rpm when the turbine is shut down. It is observed that 3.5 minutes is required for the rotor to coast to rest. Assuming uniformly accelerated motion, determine the number of revolutions that the rotor executes before coming to rest. Hint: there will be a large number of rotations.
Answer:
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
Explanation:
Given that rotor of gas turbine is decelerating at constant rate, it is required to obtained the value of angular acceleration as a function of time, as well as initial and final angular speeds. That is:
[tex]\dot n = \dot n_{o} + \ddot n \cdot t[/tex]
Where:
[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.
[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.
[tex]t[/tex] - Time, measured in minutes.
[tex]\ddot n[/tex] - Angular acceleration, measured in revoiutions per square minute.
The angular acceleration is now cleared:
[tex]\ddot n = \frac{\dot n - \dot n_{o}}{t}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]t = 3.5\,min[/tex], the angular acceleration is:
[tex]\ddot n = \frac{0\,\frac{rev}{min}-7000\,\frac{rev}{min} }{3.5\,min}[/tex]
[tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex]
Now, the final angular speed as a function of initial angular speed, angular acceleration and the change in angular position is represented by this kinematic equation:
[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot (n-n_{o})[/tex]
Where [tex]n[/tex] and [tex]n_{o}[/tex] are the initial and final angular position, respectively.
The change in angular position is cleared herein:
[tex]n-n_{o} = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex], the change in angular position is:
[tex]n-n_{o} = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(7000\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-2000\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n-n_{o} = 12250\,rev[/tex]
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
Suppose you have a lens system that is to be used primarily for 775-nm light. What is the second thinnest coating of fluorite (calcium fluoride) that would be non-reflective for this wavelength?
Answer:
406 nm
Explanation:
We are given;
Wavelength; λ = 775 nm
Refractive index of Calcium fluoride with wavelength of 775 nm as seen in the graph attached is approximately 1.4308.
n = 1.4308
Formula for the thickness of the film that would destruct the light is;
t = (m + 0.5)(λ/2n)
Where m is the order of the thickness.
The first smallest thickness is at m = 0 while the second smallest thickness is at m = 1.
Thus;
t = (1 + 0.5)(775/(2 × 1.4308))
t ≈ 406 nm
g How many rpm would a 25 m diameter Ferris wheel need to travel if a 75 kg person were to experience an effective weight of 810 N at the lower-most point of the ride
Answer:
2.52 rpm
Explanation:
given that
diameter of the wheel, d = 25 m
Mass of the person, m = 75 kg
Weight experienced, N = 810 N
Since diameter is 25, radius then is 25/2 = 12.5 m
We all know that,
v = rw
Also, the passengers weight is equal to the centripetal acceleration, and thus
mg = mv²/r
Substitute for v, we have
mg = m/r * (rw)²
mg = mr²w²/r
g = rw²
If we make w the subject of formula, we have
w² = g/r
w = √(g/r)
mg = 810
75 * g = 810
g = 810 / 75
g = 1.08 m/s²
w = √(g/r)
w = √(1.08 / 12.5)
w = √0.0864
w = 0.294 rad/s
Since the question asked us in rpm, we convert to rpm
0.294 * (60 / 2π)
2.52 revolution per minute.
A 1.8 kg microphone is connected to a spring and is oscillating in simple harmonic motion up and down with a period of 3s. Below the microphone is 1.8 hz, calculate the spring constant
Answer:
230N/m
Explanation:
Pls see attached file
Sergio has made the hypothesis that "the more time that passes, the farther away a person riding a bike will be." Do the data in the table below support his hypothesis? A. Yes, the data support the hypothesis. B. No, the data support the opposite of the hypothesis. C. The data show no relationship between the time passed and the distance.
Answer:
Option A
Explanation:
Given that
Distance = Speed / Time
So, they are in inverse relation.
Such that when the time passes, the distance from the reacing point will become less and vice versa.
So, Yes! The more time that passes, the farther away a person riding a bike will be.
Two parallel wires run in a north-south direction. The eastern wire carries 15.0 A northward while the western wire carries 6.0 A northward. If the wires are separated by 30 cm, what is the magnetic field magnitude and direction at a point between the wires at a distance of 10 cm from the western wire?
Answer:
The magnitude and direction of the magnetic field is 2.7 x 10⁻⁵ T upwards
Explanation:
Given;
current in the eastern wire, [tex]I_e[/tex] = 15 A
current in the western wire, [tex]I_w[/tex] = 6 A
distance between the wires, d = 30 cm = 0.3 m
The magnetic field at a distance R from a line current I, is given as;
[tex]B = \frac{\mu_o I }{2 \pi R}[/tex]
The magnetic field between the wires, are in opposite directions, and since the currents are also in opposite directions, the magnetic fields of the wires will be added.
The total field = magnetic field (east) + magnetic field (west);
[tex]B = \frac{\mu_o I_e}{2 \pi R_e} + \frac{\mu_0 I_w}{2 \pi R_w} \\\\B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})[/tex]
where;
[tex]R_w[/tex] is the distance of the field from west = 10cm = 0.1 m
[tex]R_e[/tex] is the distance of the field on east from west = d - 10cm = 30cm - 10cm = 20cm = 0.2 m
The total magnetic field is;
[tex]B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})\\\\B = \frac{4\pi *10^{-7}}{2\pi} (\frac{15}{0.2} + \frac{6}{0.1})\\\\B = 2*10^{-7}(75 + 60)\\\\B = 2*10^{-7}(135)\\\\B = 2.7*10^{-5} \ T[/tex]
Since total magnetic field is positive, the direction of the field is upwards (positive y direction)
Therefore, the magnitude and direction of the magnetic field is 2.7 x 10⁻⁵ T upwards
A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium:__________.
1. the electric fields at the surfaces of the two spheres are equal.
2. the amount of charge on each sphere is q/2.
3. both spheres are at the same potential. the potentials are in the ratio V2/V1 = q2/q1.
4. the potentials are in the ratio V2/V1 = r2/r1 .
Answer:
Option 3 = both spheres are at the same potential.
Explanation:
So, let us complete or fill the missing gap in the question above;
" A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium BOTH SPHERES ARE AT THE SAME POTENTIAL"
The reason both spheres are at the same potential after the charges on the spheres are in equilibrium is given below:
=> So, if we take a look at the Question again, the kind of connection described in the question above (that is a charged sphere, say X is connected another charged sphere, say Y by a conducting wire) will eventually cause the movement of charges(which initially are not of the same potential) from X to Y and from Y to X and this will continue until both spheres are at the same potential.
A department store expects to have 225 customers and 20 employees at peak times in summer. Determine the contribution of people to the total cooling load of the store. The average rate of heat generation from people doing light work is 115 W, and 70% of it is in sensible form.
Answer:
The contribution of people to the cooling load of the store is 19722.5 W
Explanation:
Total amount of customers = 225
Total amount of employees = 20
Total amount of people in the store at that instant n = 245 people
Average rate of heat generation Q = 115 W
percentage of these heat generated that is sensible heat = 70%
Sensible heat raises the surrounding temperature. Latent heat only causes a change of state.
The total heat generated by all the people in the store = n x Q
==> 245 x 115 = 28175 W
but only 70% of this heat is sensible heat that raises the temperature of the store, therefore, the contribution of people to the cooling load of the store = 70% of 28175 W
==> 0.7 x 28175 = 19722.5 W
A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.60 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.70
Answer:
The induced current is [tex]I = 6.25*10^{-4} \ A[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 1[/tex]
The cross-sectional area is [tex]A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2[/tex]
The initial magnetic field is [tex]B_i = 0.500 \ T[/tex]
The magnetic field at time = 1.02 s is [tex]B_t = 2.60 \ T[/tex]
The resistance is [tex]R = 2.70\ \Omega[/tex]
The induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{ d\phi }{dt}[/tex]
The negative sign tells us that the induced emf is moving opposite to the change in magnetic flux
Here [tex]d\phi[/tex] is the change in magnetic flux which is mathematically represented as
[tex]d \phi = dB * A[/tex]
Where dB is the change in magnetic field which is mathematically represented as
[tex]dB = B_t - B_i[/tex]
substituting values
[tex]dB = 2.60 - 0.500[/tex]
[tex]dB = 2.1 \ T[/tex]
Thus
[tex]d \phi = 2.1 * 8.20 *10^{-4}[/tex]
[tex]d \phi = 1.722*10^{-3} \ weber[/tex]
So
[tex]|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}[/tex]
[tex]|\epsilon| = 1.69 *10^{-3} \ V[/tex]
The induced current i mathematically represented as
[tex]I = \frac{\epsilon}{ R }[/tex]
substituting values
[tex]I = \frac{1.69*10^{-3}}{ 2.70 }[/tex]
[tex]I = 6.25*10^{-4} \ A[/tex]
Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
The object with the twice the area of the other object, will have the larger drag coefficient.
Explanation:
The equation for drag force is given as
[tex]F_{D} = \frac{1}{2}pu^{2} C_{D} A[/tex]
where [tex]F_{D}[/tex] IS the drag force on the object
p = density of the fluid through which the object moves
u = relative velocity of the object through the fluid
p = density of the fluid
[tex]C_{D}[/tex] = coefficient of drag
A = area of the object
Note that [tex]C_{D}[/tex] is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid
The above equation can also be broken down as
[tex]F_{D}[/tex] ∝ [tex]P_{D}[/tex] A
where [tex]P_{D}[/tex] is the pressure exerted by the fluid on the area A
Also note that [tex]P_{D}[/tex] = [tex]\frac{1}{2}pu^{2}[/tex]
which also clarifies that the drag force is approximately proportional to the abject's area.
In this case, the object with the twice the area of the other object, will have the larger drag coefficient.
A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm light from a He-Ne laser. Because the lecture hall is very large, the interference pattern will be projected on a wall that is 4.0 m from the slits. For easy viewing by all students in the class, the professor wants the distance between the m
What slit separation is required in order to produce the desired interference pattern?
d=________m
Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm
Answer:
d = 1.0128×10⁻⁵m
Explanation:
given:
length L = 4.0m
maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m
wavelength λ = 633nm = 633×10⁻⁹m
note:
dsinθ = mλ (constructive interference)
where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength
for small angle
sinθ ≈ tanθ
[tex]d (\frac{y}{L}) =[/tex] mλ
[tex]d (\frac{y}{L}) = (1)(633nm)[/tex]
[tex]d(\frac{0.25}{4} ) = (1)(633nm)[/tex]
d = 1.0128×10⁻⁵m
water and air are both fluids. why is it easier to lift a rock in water rather thatn lifting a rock in air? a the force of gravity. b the bouyant force is greater on the rock in water. c the bouyant force is greater on the rock in air. d the force of gravity on the rock is less in water.
Answer:
The answer is option b.the buoyant force is greater on the rock in water.
In a polar coordinate system, the velocity vector can be written as . The term theta with dot on top is called _______________________ angular velocity transverse velocity radial velocity angular acceleration
Answer:
I believe it's called rapid growth
Explanation:
that is my answer no matter what
A bowling ball of mass 5 kg rolls off the edge of a building 20 meters tall. What is the work done by gravity during the fall, in Joules
Answer:
1000j
Explanation:
work done = force x distance
w = 5 x 10 x 20 = 1000joules
A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval
Answer:
The time interval is [tex]t = 3 \ s[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = 4.0 \ rad/s^2[/tex]
The time taken is [tex]t = 4.0 \ s[/tex]
The angular displacement is [tex]\theta = 80 \ radians[/tex]
The angular displacement can be represented by the second equation of motion as shown below
[tex]\theta = w_i t + \frac{1}{2} \alpha t^2[/tex]
where [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval
So substituting values
[tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]
=> [tex]w_i = 12 \ rad/s[/tex]
Now considering this motion starting from the start point (that is rest ) we have
[tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]
Where [tex]w__{0}}[/tex] is the angular velocity at rest which is zero and [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s
[tex]12 = 0 + 4 t[/tex]
=> [tex]t = 3 \ s[/tex]
Following are the response to the given question:
Given:
[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]
[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]
To find:
[tex]\to \omega=?\\\\\to t=?\\\\[/tex]
Solution:
Using formula:
[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]
It would be the angle for rotation at the start of the 4-second interval.
This duration can be estimated by leveraging the fact that the wheel begins from rest.
[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]
Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".
Learn more:
brainly.com/question/7464119
Equal charges, one at rest, the other having a velocity of 104 m/s, are released in a uniform magnetic field. Which charge has the largest force exerted on it by the magnetic field
Answer:
case 1 of physics is the answer
Two uncharged metal spheres, #1 and #2, are mounted on insulating support rods. A third metal sphere, carrying a positive charge, is then placed near #2. Now a copper wire is momentarily connected between #1 and #2 and then removed. Finally, sphere #3 is removed.
In this final state
a) spheres #1 and #2 are still uncharged.
b) sphere #1 carries negative charge and #2 carries positive charge.
c) spheres #1 and #2 both carry positive charge.
d) spheres #1 and #2 both carry negative charge.
e) sphere #1 carries positive charge and #2 carries negative charge
Answer:
sphere #1 carries positive charge and #2 carries negative charge
This is because from the laws of static electricity, disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charged
Two people, one of mass 85 kg and the other of mass 50 kg, sit in a rowboat of mass 90 kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat, 3.5 m apart from each other, now exchange seats. How far does the boat move?
Answer:
0.11m
Explanation:
let's assume the boat is of uniform construction
Ignoring friction losses
Also assume the origin is at the end of the boat originally with the heavier person
the center of mass of the whole system will not change relative to the water when the two swap ends
Originally, the center of mass is
85[0] + 90[3.5/2] + 50[3.5] / (85 + 90 + 50) = 1.14m from the origin
after the swap, the center of mass is
50[0] + 90[3.5/2] + 85[3.5] / (85 + 90+ 50) = 1.030m from the origin
The center of mass has shifted
1.14-1.030 = 0.11m
as no external force acted on the system, the center of mass relative to the water will not change. The boat will therefore shift towards the end where the heavier person originally sat
An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
a. The radius of the circular path the electron travels
b. The magnitude of the electron's acceleration inside the field
c. The time the electron is in the magnetic field
d. The magnitude of the net force acting on the electron inside the field
Answer:
Explanation:
For circular path in magnetic field
mv² / R = Bqv ,
m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .
a )
R = mv / Bq
If v is changed to 2v , keeping other factors unchanged , R will be doubled
b )
magnitude of acceleration inside field
= v² / R
= Bqv / m
As v is doubled , acceleration will also be doubled
c )
If T be the time inside the magnetic field
T = π R / v
= π / v x mv / Bq
= π m / Bq
As is does not contain v that means T remains unchanged .
d )
Net force acting on electron
= m v² / R = Bqv
Net force = Bqv
As v becomes twice force too becomes twice .
So a . b , d are correct answer.
A conventional current of 3 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.093 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero
Answer:
The current in the small radius loop must be 0.9677 A
Explanation:
Recall that the formula for the magnetic field at the center of a loop of radius R which runs a current I, is given by:
[tex]B=\mu_0\,\frac{I}{2\,R}[/tex]
therefore for the first loop in the problem, that magnetic field strength is:
[tex]B=\mu_0\,\frac{I}{2\,R} =\mu_0\,\frac{3}{2\,(0.093)} =16.129\,\mu_{0}\,[/tex]
with the direction of the magnetic field towards the plane.
For the second smaller loop of wire, since the current goes counterclockwise, the magnetic field will be pointing coming out of the plane, and will subtract from the othe field. In order to the addition of these two magnetic fields to be zero, the magnitudes of them have to be equal, that is:
[tex]16.129\,\,\mu_{0}=\mu_0\,\frac{I'}{2\,R'} =\mu_{0}\,\frac{I'}{2\,(0.03)} \\I'=16.129\,(2)\,(0.03)=0.9677\,\,Amps[/tex]
The phenomenon of magnetism is best understood in terms ofA) the existence of magnetic poles.B)the magnetic fields associated with the movement of charged particles.C)gravitational forces between nuclei and orbital electrons.D) electrical fluid
Answer:
A) the existence of magnetic poles.Explanation:
Magnetism is defined as the ability of a magnet to attract magnetic substance to itself. Such magnet has the ability of being magnetized. A magnet is known to possess poles which are the north poles and south poles. The presence of this poles is what makes them possess the properties of a magnet. An ordinary steel bar doesn't have the properties of a magnet unless it is magnetized and when you are trying to magnetize a steel bar, you are invariably introducing the magnetic poles.
According to the law of magnetism, like poles repel but unlike poles attract. From the above explanation, it can be concluded that the phenomenon of magnetism is best understood interns of existence of magnetic poles. This poles are called the north and the south poles.
Find the rms (a) electric and (b)magnetic fields at a point 2.00 m from a lightbulb that radiates 90.0 W of light uniformly in all directions.
Answer:
a) rms of electric field =
[tex]E_{rms}[/tex]= 25.97 V/m
b) rms of magnetic field
[tex]B_{rms}[/tex] = 8.655 × 10⁻⁸
[tex]B_{rms}[/tex] = 86.55nT
Explanation:
given
power p = 90.0W
distance d = 2.00m
Intensity = [tex]\frac{power}{area}[/tex]
I = [tex]\frac{p}{A}[/tex]
A = [tex]4\pi d^{2}[/tex]
I = [tex]\frac{p}{4\pi d^{2} }[/tex]
I = [tex]\frac{90}{4\pi(2^{2}) }[/tex]
I = 1.79 W/m²
a) [tex]I_{ave}[/tex] = ε₀ × [tex]E^{2} _{rms}[/tex] × c
where ε₀ is permittivity of free space = 8.85×10⁻¹², [tex]E^{2} _{rms}[/tex] is the root mean value and c is speed of light = 3×10⁸m/s
1.79 = 8.85×10⁻¹² × [tex]E^{2} _{rms}[/tex] × 3×10⁸
[tex]E^{2} _{rms}[/tex] = [tex]\frac{1.79}{8.85x10^{-12} x 3x10^{8} }[/tex]
[tex]E^{2} _{rms}[/tex]= 674.1996
[tex]E_{rms}[/tex]= 25.97 V/m
b)for rems magnetic field
[tex]E_{rms}[/tex]= c [tex]B_{rms}[/tex]
[tex]B_{rms}[/tex] = [tex]\frac{E_{rms} }{c}[/tex]
[tex]B_{rms}[/tex] = [tex]\frac{25.97 V/m}{3x10^{8} }[/tex]
[tex]B_{rms}[/tex] = 8.655 × 10⁻⁸
[tex]B_{rms}[/tex] = 86.55nT
The number of neutrons in the nucleus of zinc 65 Zn 30 is:
35
Need more data to answer
65
30
Explanation:
proton number + neutron number = atomic mass
30 + 35 = 65
If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will
If radio waves were used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, Earth would receive their signals at a speed of
Answer:
Explanation:
speed of alien spaceship = .1 c
We shall apply formula of relativistic mechanics to solve the problem
relative velocity =
[tex]\frac{v+v_1}{1 -\frac{v\times v }{c^2} }[/tex]
Here v = v₁ = .1 c
relative velocity = .1c + .1 c / 1 - .1²
= .2 c / .99
= .202 c
The earth would receive the signal at the speed of .202 c .
You are given two infinite, parallel wires, each carrying current.The wires are separated by a distance, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.
A. Is the force between the wires attractive orrepulsive?
B. What is the force per unit length between the two wires?
Answer:
A. Attractive
B. ( μ₀I² ) / ( 2πd )
Explanation:
A. We know that currents in the same direction attract, and currents in the opposite direction repel, according to ampere's law. In this case the current in the two wires are flowing in the same direction, and hence the force between the two wires are attractive.
B. Suppose that two wires of length [tex]l_1[/tex] and [tex]l_2[/tex] both carry the current [tex]I[/tex] in the same direction ( given ). In the presence of a magnetic field produced by wire 1, a force of magnitude m say, is experienced by wire 2. The magnitude of the magnetic field produced by wire 1 at distance say d, from it's axis, should thus be the following -
[tex]B_1[/tex] = μ₀I / 2πd
The force experienced by wire 2 should thus be -
[tex]F_2[/tex] = I( [tex]l_2[/tex] [tex]*[/tex] [tex]B_1[/tex] )
= I [tex]*[/tex] [tex]l_2 * B_1 *[/tex] Sin( 90 )
= I [tex]*[/tex] [tex]l_2[/tex] ( μ₀I / 2πd )
Therefore the force per unit length experienced by wire 2 toward wire 1 should be ...
( [tex]F_2[/tex] / [tex]l_2[/tex] ) = ( μ₀I² ) / ( 2πd ) ... which is our solution
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
Answer:
Option A
Explanation:
From the graph, we came to know that Force and acceleration are in direct relationship.
Also,
Force = 0 when Acceleration = 0
Because Both are 0 at the origin.
Answer:
A. It will be 0 meters per second per second.
Explanation:
The force and acceleration is in a proportional relationship, that means the line goes through the origin.
On the graph, when the force is at 0, the acceleration is 0. The line passes through the origin.
g Question 11 pts Consider two masses connected by a string hanging over a pulley. The pulley is a uniform cylinder of mass 3.0 kg. Initially m1 is on the ground and m2 rests 2.9 m above the ground. After the system is released, what is the speed of m2 just before it hits the ground? m1= 30 kg and m2= 35 kg Group of answer choices 2.1 m/s 1.4 m/s 9.8 m/s 4.3 m/s 1.9 m/s
Answer:
The speed of m2 just before it hits the ground is 2.1 m/s
Explanation:
mass on the ground m1 = 30 kg
mass oat rest at the above the ground m2 = 35 kg
height of m2 above the ground =2.9 m
Let the tension on the string be taken as T
for the mass m2 to reach the ground, its force equation is given as
[tex]m_{2} g - T = m_{2}a[/tex] ....equ 1
where g is acceleration due to gravity = 9.81 m/s^2
and a is the acceleration with which it moves down
For mass m1 to move up, its force equation is
[tex]T - m_{1} g = m_{1} a[/tex]
[tex]T = m_{1}a + m_{1}g[/tex]
[tex]T = m_{1}(a + g)[/tex] ....equ 2
substituting T in equ 1, we have
[tex]m_{2} g - m_{1}(a+g) = m_{2}a[/tex]
imputing values, we have
[tex](35*9.81) - 30(a+9.81) = 35a[/tex]
[tex]343.35 - 30a-294.3 = 35a[/tex]
[tex]343.35 -294.3 = 35a+ 30a[/tex]
[tex]49.05 = 65a[/tex]
a = 49.05/65 = 0.755 m/s^2
The initial velocity of mass m2 = u = 0
acceleration of mass m2 = a = 0.755 m/s^2
distance to the ground = d = 2.9 m
final velocity = v = ?
using Newton's equation of motion
[tex]v^{2}= u^{2} + 2ad[/tex]
substituting values, we have
[tex]v^{2}= 0^{2} + 2*0.755*2.9[/tex]
[tex]v^{2}= 2*0.755*2.9 = 4.379\\v = \sqrt{4.379}[/tex]
v = 2.1 m/s