Complete Question
A 590-turn solenoid is 12 cm long. The current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?
Answer:
The force is [tex]F = 0.1602 \ N[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 590 \ turns[/tex]
The length of the solenoid is [tex]L = 12 \ cm = 0.12 \ m[/tex]
The current is [tex]I = 36 \ A[/tex]
The diameter is [tex]D = 4.5 \ cm = 0.045 \ m[/tex]
The current carried by the wire is [tex]I = 27 \ A[/tex]
The length of the wire is [tex]l = 2 cm = 0.02 \ m[/tex]
Generally the magnitude of the force on this wire assuming the solenoid's field points due east is mathematically represented as
[tex]F = B * I * l[/tex]
Here B is the magnetic field which is mathematically represented as
[tex]B = \frac{\mu_o * N * I }{L}[/tex]
Here [tex]\mu _o[/tex] is permeability of free space with value [tex]\mu_ o = 4\pi *10^{-7} \ N/A^2[/tex]
substituting values
[tex]B = \frac{4 \pi *10^{-7} * 590 * 36 }{ 0.12}[/tex]
[tex]B = 0.2225 \ T[/tex]
So
[tex]F = 0.2225 * 36 * 0.02[/tex]
[tex]F = 0.1602 \ N[/tex]
Which does not account for the fact that fish can survive the winter in ponds in temperate climate zones? 1. the density of ice versus that of water 2. the unique properties of water 3. the intermolecular bonding of water 4. the tendency for water to freeze from the bottom up
Answer:
3. the intermolecular bonding of water
Explanation:
Anomalous behavior of water is an advantage in aquatic habitat during winter. Because of some unique properties of water, it behaves irregularly. Thus, a pond or river does not freeze completely during winter.
Water has its highest density when temperature is 4[tex]^{0}C[/tex] , and lowest volume at 4[tex]^{0}C[/tex]. Thus, the denser layers of water sink accordingly until the upper layer is the least dense during winter. This layer then freeze leaving the layers below it unfrozen.
Answer:
D. The tendency for water to freeze from the bottom up.
Explanation:
Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the concentric spheres. When compared to the number of field lines N1 going through the sphere of radius R, the number of electric field lines N2 going through the sphere of radius 2R is
Answer:
N2 = ¼N1
Explanation:
First of all, let's define the terms;
N1 = number of electric field lines going through the sphere of radius R
N2 = number of electric field lines going through the sphere of radius 2R
Q = the charge enclosed at the centre of concentric spheres
ε_o = a constant known as "permittivity of the free space"
E1 = Electric field in the sphere of radius R.
E2 = Electric field in the sphere of radius 2R.
A1 = Area of sphere of radius R.
A2 = Area of sphere of radius 2R
Now, from Gauss's law, the electric flux through the sphere of radius R is given by;
Φ = Q/ε_o
We also know that;
Φ = EA
Thus;
E1 × A1 = Q/ε_o
E1 = Q/(ε_o × A1)
Where A1 = 4πR²
E1 = Q/(ε_o × 4πR²)
Similarly, for the sphere of radius 2R,we have;
E2 = Q/(ε_o × 4π(2R)²)
Factorizing out to get;
E2 = ¼Q/(ε_o × 4πR²)
Comparing E2 with E1, we arrive at;
E2 = ¼E1
Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;
N2 = ¼N1
What is the density of the unknown fluid in Figure below? ρwater = 1000 kgm−3
Answer:
2500 kg/m³
Explanation:
P = P
ρgh = ρgh
ρh = ρh
(1000 kg/m³) (8.9 cm) = ρ (3.5 cm)
ρ ≈ 2500 kg/m³
A small glass bead charged to 5.0 nC is in the plane that bisects a thin, uniformly charged, 10-cm-long glass rod and is 4.0 cm from the rod's center. The bead is repelled from the rod with a force of 910 N. What is the total charge on the rod?
Answer:
Explanation:
Let B= bead
Q = rod
the electric field at the glass bead pocation is
(Gauss theorem)
E = Q / (2 π d L εo)
the force is
F = q E = q Q / (2 π d L εo)
then
Q = 2 π d L εo F / q
Q = 2*3.14*4x10^-2*10^-1*8.85x10^-12*910x10^-4 / 5x10^-9 = 2.87x10^-8 C = 40.5 nC
1. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
2. Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling with slipping up an incline with the same initial linear (translational) speed. Which goes farthest up
the incline?
a. the ball
b. the disk
c. the hoop
d. the hoop and the disk roll to the same height, farther
than the ball
e. they all roll to the same height
Answer:
The hoop
Explanation:
Because it has a smaller calculated inertia of 2/3mr² compares to the disc
P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism articulated to the ground at its vertex A, while vertex C is attached to the vertical cord fixed to the ground. If the coefficient of friction between the prism and the blocks is 0.4; determine the maximum angle that measures the inclination of the AC face with respect to the horizontal so that the system remains in equilibrium.
Answer:
21.8°
Explanation:
Let's call θ the angle between BC and the horizontal.
Draw a free body diagram for each block.
There are 4 forces acting on block D:
Weight force P pulling down,
Normal force N₁ pushing perpendicular to AB,
Friction force N₁μ pushing parallel up AB,
and tension force T pushing parallel up AB.
There are 4 forces acting on block E:
Weight force P pulling down,
Normal force N₂ pushing perpendicular to BC,
Friction force N₂μ pushing parallel to BC,
and tension force T pulling parallel to BC.
Sum of forces on D in the perpendicular direction:
∑F = ma
N₁ − P sin θ = 0
N₁ = P sin θ
Sum of forces on D in the parallel direction:
∑F = ma
T + N₁μ − P cos θ = 0
T = P cos θ − N₁μ
T = P cos θ − P sin θ μ
T = P (cos θ − sin θ μ)
Sum of forces on E in the perpendicular direction:
∑F = ma
N₂ − P cos θ = 0
N₂ = P cos θ
Sum of forces on E in the parallel direction:
∑F = ma
N₂μ + P sin θ − T = 0
T = N₂μ + P sin θ
T = P cos θ μ + P sin θ
T = P (cos θ μ + sin θ)
Set equal:
P (cos θ − sin θ μ) = P (cos θ μ + sin θ)
cos θ − sin θ μ = cos θ μ + sin θ
1 − tan θ μ = μ + tan θ
1 − μ = tan θ μ + tan θ
1 − μ = tan θ (μ + 1)
tan θ = (1 − μ) / (1 + μ)
Plug in values:
tan θ = (1 − 0.4) / (1 + 0.4)
θ = 23.2°
∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.
The hydrogen spectrum has a red line at 656 nm, and a blue line at 434 nm. What is the first order angular separation between the two spectral lines obtained with a diffraction grating with 5000 rulings/cm?
Answer:
Explanation:
grating element or slit width a = 1 x 10⁻² / 5000
= 2 x 10⁻⁶ m
angular width of first order spectral line of wavelength λ
= λ / a
for blue line angular width
= 434 x 10⁻⁹ / 2 x 10⁻⁶ radian
= 217 x 10⁻³ radian
for red line angular width
= 656 x 10⁻⁹ / 2 x 10⁻⁶ radian
= 328 x 10⁻³ radian
difference of their angular width
= 328 x 10⁻³ - 217 x 10⁻³
= 111 x 10⁻³ radian
Ans .
g A projectile is fired from the ground at an angle of θ = π 4 toward a tower located 600 m away. If the projectile has an initial speed of 120 m/s, find the height at which it strikes the tower
Answer:
The projectile strikes the tower at a height of 354.824 meters.
Explanation:
The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:
Horizontal motion
[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]
Vertical motion
[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]
Where:
[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.
[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
The time spent for the projectile to strike the tower is obtained from first equation:
[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]
If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:
[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]
[tex]t \approx 7.071\,s[/tex]
Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])
[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]
[tex]y \approx 354.824\,m[/tex]
The projectile strikes the tower at a height of 354.824 meters.
1. A 0.430kg baseball comes off a bar and goes straight up in the air. At a height of 10.0m, the baseball has a speed of 25.3m/s. Determine the mechanical energy at the height. Show all your work. 2. What is the baseball's mechanical energy when it is at a height of 8.0m? Explain?
Answer:
180 J
Explanation:
Mechanical energy = kinetic energy + potential energy
ME = KE + PE
ME = ½ mv² + mgh
ME = ½ (0.430 kg) (25.3 m/s)² + (0.430 kg) (9.8 m/s²) (10.0 m)
ME = 180 J
Mechanical energy is conserved, so it is 180 J at all points of the trajectory.
The baseball's mechanical energy when it is at a height of 8.0m is 180 J.
What is mechanical energy?The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time. Mechanical energy is always conserved.
Mechanical energy = kinetic energy + potential energy
Given is the mass of baseball m= 0.430 kg, height h =10m, speed v= 25.3m/s.
ME = KE + PE
ME = ½ mv² + mgh
Substitute the values, we get
ME = ½ (0.430 kg) (25.3 m/s)² + (0.430 kg) (9.8 m/s²) (10.0 m)
ME = 180 J
Thus, the baseball's mechanical energy when it is at a height of 8.0m is 180 J.
Learn more about mechanical energy.
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of
The radii a wheel are 25 cm
and 5cm respectively, it is found
that an effort of 40N is required
to raise slowly a load 16ON
160 N. Find the Mechanical
Adventage and Effeciency,
Answer:
Explanation:
Given that
Effort = 40N
Load = 16ON
M.A = load/effort
M.A= 160N/40N
M.A = 4
Velocity ratio = V.R =radius of the wheel/radius of the axel
= 25cm/5cm
= 5
Efficiency = mechanical advantage/velocity ratio × 100/1
= 4/5 × 100/1
= 0.8×100/1
= 80%
Hence, the mechanical advantage of the machine is 4 while the efficiency is 80%.
y=k/x, x is halved.
what happens to the value of y
Answer:
y is doubled
Explanation:
If x is halved, that means the value is doubled. Here is an exmaple:
y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.
A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle. What is the y-component of the velocity of the second ball?
Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
[tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s
Seismic attenuation and how spherical spreading affect amplitude, can anyone explain this please!
Answer:
Hey there!
This can be a confusing topic, so it's totally fine if you get confused...
First, Seismic Attenuation is how seismic waves lose energy as they expand and spread.
Secondly, when distance increases, amplitude decreases. This is because the distance (spherical spreading would mean radius) is inversely proportional to amplitude.
Let me know if this helps :)
A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete swing?
Answer:
Time taken for 1 swing = 3.81 second
Explanation:
Given:
Time taken for 1 swing = 2.20 Sec
Find:
Time taken for 1 swing , when triple the length(T2)
Computation:
Time taken for 1 swing = 2π[√l/g]
2.20 = 2π[√l/g].......Eq1
Time taken for 1 swing , when triple the length (3L)
Time taken for 1 swing = 2π[√3l/g].......Eq2
Squaring and dividing the eq(1) by (2)
4.84 / T2² = 1 / 3
T2 = 3.81 second
Time taken for 1 swing = 3.81 second
What is the current in milliamperes produced by the solar cells of a pocket calculator through which 4.2 C of charge passes in 2.7 h
Answer:
0.432mAExplanation:
Current produced by the solar cells of the pocket calculator is expressed using the formula I = Q/t where;
Q is the charge (in Columbs)
t is the time (in seconds)
Given parameters
Q = 4.2C
t = 2.7 hrs
t = 2.7*60*60
t = 9720 seconds
Required
Current produced by the solar cell I
Substituting the given values into the formula;
I = 4.2/9720
I = 0.000432A
I = 0.432mA
Hence, the current in milliamperes produced by the solar cells of a pocket calculator is 0.432mA
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that carries a 4.5-A current
Answer:
Explanation:
From the question we are told that
The radius is [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]
The current is [tex]I = 4.5 \ A[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{J}{\sigma }[/tex]
Where [tex]\sigma[/tex] is the conductivity of aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]
J is the current density which mathematically represented as
[tex]J = \frac{I}{A}[/tex]
Here A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (1.4*10^{-3})^2[/tex]
[tex]A = 6.158*10^{-6} \ m^2[/tex]
So
[tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]
[tex]J = 730757 A/m^2[/tex]
So
[tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]
[tex]E = 0.021 \ N/C[/tex]
1. What does the acronym LASER stand for? What characteristic of a laser makes it suitable for today's experiment?
Answer:Light Amplification by Stimulated Emission of Radiation. It is able to convert light or electrical energy into focused high energy beam to treat some sickness and diseases.
Explanation:
Answer:
Light amplification by stimulated emission of radiation
A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the coefficient of contraction is 0.68 and the coefficient of velocity is 0.92, what is the discharge?
Answer:
The discharge rate is [tex]Q = 0.0192 \ m^3 /s[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 60 \ mm = 0.06 \ m[/tex]
The head is [tex]h = 6 \ m[/tex]
The coefficient of contraction is [tex]Cc = 0.68[/tex]
The coefficient of velocity is [tex]Cv = 0.92[/tex]
The radius is mathematically evaluated as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{ 0.06 }{2}[/tex]
[tex]r = 0.03 \ m[/tex]
The area is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * (0.03)^2[/tex]
[tex]A = 0.00283 \ m^2[/tex]
The discharge rate is mathematically represented as
[tex]Q = Cv *Cc * A * \sqrt{ 2 * g * h}[/tex]
substituting values
[tex]Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}[/tex]
[tex]Q = 0.0192 \ m^3 /s[/tex]
a transformer changes 95 v acorss the primary to 875 V acorss the secondary. If the primmary coil has 450 turns how many turns does the seconday have g
Answer:
The number of turns in the secondary coil is 4145 turns
Explanation:
Given;
the induced emf on the primary coil, [tex]E_p[/tex] = 95 V
the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V
the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns
the number of turns in the secondary coil, [tex]N_s[/tex] = ?
The number of turns in the secondary coil is calculated as;
[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]
[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]
Therefore, the number of turns in the secondary coil is 4145 turns.
A long, horizontal hose of diameter 5.4 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1.2 cm. Water squirts from the nozzle at velocity 20 m/sec. Assume that the water has no viscosity or other form of energy dissipation.
a) What is the velocity of the water in the hose?
b) What is the pressure differential between the water in the hose and water in the nozzle?
c) How long will it take to fill a tub of volume 120 liters with the hose?
Answer:
a) 0.988 m/s
b) 199512 Pa
c) 57.52 s
Explanation:
given that
A
A1 v1 = A2 v2
d1² v1 = d2² v2
v2 = [d1/d2]² v1
v2 = (1.2/5.4)² * 20
v2 = 0.049 * 20
v2 = 0.988 m/s
B
P + 1/2 ρ v² = K.
[p2 - p1] = 1/2 ρ [v1² - v2²]
[p2 - p1] = 1/2 * 1000 [20² - 0.988²]
[p2 - p1] = 500 * (400 - 0.976)
[p2 - p1] = 500 * 399
[p2 - p1] = 199512 Pa
C
Flow rate = AV = π [d²/ 4 ] * v
= π [0.012² / 4 ] * 20 = 0.00226 m³ /s
= π [0.054² / 4 ] * 0.988 = 0.00226 m³ /s
130 liters = 0.13 m³
t = 0.13/ 0.00226 = 57.52 s
A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision.
(a) Is she nearsighted or farsighted?
(b) What type of lens (converging or diverging) is needed to correct her vision?
(c) What focal length contact lens is needed, and what is its power in diopters?
Answer:
(a) nearsighted
(b) diverging
(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D
Explanation:
(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.
(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)
(c) the absolute value of the focal length (f) is given by the formula:
[tex]f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D[/tex]
So it would normally be written with a negative signs in front indicating a divergent lens.
Suppose you are playing hockey on a new-age ice surface for which there is no friction between the ice and the hockey puck. You wind up and hit the puck as hard as you can. After the puck loses contact with your stick, the puck will
Answer:
Not slow down or speed up.
Explanation:
Hitting the puck accelerates the speed of the puck from zero to the speed with which it leaves at the instance they lose contact. Since there is no friction between the puck and the ice, there will be no force decelerating or accelerating the hockey puck, allowing the puck to move away and remain in motion without speeding up or slowing down indefinitely theoretically.
Convert 76.2 kilometers to meters?
Answer
76200meters
Explanation:
we know that 1km=1000meters
to convert km into meters we we divide km by meters
=76.2/1000
=76200meters
A square loop, length l on each side, is shot with velocity v0 into a uniform magnetic field B. The field is perpendicular to the plane of the loop. The loop has mass m and resistance R, and it enters the field at t = 0s. Assume that the loop is moving to the right along the x-axis and that the field begins at x = 0m.
Required:
Find an expression for the loop's velocity as a function of time as it enters the magnetic field.
Answer:
v₀(1 + B²L²t/mR)
Explanation:
We know that the force on the loop is F = BIL where B = magnetic field strength, I = current and L = length of side of loop. Now the current in the loop I = ε/R where ε = induced e.m.f in the loop = BLv₀ where v₀ = velocity of loop and r = resistance of loop
F = BIL = B(BLv₀)L/R = B²L²v₀/R
Since F = ma where a = acceleration of loop and m = mass of loop
a = F/m = B²L²v₀/mR
Using v = u + at where u = initial velocity of loop = v₀, t = time after t = 0 and v = velocity of loop after time t = 0
Substituting the value of a and u into v, we have
v = v₀ + B²L²v₀t/mR
= v₀(1 + B²L²t/mR)
So the velocity of the loop after time t is v = v₀(1 + B²L²t/mR)
The expression for the loop's velocity as a function of time as it enters the magnetic field is v = v₀(1 + B²L²t/mR).
Calculation of the loop velocity:As we know that
Force on the loop
F = BIL
here
B = magnetic field strength,
I = current
and L = length of side of loop.
Now
the current in the loop I = ε/R
where
ε = induced e.m.f in the loop = BLv₀
where v₀ = velocity of loop
and r = resistance of loop
So,
F = BIL = B(BLv₀)L/R = B²L²v₀/R
Also, F = ma where a = acceleration of loop and m = mass of loop
Now
a = F/m = B²L²v₀/mR
We have to use
v = u + at
where
u = initial velocity of loop = v₀,
t = time after t = 0
and v = velocity of loop after time t = 0
So, it be like
v = v₀ + B²L²v₀t/mR
= v₀(1 + B²L²t/mR)
Learn more about velocity here: https://brainly.com/question/332163
A rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing down. What is the direction of the induced current in the coil?
Answer:
There is no induced current on the coil.
Explanation:
Current is induced in a coil or a circuit, when there is a break of flux linkage. A break in flux linkage is caused by a changing magnetic field, and must be achieved by a relative motion between the coil and the magnet. Holding the magnet above the center of the coil will cause no changing magnetic filed since there is no relative motion between the coil and the magnet.
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.
What is the observed wavelength of the 656.3 nm (first Balmer) line of hydrogen emitted by a galaxy at a distance of 2.40 x 108 ly
Answer:
λ = 667.85 nm
Explanation:
Let f be the frequency detected by the observer
Let v be the speed at which the observer is moving.
Now, when the direction at which the observer is moving is away from the source, we have the frequency as;
f = f_o√((1 - β)/(1 + β))
From wave equations, we know that the wavelength is inversely proportional to the frequency. Thus, wavelength is now;
λ = λ_o√((1 + β)/(1 - β))
Where, β = Hr/c
H is hubbles constant which has a value of 0.0218 m/s • ly
c is speed of light = 3 × 10^(8) m/s
r is given as 2.40 x 10^(8) ly
Thus,
β = (0.0218 × 2.4 x 10^(8))/(3 × 10^(8))
β = 0.01744
Since we are given λ_o = 656.3 nm
Then;
λ = 656.3√((1 + 0.01744)/(1 - 0.01744))
λ = 667.85 nm
3. Which of the following accurately describes circuits?
O A. In a parallel circuit, the same amount of current flows through each part of the circuit
O B. In a series circuit, the amount of current passing through each part of the circuit may vary
O C. In a series circuit, the current can flow through only one path from start to finish
O D. In a parallel circuit, there's only one path for the current to travel.
Answer:
Option (c)
Explanation:
In a Series circuit, as the components are connected end-to-end ,the current can flow through only one path from start to finish.
(C.) is the only correct statement in the list of choices.
In a series circuit, the current can flow through only one path from start to finish.
A steel bridge is 1000 m long at -20°C in winter. What is the change in length when the temperature rises to 40°C in summer? The average coefficient of linear expansion of this steel is 11 × 10-6 C-1.
Answer:
ΔL = 0.66 m
Explanation:
The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:
ΔL = αLΔT
where,
ΔL = Change in Length of the bridge = ?
α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹
L = Original Length of the Bridge = 1000 m
ΔT = Change in Temperature = Final Temperature - Initial Temperature
ΔT = 40°C - (-20°C) = 60°C
Therefore,
ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)
ΔL = 0.66 m
Find the momentum of a particl with a mass of one gram moving with half the speed of light.
Answer:
129900
Explanation:
Given that
Mass of the particle, m = 1 g = 1*10^-3 kg
Speed of the particle, u = ½c
Speed of light, c = 3*10^8
To solve this, we will use the formula
p = ymu, where
y = √[1 - (u²/c²)]
Let's solve for y, first. We have
y = √[1 - (1.5*10^8²/3*10^8²)]
y = √(1 - ½²)
y = √(1 - ¼)
y = √0.75
y = 0.8660, using our newly gotten y, we use it to solve the final equation
p = ymu
p = 0.866 * 1*10^-3 * 1.5*10^8
p = 129900 kgm/s
thus, we have found that the momentum of the particle is 129900 kgm/s
