Answer:
5 V
Explanation:
The maximum voltage the voltmeter can read will be the voltage drop across the 20 Ω resistance and the voltage drop across the 4980 Ω resistance.
V' = Ir +IR.................... equation 1
Where V' = Maximum voltage the voltmeter can read, I = current, r = resistance of the galvanometer coil, R = The resistance connected in series to the galvanometer.
Given: I = 1 mA = 0.001 A, r = 20Ω, R = 4980Ω
Substitute into equation 1
V' = 20(0.001)+4980(0.001)
V' = 0.02+4.98
V' = 5 V
Answer:
Explanation:
E = IR + Ir
Where,
I = current
= 1 mA
= 0.001 A
R = 4980 Ω
Internal resistance, r = 20 Ω
Maximum voltage, E = 0.001 × 4980 + 0.001 × 20
= 4.98 + 0.02
= 5 V
You are operating an 80kg reciprocating machine. The manufacturer notified you thatthere is an imbalance mass of 3kg on the rotating shaft, which has a 10cm diameter.The system was designed to have negligible damping.P.1.1What is the steady state amplitude of the machine’s displacement if you are operatingat very high frequencies?
Answer:
Explanation:
The system can be modeled as,
Using magnitude relationship for imbalance system.
Check attachment for solution
An electron moving at4.00 × 103m/sin a 1.25-Tmagnetic field experiences a magnetic force of1.40 × 10−16N.What angle does the velocity of theelectron make with the magnetic field? There are twoanswers
Answer:
Explanation:
velocity of electron V = 4 x 10³ m/s
magnetic field B = 1.25 T .
magnetic force = 1.4 x 10⁻¹⁶ N.
If direction of velocity makes angle θ with magnetic field
magnetic force = magnetic field x charge on electron x velocity x sinθ
1.4 x 10⁻¹⁶ = 1.25 x 1.6 x 10⁻¹⁹ x 4 x 10³ x sinθ
1.4 x 10⁻¹⁶ = 8 x 10⁻¹⁶ x sinθ
sinθ = 1.4 / 8
= .175
θ = 10 degree
or 180 - 10
= 170 degree
because
sinθ = sin (180 - θ)
Sound with frequency 1240 Hz leaves a room through a doorway with a width of 1.13 m . At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344 m/s for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections. Express your answer in radians.
Given Information:
frequency = 1240 Hz
width = a = 1.13 m
speed of sound = c = 344 m/s
Required Information:
angle = θ = ?
Answer:
θ = 14.18 rad
Explanation:
We can find out the angle relative to the centerline perpendicular to the doorway by using the following relation
sin(θ) = λ/a
Where λ is the wavelength of the sound wave and a is width
λ = c/f
Where c is the speed of the sound and f is the frequency
λ = 344/1240
λ = 0.277
sin(θ) = λ/a
θ =sin⁻¹(λ/a)
θ =sin⁻¹(0.277/1.13)
θ =sin⁻¹(0.277/1.13)
θ = 14.18 rad
For a two-level system, the weight of a given energy distribution can be expressed in terms of the number of systems, N, and the number of systems occupying the excited state, n1. What is the expression for weight in terms of these quantities
Answer:
W = N!/(n0! * n1!)
Explanation:
Let n0 = number of particles in the lowest energy state
n1 = number of particles in the excited energy state.
Using this, we can say that N = n0 + n1
From this we can then express the weight, W of the close system by finding the factorials of each particles
W = N!/(n0! * n1!)
Hence, the weight W is expressed as W = N!/(n0! * n1!)
A steel cable supports an actor as he swings onto the stage. The weight of the actor stretches the steel cable. To describe the relationship between stress and strain for the steel cable, you would use:
a. both shear modulus and bulk modulus.
b. Young's modulus
c. bulk modulus.
d. shear modulus
e. both Young's modulus and bulk modulus.
Answer:
Option (b)
Explanation:
As the actor swings on a steel cable, the cable stretches due to the weight of the actor. So, the stress produced by the weight of the actor is longitudinal stress and the strain is also longitudinal in nature.
The modulus of elasticity associated to the normal stress and the longitudinal strain is young's modulus.
By clicking the two-slit barrier and dragging it, you can change its position. Always allow a few seconds for the simulation to catch up with the changes you are making before analyzing the results. When the two-slit barrier is brought closer to the screen (and farther away from the source), what happens to the distance between two adjacent bright (or dark) fringes?
A. It becomes larger.
B. It becomes smaller.
C. It stays the same.
Answer:
the answer the correct one is B
Explanation:
The interference phenomenon is described by
d sin θ = m λ
The spectrum is recorded on a screen, so we can use trigonometry
tan θ = y / L
In this experiment the angles are very small, so
tan θ = sin θ /cos θ = sin θ
We replace
d y / L = m La m
y = m Lam L / d
When the slits approach the screen the value of L decreases,
Therefore the value of the separation between the slits must also decrease
When reviewing the answer the correct one is B
A real battery with internal resistance 0.460 Ω and emf 9.00 V is used to charge a 56.0-µF capacitor. A 21.0-Ω resistor is put in series with the battery and the capacitor when charging. (a) What is the time constant for this circuit?
Answer: 1.176×10^-3 s
Explanation: The time constant formulae for an RC circuit is given below as
t =RC
Where t = time constant , R = magnitude of resistance = 21 ohms , C = capacitance of capacitor = 56 uf = 56×10^-6 F
t = 56×10^-6 × 21
t = 1176×10^-6
t = 1.176×10^-3 s
Given Information:
Internal resistance of battery = 0.460
Resistance = 21.0 Ω
Capacitance = 56.0 µF
Required Information:
time constant = τ = ?
Answer:
τ = 0.0012
Explanation:
The time constant τ provides the information about how long it will take to charge the capacitor up to certain level.
τ = Req*C
Where Req is the equivalent resistance and C is the capacitance.
Req = R + r
Where R is the resistance of the resistor and r is the internal resistance of the battery.
Req = 21.0 + 0.460 = 21.460 Ω
τ = Req*C
τ = 21.46*56x10⁻⁶
τ = 0.0012
A capacitor approximately charges to 63% in one τ and about 99% in 5τ
A simply supported wood beam with a span of L = 15 ft supports a uniformly distributed load of w0 = 270 lb/ft. The allowable bending stress of the wood is 1.95 ksi. If the aspect ratio of the solid rectangular wood beam is specified as h/b = 1.75, calculate the minimum width b that can be used for the beam.
The minimum width b that can be used for the beam is 4.32 in
To determine the minimum width (b) of the solid rectangular wood beam, we employ principles of structural engineering.
The maximum bending moment (M) for a simply supported beam with a uniformly distributed load (\(w_0\)) occurs at the center and is given by [tex]\(M = \frac{w_0 L^2}{8}\).[/tex]
The section modulus (S) for a rectangular cross-section is [tex]\(S = \frac{b \times h^2}{6}\).[/tex]
The bending stress (\(σ_b\)) is given by[tex]\(σ_b = \frac{M}{S}\)[/tex].
Setting \(σ_b\) equal to the allowable bending stress [tex](\(1.95 \, \text{ksi}\))[/tex]and using the specified aspect ratio (h/b = 1.75), we can solve for b.
This ensures that the wood beam meets structural safety criteria.
The calculation yields a minimum width b of 4.32 in, ensuring that the beam is structurally sound, withstanding the specified uniformly distributed load and adhering to the aspect ratio constraint in compliance with the allowable bending stress.
A couple is defined as two parallel forces, separated by a distance, that have equal magnitudes but opposite directions. A couple only produces a rotation in a specified direction. The moment produced by a couple is called a couple moment.
A submarine hatch door is to be opened by applying two oppositely oriented forces of equal magnitude F=265N at points A and B on the hatch door wheel. The radii of the wheel's inner and outer rings are r1 = 0.470 m and r2 = 0.200 m, respectively.
Calculate the moments MA and MB about point D for the forces applied at points A and B. Then, determine the resulting couple moment MR. Assume that a positive moment produces a counterclockwise rotation whereas a negative moment produces a clockwise rotation.
Answer:
MA = 178 Nm
MB = 72 Nm
MR = 249 Nm
For the moments MA the lever arm was taken as r1 + r2 which is the distance of the point of application of the force at point A to point D.
For the moment MB, the distance of the point of application of force is r1 - r2 which is the distance from the outer ring to the inner ring.
The couple moment is given by F × r1. Which is basically a sum of the moments of both forces applied on the wheel.
Explanation:
See the attachment for detail of the calculation.
The moment MA for the force at point A is 124.55 N*m, the moment MB for the force at point B is 53 N*m, and the resulting couple moment MR is 71.55 N*m.
Explanation:To calculate the moments MA and MB about point D for the forces applied at points A and B, we can use the formula:
M = F * r
Where M is the moment, F is the force, and r is the radius. For point A, the radius is r1 = 0.470 m and the force is F = 265 N. So, MA = 265 N * 0.470 m = 124.55 N*m (counterclockwise).
Similarly, for point B, the radius is r2 = 0.200 m and the force is F = 265 N. So, MB = 265 N * 0.200 m = 53 N*m (clockwise).
To determine the resulting couple moment MR, we can subtract the clockwise moment (MB) from the counterclockwise moment (MA). MR = MA - MB = 124.55 N*m - 53 N*m = 71.55 N*m (counterclockwise).
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Two protons are released from rest when they are 0.750 {\rm nm} apart.
a) What is the maximum speed they will reach?
b) What is the max acceleration they will reach?
Explanation:
Given:
m = 1.673 × 10^-27 kg
Q = q = 1.602 × 10^-19 C
r = 0.75 nm
= 0.75 × 10^-9 m
A.
Energy, U = (kQq)/r
Ut = 1/2 mv^2 + 1/2 mv^2
1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9
v = 1.356 × 10^4 m/s
B.
F = (kQq)/r^2
F = m × a
1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2
a = 2.45 × 10^17 m/s^2.
Light of wavelength 550 nm comes into a thin slit and produces a diffraction pattern on a board 8.0 m away. The first minimum dark fringe appears 3 mm from the central maximum. What is the width of the slit?
Answer:
Width of the slit will be equal to 1.47 mm
Explanation:
We have given wavelength of the light [tex]\lambda =550nm=550\times 10^{-9}m[/tex]
Distance D = 8 m
Distance between first minimum dark fringe and the central maximum is 2 mm
So [tex]x=3\times 10^{-3}m[/tex]
We have to find the width of the slit
For the first order wavelength is equal to [tex]\lambda =\frac{x}{D}\times a[/tex], here a width of slit
So [tex]a=\frac{\lambda D}{x}=\frac{550\times 10^{-9}\times 8}{3\times 10^{-3}}=1466.666\times 10^{-6}=1.47mm[/tex]
So width of the slit will be equal to 1.47 mm
The inner and outer surfaces of a cell membrane carry a negative and positive charge respectively. Because of these charges, a potential difference of about 70 mV exists across the membrane. The thickness of the membrane is 8 nm. If the membrane were empty (filled with air), what would the magnitude of the electric field inside the membrane
Answer:
The magnitude of the electric field inside the membrane is 8.8×10⁶V/m
Explanation:
The electric field due to electric potential at a distance Δs is given by
E=ΔV/Δs
We have to find the magnitude electric field in the membrane
Ecell= -ΔV/Δs
[tex]E_{cell}=-\frac{V_{in}-V_{out}}{s} \\E_{cell}=\frac{V_{out}-V_{in}}{s}\\E_{cell}=\frac{0.070V}{8*10^{-9}m } \\E_{cell}=8.8*10^{6}V/m[/tex]
The magnitude of the electric field inside the membrane is 8.8×10⁶V/m
The magnitude of the electric field inside the membrane, if it were empty, would be approximately [tex]\( 8.75 \times 10^6 \) V/m.[/tex]
To find the magnitude of the electric field inside the membrane, we can use the formula for electric field strength ( E ) due to a uniform field between two parallel plates, which is given by:
[tex]\[ E = \frac{V}{d} \][/tex]
where ( V ) is the potential difference across the plates (in volts) and ( d ) is the separation between the plates (in meters). In this case, the potential difference ( V ) is given as 70 mV, which we need to convert to volts:
[tex]\[ V = 70 \text{ mV} = 70 \times 10^{-3} \text{ V} \][/tex]
The thickness of the membrane ( d ) is given as 8 nm, which we need to convert to meters:
[tex]\[ d = 8 \text{ nm} = 8 \times 10^{-9} \text{ m} \][/tex]
Now we can plug these values into the formula for the electric field:
[tex]\[ E = \frac{70 \times 10^{-3} \text{ V}}{8 \times 10^{-9} \text{ m}} \][/tex]
[tex]\[ E = \frac{70}{8} \times 10^{(3 - (-9))} \text{ V/m} \][/tex]
[tex]\[ E = 8.75 \times 10^{6} \text{ V/m} \][/tex]
"Comparing microwaves and visible light, which of the following is true? 1. Microwaves have higher frequency, same speed, and longer wavelength than visible light. 2. Microwaves have lower frequency, same speed, and longer wavelength than visible light. 3. Microwaves have lower frequency, slower speed, and longer wavelength than visible light. 4. Microwaves have lower frequency, faster speed, and shorter wavelength than visible light."
Answer:
2. Microwaves have lower frequency, same speed, and longer wavelength than visible light.
Explanation:
Microwaves are a form of electromagnetic radiation. Most people are familiar with this type of waves because they are used in microwave ovens. When compared to visible light, microwaves have lower frequency, same speed and longer wavelength than visible light. The prefix "micro" is used to indicate that microwaves are smaller (shorter wavelengths) than radio waves.
"The correct option is 2. Microwaves have lower frequency, same speed, and longer wavelength than visible light.
To understand why this option is correct, let's consider the relationship between frequency, wavelength, and speed for electromagnetic waves, which is given by the equation:
[tex]\[ c = f \times \lambda \][/tex]
1. Speed of light (in a vacuum) is constant for all electromagnetic waves, including microwaves and visible light. Therefore, the speed of microwaves and visible light is the same.
2. Microwaves have a lower frequency than visible light. The frequency of microwaves typically ranges from 300 MHz to 300 GHz, while the frequency of visible light ranges from approximately 430 THz to 750 THz. Since microwaves have a lower frequency, they also have a longer wavelength according to the equation [tex]\( c = f \times \lambda \)[/tex].
3. Since the speed of all electromagnetic waves is the same in a vacuum, and microwaves have a lower frequency, they must have a longer wavelength to maintain the constant speed. This is consistent with the relationship , where a lower frequency requires a longer wavelength to keep the speed constant.
In summary, microwaves have a lower frequency and longer wavelength than visible light, but they travel at the same speed. This makes option 2 the correct choice."
A 2.0-kg ball is moving with a constant speed of 5.0 m/s in a horizontal circle whose diameter is 1.0 m. What is the magnitude of the net force on the ball
Answer:
F = 100.0 N
Explanation:
in order to the ball keeps moving at a constant speed in a circle, instead of moving along a straight line, there must be an acceleration that accounts for the change in direction of the ball.This acceleration, called centripetal, is directed at any time, towards the center of the circle.As any acceleration, as dictated by Newton's 2nd law, it must be produced by a force, called centripetal force.The magnitude of this force is related with the mass, the speed and the radius of the circle, as follows:[tex]F_{c} = m *\frac{v^{2} }{r} = 2.0 kg *\frac{(5.0m/s)^{2} }{0.5m} =100.0 N[/tex]
The magnitude of the net force on the ball is 100.0 N100.0N
Explanation:As the ball moves round the circle, a centripetal acceleration which is directed towards the center of the circle will keep the ball from falling off. This acceleration produces a force called centripetal force (F).
Since this is the only force acting on the ball, then the net force acting on the ball to keep it moving round the circle is the centripetal force.
F = m a [according to Newton's second law of motion] --------------(i)
Where;
m = mass of the ball.
a = centripetal acceleration. = [tex]\frac{v^2}{r}[/tex]
v = speed of the ball.
r = radius of the circle.
Substitute a = [tex]\frac{v^2}{r}[/tex] into equation (i) as follows;
F = m x [tex]\frac{v^2}{r}[/tex] --------------------(ii)
From the question;
m = 2.0kg
v = 5.0m/s
r = diameter / 2 [diameter = 1.0m]
r = 1.0 / 2
r = 0.5m
Substitute these values into equation (ii) as follows;
F = 2.0 x [tex]\frac{5.0^{2} }{0.5}[/tex]
F = 2.0 x [tex]\frac{25.0}{0.5}[/tex]
F = 2.0 x 50.0
F = 100.0N
Therefore, the magnitude of the net force on the ball is 100.0N
In each case the momentum before the collision is: (2.00 kg) (2.00 m/s) = 4.00 kg * m/s
1. In each of the three cases above show that momentum is conserved by finding the total momentum after the collision.
2. In each of the three cases, find the kinetic energy lost and characterize the collision as elastic, partially inelastic, or totally inelastic. The kinetic energy before the collision is (1/2)(2.00 kg)(2.00 m/s)^2 = 4.00 kg * m^2/s^2 = 4.00 J.
3. An impossible outcome of such a collision is that A stocks to B and they both move off together at 1.414 m/s. First show that this collision would satisfy conservation of kinetic energy and then explain briefly why it is an impossible result.
Answer:
Check Explanation.
Explanation:
Momentum before collision = (2)(2) + (2)(0) = 4 kgm/s
a) Scenario A
After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s
Momentum after collision = (sum of the masses) × (common velocity) = (2+2) × (1) = 4 kgm/s
Which is equal to the momentum before collision, hence, momentum is conserved.
Scenario B
They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s
Momentum after collision = (2)(0.5) + (2)(1.5) = 1 + 3 = 4.0 kgm/s
This is equal to the momentum before collision too, hence, momentum is conserved.
Scenario C
Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s
Momentum after collision = (2)(0) + (2)(2) = 0 + 4 = 4.0 kgm/s
This is equal to the momentum before collision, hence, momentum is conserved.
b) Kinetic energy is normally conserved in a perfectly elastic collision, if the two bodies do not stick together after collision and kinetic energy isn't still conserved, then the collision is termed partially inelastic.
Kinetic energy before collision = (1/2)(2.00)(2.00²) + (1/2)(2)(0²) = 4.00 J.
Scenario A
After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s
Kinetic energy after collision = (1/2)(2+2)(1²) = 2.0 J
Kinetic energy lost = (kinetic energy before collision) - (kinetic energy after collision) = 4 - 2 = 2.00 J
Kinetic energy after collision isn't equal to kinetic energy before collision. This collision is evidently totally inelastic.
Scenario B
They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s
Kinetic energy after collision = (1/2)(2)(0.5²) + (1/2)(2)(1.5²) = 0.25 + 3.75 = 4.0 J
Kinetic energy lost = 4 - 4 = 0 J
Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.
Scenario C
Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s
Kinetic energy after collision = (1/2)(2)(0²) + (1/2)(2)(2²) = 4.0 J
Kinetic energy lost = 4 - 4 = 0 J
Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.
c) An impossible outcome of such a collision is that A stocks to B and they both move off together at 1.414 m/s.
In this scenario,
Kinetic energy after collision = (1/2)(2+2)(1.414²) = 4.0 J
This kinetic energy after collision is equal to the kinetic energy before collision and this satisfies the conservation of kinetic energy.
But the collision isn't possible because, the momentum after collision isn't equal to the momentum before collision.
Momentum after collision = (2+2)(1.414) = 5.656 kgm/s
which is not equal to the 4.0 kgm/s obtained before collision.
This is an impossible result because in all types of collision or explosion, the second law explains that first of all, the momentum is always conserved. And this evidently violates the rule. Hence, it is not possible.
A sports car moves around a banked curve at just the right constant speed v so that no friction is needed to make the turn. During the turn, the driver (mass m) feels as though she weighs x times her actual weight. Find the magnitude of the net force on the driver during the turn in terms of m, g, and x.
The magnitude of the net force on the driver during a turn on a banked curve, when no friction is required, is equal to the driver's mass multiplied by gravity and the factor x, which represents the perceived increase in weight. Therefore, the magnitude of the net force is m × g × x.
Explanation:If a sports car moves around a banked curve at a constant speed such that no friction is needed, this means that the net force is providing the necessary centripetal force for the turn. According to the problem, the driver feels as though she weighs x times her actual weight. This perception of increased weight is due to the normal force provided by the banked road, which has both vertical and horizontal components.
Using Newton's second law, the net force on the driver can be expressed as net force = mass × acceleration. In this case, the centripetal acceleration is due to the net horizontal force, which is the horizontal component of the normal force.
The vertical component of the normal force is balancing the driver's actual weight, and the horizontal component of the normal force equals the centripetal force necessary for circular motion. Therefore, the normal force experienced by the driver (which is responsible for the feeling of increased weight) is FN = m × g × x. Since this is the net force and it is providing the centripetal force, the magnitude of the net force on the driver is also FN = m × g × x.
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A bar magnet is held above the center of a conducting ring in the horizontal plane. The magnet is dropped so it falls lengthwise toward the center of the ring. Will the falling magnet be attracted toward the ring or be repelled by the ring due to the magnetic interaction of the magnet and the ring?
Explanation:
Since, it is given that the magnet drops and falls lengthwise towards the canter of the ring. As a result, change in magnetic flux will occur which tends to induce an electric current in the ring.
Therefore, a magnetic field is also produced by the ring itself which will actually oppose or repel the magnet.
Thus, we can conclude that the falling magnet be repelled by the ring due to the magnetic interaction of the magnet and the ring.
A long solenoid that has 1,140 turns uniformly distributed over a length of 0.415 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur?
Answer:
Therefore,
Current required is , I
[tex]I = 0.0289\ Ampere[/tex]
Explanation:
Given:
Turns = N = 1140
length of solenoid = l = 0.415 m
Magnetic Field,
[tex]B = 1.00\times 10^{-4}\ T[/tex]
To Find:
Current , I = ?
Solution:
If N is the number of turns in the length, the total current through the rectangle is NI. Therefore, Ampere’s law applied to this path gives
[tex]\int {B} \, ds= Bl=\mu_{0}NI[/tex]
Where,
B = Strength of magnetic field
l = Length of solenoid
N = Number of turns
I = Current
[tex]\mu_{0}=Permeability\ in\ free\ space=4\pi\times 10^{-7}\ Tm/A[/tex]
Therefore,
[tex]I =\dfrac{Bl}{\mu_{0}N}[/tex]
Substituting the values we get
[tex]I =\dfrac{1.00\times 10^{-4}\times 0.415}{4\times 3.14\times 10^{-7}\times 1140}=0.0289\ Ampere[/tex]
Therefore,
Current required is , I
[tex]I = 0.0289\ Ampere[/tex]
A projectile is fired with an initial speed of 230 m/s and an angle of elevation 60°. The projectile is fired from a position 100 m above the ground. (Recall g = 9.8 m/s². Round your answers to the nearest whole number.)
(a) Find the range of the projectile.
(b) Find the maximum height reached.
(c) Find the speed at impact.
Answer:
Explanation:....
The range of the projectile is 4675 m. The maximum height reached by the projectile is 2124 m. The speed of the projectile at impact is 234 m/s.
(a) Range is defined as the maximum horizontal distance covered by a projectile during its motion. It is given by the formula:
[tex]R = \frac{u^2 sin2\theta}{g}[/tex],
where u = initial speed of the projectile = 230 m/s
θ = angle of projection = 60°
g = acceleration due to gravity = 9.8 m/s²
Using these in formulas, we get:
[tex]R = \frac{(230 \hspace{0.8mm} m/s)^2 \hspace{0.8mm} sin2(60)}{9.8 \hspace{0.8mm} m/s^2}[/tex]
or, R = 4674.77 m ≈ 4675 m
(b) The maximum height reached by the projectile is given by the formula:
[tex]H = \frac{u^2 \hspace{0.8mm} sin^2\theta}{2g}[/tex]
using the numeric values, we get:
[tex]H = \frac{(230 \hspace{0.8mm}m/s)^2 sin^2(60)}{2 \times (9.8 \hspace{0.8mm} m/s^2)}[/tex]
or, H = 2024.23 m
However, since the projectile was fired 100 m above the ground, hence, the maximum height would be:
h = 100 m + H = 100 m + 2024.23 m = 2124.23 m ≈ 2124 m
(c) The speed of the projectile comprises of two parts, a horizontal velocity and a vertical velocity.
The horizontal velocity is given as [tex]v_x[/tex]. It is constant throught the motion and its magnitude can be determined by the formula:
[tex]v_x = v cos\theta = 230 \hspace{0.8 mm} m/s \times cos(60)[/tex]
or, [tex]v_x = 115 \hspace{0.8mm} m/s[/tex]
The vertical velocity is given as [tex]v_y[/tex]. It is not constant throught the motion. At the highest point, the magnitude of vertical velocity is zero. Using the kinetic equation,
[tex]v_y^2[/tex] = u² + 2gs, where s = h = 2124 m, we get:
[tex]v_y^2[/tex] = 0 + (2 × 9.8 m/s² × 2124 m)
or, [tex]v_y^2[/tex] = 41630.4 m²/s²
or, [tex]v_y[/tex] = 204.03 m/s ≈ 204 m/s
Now, the magnitude of speed will be equal to the resultant of both the horzontal and vertical velocities.
[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]
or, [tex]v = \sqrt{204^2 + 115^2} \hspace{0.8 mm} m/s[/tex]
or, v = 234.21 m/s ≈ 234 m/s
Star #1 is approaching the Earth with speed v. Star #2 is receding from the Earth with the same speed v. Measurements of the same spectral line from each of the stars show Doppler shifts in frequency. The light from which star will have the larger magnitude shift in frequency? a. star #1 b. Both stars will have the same shift. c. The value of the speed must be known before an answer can be found. d. star #2
Answer:
b. Both stars will have the same shift.
Explanation:
It's a very simple problem to solve. Star 1 is approaching toward Earth with a speed v, so let's assume that the change in Doppler Shift is +F and Star 2 is moving away so the change in Doppler shift is -F. But it's time to notice the speed of both stars and that is same but only directions are different. speed is the main factor here. The magnitude of both shifts is F as we can see and + and - are showing there direction of motion. So, because of same amount of speed, both stars will have same shift magnitude. (Just the directions are different)
A fixed-geometry supersonic inlet starts at a Mach number of 3. After starting, the cruise Mach number is 2, and the operating shock is positioned at a location where the area is 10% larger than the throat. a) Assuming the flow is ideal except for shock losses, find the inlet stagnation b) During cruise, the Mach number at the exit of the diffuser M2 is required to be 0.3. Determine the ratio of the areas at the diffuser exit to that at the inlet (find AJA, and the static pressure ratio p./p. A,JA), and the static pressure ratio p./p
Answer:
Explanation: see attachment below
The six metals have the work functions, W.
Part A Rank these metals on the basis of their cutoff frequency. Rank from largest to smallest. To rank items as equivalent, overlap them.
Part B Rank these metals on the basis of the maximum wavelength of light needed to free electrons from their surface. Rank from largest to smallest. To rank items as equivalent, overlap them.
Part C Each metal is illuminated with 400 nm (3.10 eV) light. Rank the metals on the basis of the maximum kinetic energy of the emitted electrons. (If no electrons are emitted from a metal, the maximum kinetic energy is zero, so rank that metal as smallest.) Rank from largest to smallest. To rank items as equivalent, overlap them.
Cesium= w= 2.1 eV Aliminium= w= 4.1 eV Beryllium= 5.0 eV Potassium= 2.3 eV Platinium= w= 6.4 eV Magnisium=w= 3.7 eV
Answer:
Explanation:
W (Ce) = 2.1 eV
W (Al) = 4.1 eV
W (Be) = 5 eV
W (K) = 2.3 eV
W (Pt) = 6.4 eV
W (Mg) = 3.7 eV
Part A:
Work function is directly proportional to the cut off frequency.
let f denotes the cut off frequency.
So, f (Be) > f (Be) > f (Al) > f (Mg) > f (K) > f (Ce)
Part B:
Maximum wavelength for the emission is inversely proportional to the cut off frequency
So, λ (Ce) > λ (K) > λ (Mg) > λ (Al) > λ (Be) > λ (Pt)
Part C:
E = 3.10 eV
Let K is the maximum kinetic energy
K = E - W
K (Ce) = 3.1 - 2.1 = 1 eV
K (Al) 3.1 - 4.1 = not possible
K (Be) = 3.1 - 5 = not possible
K (K) = 3.1 - 2.3 = 0.8 eV
K (Pt) = .1 - 6.4 = not possible
K (Mg) = 3.1 - 3.7 = not possible
So, K (Ce) > K (K) > K (Al) = K (Be) = K (Pt) = K (Mg)
Metals can be ranked based on their work functions, which determine cutoff frequency, the maximum wavelength of light needed to free electrons, and maximum kinetic energy of emitted electrons under specific light. Lower work functions result in higher cutoff frequencies and longer required wavelengths, and more energetic emitted electrons when under 400 nm light.
Explanation:The ranking of the metals, in terms of cutoff frequency, maximum wavelength of light, and maximum kinetic energy, is based on the work function of each metal, which is the minimum energy required to free an electron from the metal's surface. The cutoff frequency is inversely related to the work function, so the metal with the lowest work function (cesium, 2.1 eV) will have the highest cutoff frequency, and the metal with the highest work function (platinum, 6.4 eV) will have the lowest cutoff frequency.
Conversely, the maximum wavelength of light needed to free electrons is directly related to the work function, so cesium will require the longest wavelength and platinum will require the shortest wavelength. When illuminated with 400 nm (3.10 eV) light, the kinetic energy of the emitted electrons will be the difference between the energy of the light and the work function, so cesium (with the smallest work function) will have the most energetic electrons, and metals with a work function larger than 3.10 eV (like aluminum, beryllium, and platinum) will not emit electrons at all.
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A segment of wire carries a current of 25 A along the x axis from x = −2.0 m to x = 0 and then along the y axis from y = 0 to y = 3.0 m. In this region of space, the magnetic field is equal to 40 mT in the positive z direction. What is the magnitude of the force on this segment of wire?
Final answer:
The magnitude of force on the wire segment along the y-axis is 3.0N and directed along the negative x-axis, while there's no force on the segment along the x-axis.
Explanation:
To find the magnitude of the force on the wire, we can use the formula F = I * (L x B), where F is the magnetic force, I is the current, L is the length vector of the wire segment, and B is the magnetic field. The force will be perpendicular to both the current direction and the magnetic field direction, according to the right-hand rule.
The wire is in two segments, one along the x-axis and one along the y-axis. First, we calculate the force on the section of the wire along the x-axis: Fx = I * Lx * B, which is 0 since the magnetic field is in the z-direction and Lx is in the x-direction, making the angle between them 90 degrees. So, there's no force for this segment.
Second, we calculate the force on the section of the wire along the y-axis: Fy = I * Ly * B. The force experienced by this segment can be calculated using the formula with the values provided (I=25A, Ly=3.0m, B=40mT). Therefore, Fy = 25A * 3.0m * 40mT = 3000mN or 3.0N, which will be directed along the negative x-axis, under the right-hand rule.
A bicycle is rolling down a circular portion of a path; this portion of the path has a radius of 9.30 m. As the drawing illustrates, the angular displacement of the bicycle is 1.130 rad. What is the angle (in radians) through which each bicycle wheel (radius = 0.300 m) rotates?
Answer:
The angle through which each bicycle wheel rotates is 35.03 rad.
Explanation:
Given;
the radius of the circular path, R = 9.30 m
the angular displacement of the bicycle, θ = 1.130 rad
the radius of the bicycle wheel, r = 0.3
S = θR
where;
s is the distance of the circular path
S = 1.13 x 9.3 = 10.509 m
The angle (in radians) through which each bicycle wheel of radius 0.300 m rotates is given as;
θr = 10.509 m
θ = 10.509 / 0.3
θ = 35.03 rad.
Therefore, the angle through which each bicycle wheel of radius 0.300 m rotates is 35.03 rad.
A circular coil has a 18.0 cm radius and consists of 25.0 closely wound turns of wire. An externally produced magnetic field of magnitude 3.00 mT is perpendicular to the coil. (a) If no current is in the coil, what magnetic flux links its turns?
Answer:
The magnetic flux links to its turns = [tex]7.6 \times10^{-3}[/tex] Wb.
Explanation:
Given :
Radius of circular coil = [tex]18 \times 10^{-2}[/tex] m
Number of turns = 25
Magnetic field = [tex]3 \times10^{-3}[/tex] T
Magnetic flux (Φ) is a measure of the magnetic field lines passes through a given area. The unit of magnetic flux is weber (Wb).
We know that,
⇒ Φ = [tex]BA[/tex]
Where [tex]B =[/tex] ext. magnetic field, [tex]A =[/tex] area of loop or coil.
But here given in question, we have turns of wire so our above eq. modified as follows.
⇒ Φ = [tex]NBA[/tex]
Where [tex]N =[/tex] no. of turns.
∴ Φ = [tex]25 \times 3 \times 10^{-3} \pi (18 \times10^{-2} )^{2}[/tex]
Φ = [tex]7.6 \times 10^{-3} Wb[/tex]
Thus, the magnetic flux links to its turns = [tex]7.6 \times 10^{-3} Wb[/tex]
A block is placed on an inclined plane with an angle of inclination θ (in degrees) with respect to horizontal. The coefficient of static friction between the block and the inclined plane is 0.4. For what maximum value of θ will the block remain stationary on the inclined surface?
Final answer:
The maximum angle of inclination θ for which a block will remain stationary on an inclined plane, given a coefficient of static friction of 0.4, can be found using the arctangent function resulting in approximately 21.8 degrees.
Explanation:
To find the maximum angle of inclination θ at which a block will remain stationary on an inclined surface, we can relate the coefficient of static friction (0.4 in this case) to the angle using the tangent function. When static friction has reached its maximum value, the block is on the verge of sliding, and the maximum force of static friction is equal to the component of the block's weight parallel to the incline. This force can be described by the equation μsN = mg sin(θ), where μs is the static friction coefficient, N is the normal force, m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of inclination. Since the normal force is equal to mg cos(θ), substituting this into the equation and solving for θ gives us θ = tan-1(μs). Plugging in the given coefficient of static friction, the maximum angle θ can be found.
Using the given coefficient of 0.4, we can calculate the angle: θ = tan-1(0.4). Therefore, the maximum angle θ, for the block to remain stationary, is approximately 21.8 degrees.
Final answer:
The maximum angle of inclination θ at which a block will remain stationary on an inclined plane with a coefficient of static friction of 0.4 is approximately 21.8 degrees.
Explanation:
The maximum angle of inclination θ at which a block will remain stationary on an inclined plane with a coefficient of static friction of 0.4 is determined by using the relationship between the angle of inclination and the coefficient of static friction. This relationship is given by the equation tan(θ) = μ, where μ is the coefficient of static friction. To find the angle where the block starts to slide, we take the inverse tangent (arctan or tan-1) of the coefficient of static friction. Therefore, the maximum angle θ is tan-1(0.4).
Calculating this, we get θ = tan-1(0.4) ≈ 21.8°. Hence, for angles of inclination less than or equal to 21.8 degrees, the block will not slide down the incline due to static friction.
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.97 10-6 W/m2 at a distance of 113 m from the explosion, at what distance from the explosion is the sound intensity half this value?
Answer:
160 m
Explanation:
The intensity, I, of the sound is inversely proportional to the square of the distance, r, from the source.
[tex]I\propto \dfrac{1}{r^2}[/tex]
Hence,
[tex]I_1r_1^2 = I_2r_2^2[/tex]
[tex]r_2 = r_1\sqrt{\dfrac{I_1}{I_2}}[/tex]
From the question, [tex]I_2[/tex] is half of [tex]I_1[/tex]
[tex]r_2 = r_1\sqrt{\dfrac{I_1}{0.5I_1}}[/tex]
[tex]r_2 = r_1\sqrt{2}[/tex]
[tex]r_2 = 113\text{ m}\sqrt{2} = 160 \text{ m}[/tex]
Answer:
Distance ,d= 159.81m
Explanation:
The intensity, I, of the sound is inversely proportional to the square of the distance, d,from the source.
Using the equation d/do=sqrt2
d= dosqrt2
Where d=113m
d= 113sqrt2
d= 159.81m
Consider the expression below. Assume m is an integer. 6m(2m + 18) Enter an expression in the box that uses the variable m and makes the equation true. (Simplify your answer completely. If no expression exists, enter DNE.)
The expression 6m(2m + 18) simplifies to 12m² + 108m. Therefore, the equivalent expression using the variable 'm' is 12m² + 108m.
Explanation:The given expression is 6m(2m + 18). First, we distribute 6m across the terms in the parentheses, resulting in 12m² + 108m. So, an expression that uses the variable 'm' and makes the equation true when the given expression is simplified would be 12m² + 108m. This answer is fully simplified, and the variable 'm' is assumed to be an integer.
The expression 6m(2m + 18) simplifies to 12m² + 108m. Therefore, the equivalent expression using the variable 'm' is 12m² + 108m.
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If i = 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N ⋅ m , what are the lengths of the sides s of the square loop, in centimeters?
Answer:
Length of the sides of the square loop is given by
s = √[(τ)/(NIB sin θ)]
Explanation:
The torque, τ, experienced by a square loop of area, A, with N number of turns around the loop and current of I flowing in the wire, with a magnetic field presence, B, and the plane of the loop tilted at angle θ to the x-axis, is given by
τ = (N)(I)(A)(B) sin θ
If everything else is given, the length of a side of the square loop, s, can be obtained from its Area, A.
A = s²
τ = (N)(I)(A)(B) sin θ
A = (τ)/(NIB sin θ)
s² = (τ)/(NIB sin θ)
s = √[(τ)/(NIB sin θ)]
In this question, τ = 0.076 N.m, I = 1.70 A
But we still need the following to obtain a numerical value for the length of a side of the square loop.
N = number of turnsof wire around the loop
B = magnetic field strength
θ = angle to which the plane of the loop is tilted, measured with respect to the x-axis.
The question is incomplete! The complete question along with answer and explanation is provided below
Question:
A wire loop with 40 turns is formed into a square with sides of length s. The loop is in the presence of a 2.0 T uniform magnetic field B that points in the negative y direction. The plane of the loop is tilted off the x-axis by 15°. If 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N.m, what are the lengths of the sides s of the square loop, in centimeters?
Given Information:
Number of turns = N = 40 turns
Torque = τ = 0.0760 N.m
Current = I = 1.70 A
Magnetic field = B = 2 T
θ = 15°
Required Information:
Length of the sides of square loop = s = ?
Answer:
s = 4.64 cm
Explanation:
τ = NIABsin(θ)
Where N is the number of turns, I is the current flowing through the square loop, A is the area of square loop, B is the magnetic field, and θ is the angle between square loop and magnetic field strength with respect to the x-axis.
Re-arranging the equation to find out A
A = τ/ NIBsin(θ)
A = 0.0760/40*1.70*2*sin(15)
A = 0.00215 m²
We know that area of a square is
A = s²
Taking square root on both sides yields
s = √A
s = √0.00215 = 0.0464 m
in centimeters
s = 4.64 cm
As your skateboard coasts uphill, your speed changes from 3 m/s to 1 m/s in
3 seconds. What is your deceleration?
Answer:
[tex]a=-0.33\ m/s^2[/tex]
Explanation:
Accelerated Motion
The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by
[tex]\displaystyle a=\frac{\Delta v}{t}[/tex]
Or, equivalently
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
Where vf and vo are the final and initial speeds respectively. The problem gives us these values: v0 = 3 m/s, vf = 1 m/s, t = 3 seconds. Computing a
[tex]\displaystyle a=\frac{1-3}{3}=-0.33\ m/s^2[/tex]
The negative sing of a indicates there is deceleration or decreasing speed
Final answer:
The skateboard's deceleration as it coasts uphill, changing speed from 3 m/s to 1 m/s in 3 seconds, is calculated as -0.67 m/s². This indicates a decrease in speed and is consistent with deceleration.
Explanation:
If a skateboard coasts uphill and experiences a change in speed from 3 m/s to 1 m/s in 3 seconds, to find the deceleration, we use the formula for acceleration, which is the change in velocity divided by the time taken for the change. Deceleration is simply acceleration in the opposite direction to the motion (negative acceleration).
The change in velocity ({\Delta v}) is 1 m/s - 3 m/s, which equals -2 m/s (the negative sign indicates a decrease in speed). The time ({t}) is 3 seconds. Thus, the deceleration is {\Delta v / t} which is (-2 m/s) / (3 s) = -0.67 m/s². This negative sign signifies that it is indeed a deceleration.