Answer:
The answer is -x direction.
Explanation:
According to the right hand rule for electromagnetism, when the thumb points up, index finger points forward and the middle finger is perpendicular to the index finger, the thumb points in the direction of the magnetic force, the index finger in the direction of the charge movement and the middle finger in the direction of the electromagnetic lines. If the electric field is pointing in the -z direction which is into the screen and the magnetic field is in the +y direction which is upwards, then the magnetic lines are in the -x direction.
I hope this answer helps.
A kite weighs 2.50 lb and has an area of 8.00 ft2 . The tension in the kite string is 6.60 lb when the string makes an angle of 45⁰ with the horizontal. For a wind of 20 mph (29.3 ft/s), what are the coefficients of lift and drag (based on the kite’s area)? Use air density of 0.00234 slug/ft3 .
Answer:
the coefficient of lift is 3.5561
the coefficient of drag is 2.3153
Explanation:
the solution is in the attached Word file
The limit to the eye's acuity is actually related to diffraction by the pupil. What is the angle between two just‑resolvable points of light for a 3.75 mm 3.75 mm diameter pupil, assuming the average wavelength of 554 nm 554 nm ?
Answer: [tex]1.8(10)^{-4} rad[/tex]
Explanation:
This problem is related to the Rayleigh Criterion, which provides the following formula to find the acuity or limit of resolution of an optic system with circular aperture (the eye in this case):
[tex]\theta=1.22\frac{\lambda}{D}[/tex]
Where:
[tex]\theta[/tex] is the angle of resolution (related to the acuity)
[tex]\lambda=554 nm=554(10)^{-9} m[/tex] is the wavelength of the light
[tex]D=3.75 mm=3.75(10)^{-3} m[/tex] is the diameter of the pupil
Solving:
[tex]\theta=1.22 \frac{554(10)^{-9} m}{3.75(10)^{-3} m}[/tex]
[tex]\theta=1.8(10)^{-4} rad[/tex]
The human eye's acuity is limited by light's diffraction through the pupil, which determines the minimum angle, Δθ, between two resolvable points. This is calculated using the formula Δθ = 1.22 (λ / D), where λ is the light's wavelength and D the diameter of the pupil.
Explanation:The resolution of human visual acuity is related to the diffraction of light by the eye's pupil. This diffraction effect can be worked out using a formula that quantifies resolving power (Δθ), when the diameter of the aperture (in this case, the pupil size) and the wavelength of incident light are known.
In your example, a 3.75 mm diameter pupil encountering light of 554 nm wavelength can identify two points of light provided they subtend an angle (Δθ) at the pupil calculated by using the formula: Δθ = 1.22 (λ / D). For all light's range of visible wavelengths, the angle is extremely tiny - on the order of tens of thousands of a degree. This shows that the eye’s ability to distinguish two separate points of light is limited by the diffraction of light by the eye’s pupil.
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(a) For what frequencies does a 23.0-μF capacitor have a reactance below 160 Ω? f Correct: Your answer is correct. 6.95 Incorrect: Your answer is incorrect. Hz (b)What is the reactance of a 41.0-μF capacitor over this same frequency range? Xc Correct: Your answer is correct. Ω
Answer:
(a) So range of frequency [tex]f > 43.27[/tex] Hz
(b) the reactance is 89.75 Ω
Explanation:
Given:
(a)
Capacitance of a capacitor [tex]C= 23 \times 10^{-6}[/tex] F
Reactance of capacitive circuit [tex]X_{C} =[/tex] 160 Ω
From the formula of reactance,
[tex]X_{C} = \frac{1}{\omega C}[/tex]
[tex]X_{C} = \frac{1}{2\pi fC}[/tex]
[tex]f = \frac{1}{2\pi X_{C} C }[/tex]
[tex]f = \frac{1}{6.28 \times 160 \times 23 \times 10^{-6} }[/tex]
[tex]f = 43.27[/tex] Hz
So range of frequency [tex]f > 43.27[/tex] Hz
(b)
Capacitance [tex]C = 41 \times 10^{-6}[/tex] F
Frequency [tex]f = 43.27[/tex] Hz
From the formula of reactance,
[tex]X_{C} = \frac{1}{2\pi fC}[/tex]
[tex]X_{C} = \frac{1}{6.28 \times 43.27 \times 41 \times 10^{-6} }[/tex]
[tex]X_{C} =[/tex] 89.75 Ω
Therefore, the reactance is 89.75 Ω
Radar uses radio waves of a wavelength of 2.4 \({\rm m}\) . The time interval for one radiation pulse is 100 times larger than the time of one oscillation; the time between pulses is 10 times larger than the time of one pulse. What is the shortest distance to an object that this radar can detect? Express your answer with the appropriate units.
Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x [tex]10^{8}[/tex] m/s)
f= 3 x [tex]10^{8}[/tex] / 2.4
f=1.25 x [tex]10^{8}[/tex] hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x [tex]10^{8}[/tex]
T= 8 x [tex]10^{-9}[/tex] s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x [tex]10^{-9}[/tex] => 800 x [tex]10^{-9}[/tex] s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
[tex]x_{min}[/tex]= τc/2 => (800 x [tex]10^{-9}[/tex] x 3 x [tex]10^{8}[/tex])/2
[tex]x_{min}[/tex]=120m
A wire carrying a 29.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's field, and there is a 2.25 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet
Answer:
2.59 T
Explanation:
Parameters given:
Current flowing through the wire, I = 29 A
Angle between the magnetic field and wire, θ = 90°
Magnetic force, F = 2.25 N
Length of wire, L = 3 cm = 0.03 m
The magnetic force, F, is related to the magnetic field, B, by the equation below:
F = I * L * B * sinθ
Inputting the given parameters:
2.25 = 29 * 0.03 * B * sin90
2.25 = 0.87 * B
=> B = 2.25/0.87
B = 2.59 T
The magnetic field strength between the poles is 2.59 T
A 6200 line/cm diffraction grating is 3.14 cm wide. If light with wavelengths near 624 nm falls on the grating, how close can two wavelengths be if they are to be resolved in any order
Answer:
[tex]1.6026299569\times 10^{-11}\ m[/tex]
Explanation:
Grating constant
[tex]d=\dfrac{1}{6200}=0.000161\ cm=0.000161\times 10^{-2}\ m[/tex]
Number of slits
[tex]N=3.14\times 6200=19468[/tex]
Order
[tex]m=\dfrac{d}{\lambda}\\\Rightarrow m=\dfrac{0.000161\times 10^{-2}}{624\times 10^{-9}}\\\Rightarrow m\approx 2[/tex]
At m = 1
[tex]\Delta\lambda=\dfrac{\lambda}{mN}\\\Rightarrow \Delta\lambda=\dfrac{624\times 10^{-9}}{1\times 19468}\\\Rightarrow \Delta\lambda=3.2052599137\times 10^{-11}\ m[/tex]
At m = 2
[tex]\Delta\lambda=\dfrac{\lambda}{mN}\\\Rightarrow \Delta\lambda=\dfrac{624\times 10^{-9}}{2\times 19468}\\\Rightarrow \Delta\lambda=1.6026299569\times 10^{-11}\ m[/tex]
The wavelengths can be close by [tex]1.6026299569\times 10^{-11}\ m[/tex]
The gold-like mineral in the metamorphic rock pictured below is known as pyrite. When pyrite is exposed to air and water it can react
to form sulfate and an acid. The acid can discolor and eat away the surrounding rock.
Image courtesy of David Usher, USGS
Pyrite breaks down rocks in which of the following processes?
A
deposition
B. physical weathering
C.
erosion
D.
chemical weathering
Answer: C. Chemical Weathering
Explanation:
Answer:
Chemical Weathering
Explanation:
A laser beam is incident on a plate of glass that is 2.8 cm thick. The glass has an index of refraction of 1.6 and the angle of incidence is 36°. The top and bottom surfaces of the glass are parallel. What is the distance b between the beam formed by reflection off the top surface of the glass and the beam reflected off the bottom surface of the glass?
Final answer:
The distance between the reflected light beams can be found using reflection and refraction laws, applying Snell's law to calculate the angle of refraction, and subsequently using trigonometry to determine the light path within the glass and the shift caused by both reflections.
Explanation:
To determine the distance b between the light beam reflected from the top and the bottom surfaces of a glass plate, we need to use the laws of reflection and Snell's law for the refraction of light. The given information includes the thickness of the glass (2.8 cm), the index of refraction of the glass (1.6), and the angle of incidence (36°). When the laser beam strikes the top surface at an angle of incidence of 36°, it reflects at the same angle (law of reflection).
The angle of refraction can be found using Snell's law: n1 * sin(θ1) = n2 * sin(θ2), where n1 and n2 are the indices of refraction of the air and glass, respectively, and θ1 and θ2 are the angles of incidence and refraction, respectively. Once the angle of refraction is determined, the actual path of the beam within the glass can be calculated using trigonometry, and subsequently the shift b caused by the refraction at the two interfaces (again using trigonometry).
1. Is the collision between the ball and the pendulum elastic or inelastic? Justify your answer by calculating the kinetic energy of the system before collision using the value of vxo in the experiment and the kinetic energy just after collision using the experimental value of h in Eq. 9.2.
Answer:
So energy is not conserved and inelastic shock
Explanation:
In the collision between a bullet and a ballistic pendulum, characterized in that the bullet is embedded in the block, if the kinetic energy is conserved the shock is elastic and if it is not inelastic.
Let's find the kinetic energy just before the crash
K₀ = ½ m vₓ₀²
After the crash we can use the law of conservation of energy
Starting point. Right after the crash, before starting to climb
Em₀ = K = ½ (m + M) v₂²
Final point. At the maximum height of the pendulum
[tex]Em_{f}[/tex] = U = (m + M) g h
Where m is the mass of the bullet and M is the mass of the pendulum
Em₀ = Em_{f}
½ (m + M) v₂² = (m + M) g h
v₂ = √ 2g h
Now we can calculate the final kinetic energy
K_{f} = ½ (m + M) v₂²²
K_{f} = ½ (m + M) (2gh)
The relationship between these two kinetic energies is
K₀ / K_{f} = ½ m vₓ₀² / (½ (m + M) 2 g h)
K₀ / K_{f} = m / (m + M) vₓ₀² / 2 g h
We can see that in this relationship the Ko> Kf
So energy is not conserved and inelastic shock
To determine if the collision between the ball and the pendulum is elastic or inelastic, we need to calculate the kinetic energy of the system before and after the collision.
Explanation:An elastic collision is one that conserves kinetic energy, while an inelastic collision does not conserve kinetic energy. To determine if the collision between the ball and the pendulum is elastic or inelastic, we need to calculate the kinetic energy of the system before and after the collision.
In this case, we can calculate the kinetic energy of the system before the collision using the value of vxo and the kinetic energy just after the collision using the experimental value of h in Eq. 9.2.
If the kinetic energy of the system is the same before and after the collision, then it is an elastic collision. If the kinetic energy decreases after the collision, then it is an inelastic collision.
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An electron is accelerated through 1.95 103 V from rest and then enters a uniform 1.50-T magnetic field. (a) What is the maximum magnitude of the magnetic force this particle can experience?
Answer: 6.29*10^-12 N
Explanation:
given,
Potential difference of the electron, v = 1950 V
Magnetic field of the electron, B = 1.50 T
If the electron is accelerated through 19500 V from rest its Potential Energy has to be
converted to Kinetic Energy. This allows us solve for the velocity.
PE = Vq
PE = 1950 * 1.6*10^-19
PE = 3.12*10^-16 J
Also, PE = 1/2mv²
3.12*10^-16 = 1/2mv²
v = 2.62*10^7 m/s
to get F(max), we use,
F(max) = qvB
F(max) = 1.6*10^-19 * 2.62*10^7 * 1.5
F(max) = 6.29*10^-12 N
6.3 x 10⁻¹²N
Explanation:As stated by Lorentz Force law, the magnitude of a magnetic force, F, can be expressed in terms of a fixed amount of charge, q, which is moving at a constant velocity, v, in a uniform magnetic field, B, as follows;
F = qvB sin θ ------------(i)
Where;
θ = angle between the velocity and the magnetic field vectors
When the electron passes through a potential difference, V, it is made to accelerate as it gains some potential energy ([tex]P_{E}[/tex]) which is then converted to kinetic energy ([tex]K_{E}[/tex]) as it moves. i.e
[tex]P_{E}[/tex] = [tex]K_{E}[/tex] ----------------(ii)
But;
[tex]P_{E}[/tex] = qV
And;
[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex] x m x v²
Therefore substitute these into equation (ii) as follows;
qV = [tex]\frac{1}{2}[/tex] x m x v²
Make v subject of the formula;
2qV = mv²
v² = [tex]\frac{2qV}{m}[/tex]
v = [tex]\sqrt{\frac{2qV}{m} }[/tex] ---------------(iii)
From the question;
q = 1.6 x 10⁻¹⁹C (charge on an electron)
V = 1.95 x 10³V
m = 9.1 x 10⁻³¹kg
Substitute these values into equation (iii) as follows;
v = [tex]\sqrt{\frac{2*1.6*10^{-19} * 1.95*10^{3}}{9.1*10^{-31}} }[/tex]
v = [tex]\sqrt{\frac{6.24*10^{-16}}{9.1*10^{-31}} }[/tex]
v = 2.63 x 10⁷m/s
Now, from equation (i), the magnitude of the magnetic force will be maximum when the angle between the velocity and the magnetic field is 90°. i.e when θ = 90°
Substitute the values of θ, q v and B = 1.50T into equation (i) as follows;
F = qvB sin θ
F = 1.6 x 10⁻¹⁹ x 2.63 x 10⁷ x 1.50 x sin 90°
F = 6.3 x 10⁻¹²N
Therefore the maximum magnitude of the magnetic force this particle can experience is 6.3 x 10⁻¹²N
A circus act involves a trapeze artist and a baby elephant. They are going to balance on a teeter-totter that is 10 meters long and pivoted at the middle. The lady has a weight of 500 newtons and is standing on one end. The elephant has a weight of 2500 N and is walking toward her on the beam. How far does the elephant have to walk from the end toward the middle to balance the beam
Answer:
Explanation:
Balance point will be achieved as soon as the weight of the baby elephant creates torque equal to torque created by weight of woman about the pivot. torque by weight of woman
weight x distance from pivot
= 500x 5
= 2500 Nm
torque by weight of baby woman , d be distance of baby elephant from pivot at the time of balance
= 2500x d
for equilibrium
2500 d = 2500
d = 1 m
So elephant will have to walk up to 1 m close to pivot or middle point.
Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocity of 1700 ft/s. For steady-state operation, and neglecting potential energy effects, determine the exit temperature, in F
Answer:
[tex]386.2^{\circ}F[/tex]
Explanation:
We are given that
[tex]P_1=200lbf/in^2[/tex]
[tex]P_2=60lbf/in^2[/tex]
[tex]v_1=200ft/s[/tex]
[tex]v_2=1700ft/s[/tex]
[tex]T_1=500^{\circ}F[/tex]
[tex]Q=0[/tex]
[tex]C_p=1BTU/lb^{\circ}F[/tex]
We have to find the exit temperature.
By steady energy flow equation
[tex]h_1+v^2_1+Q=h_2+v^2_2[/tex]
[tex]C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}[/tex]
[tex]1BTU/lb=25037ft^2/s^2[/tex]
Substitute the values
[tex]1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}[/tex]
[tex]500+1.598=T_2+115.4[/tex]
[tex]T_2=500+1.598-115.4[/tex]
[tex]T_2=386.2^{\circ}F[/tex]
We can use the First Law of Thermodynamics and the provided initial conditions from the problem statement to calculate the exit temperature of steam in a steady-flow system, such as a nozzle. The calculation requires knowledge of thermal dynamics principles and the use of steam tables.
Explanation:This question is an application of the First Law of Thermodynamics (also known as the Law of Energy Conservation) and it pertains specifically to the flow of steam. In this case, we're working with a steady-flow system (a nozzle) where only the kinetic energy changes are important. There is no mention of the incoming steam doing work on the nozzle, nor of any heat is transferred to or from the nozzle, so, in this case, the work done (W) and heat transfer (Q) are both zero.
Furthermore, the exit pressure of the steam is given, as is the exit velocity; what we need to ascertain is the exit temperature, T2. To do this, we can use the enthalpies for steam at the given initial state, and then use the first law of thermodynamics to determine the exit enthalpy. The corresponding temperature can then be found in the steam tables associated with the exit enthalpy and pressure.
Therefore, the calculation requires an understanding of Thermal Dynamics principles and the use of Steam Tables.
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A 500 g model rocket is resting horizontally at the top edge of a 40-m-high wall when it is accidentally bumped. The bump pushes it off the edge with a horizontal speed of 0.5 m/s and at the same time causes the engine to ignite. When the engine fires, it exerts a constant 20 N horizontal thrust away from the wall. a. How far from the base of the wall does the rocket land?
Answer:
Explanation:
mass of rocket m = .5 kg
height of wall h = 40 m
initial horizontal velocity u = .5 m /s
horizontal acceleration = force / mass
= 20 / .5
a = 40 m /s²
Let rocket falls or covers 40 m vertically downwards in time t
h = 1/2 gt² , initial vertical velocity = 0
40 = 1/2 x 9.8 x t²
t = 2.8566 s
During this period it will cover horizontal distance with initial velocity of .5 m /s and acceleration a = 40m /s²
horizontal distance = ut + 1/2 at²
= .5 x 2.8566 + .5 x 40 x 2.8566²
= 1.4283 + 163.2033
= 164.63 m .
The distance from the base of the wall does the rocket land is 164.63 m .
Calculation of the distance:A 500 g model rocket is resting horizontally at the top edge of a 40-m-high wall when it is accidentally bumped. The bump pushes it off the edge with a horizontal speed of 0.5 m/s
We know that
horizontal acceleration = force / mass
a = 20 / .5
a = 40 m /s²
Now
h = 1/2 gt² ,
Here initial vertical velocity = 0
So,
40 = 1/2 x 9.8 x t²
t = 2.8566 s
Now
horizontal distance = ut + 1/2 at²
= .5 x 2.8566 + .5 x 40 x 2.8566²
= 1.4283 + 163.2033
= 164.63 m .
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Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that apply. The object can have zero acceleration and, simultaneously, nonzero velocity. The object can have zero velocity and, simultaneously, nonzero acceleration. The object can have zero velocity and, simultaneously, zero acceleration. The object can have nonzero velocity and nonzero acceleration simultaneously.
Complete Question:
An object oscillates back and forth on the end of a spring.
Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that apply. The object can have zero acceleration and, simultaneously, nonzero velocity. The object can have zero velocity and, simultaneously, nonzero acceleration. The object can have zero velocity and, simultaneously, zero acceleration. The object can have nonzero velocity and nonzero acceleration simultaneously.
Answer:
a) the object can have zero acceleration and, simultaneously, nonzero velocity
b) the object can have zero velocity and, simultaneously, nonzero acceleration
d) the object can have nonzero velocity and nonzero acceleration simultaneously
Explanation:
For an object oscillating back and forth on the end of a spring, when the object swings to the extremes, it momentarily stops and the velocity becomes zero. Rather than being zero, acceleration is maximum at these points because the net force at these extremes is maximum. This justifies option B
At the middle, which is the equilibrium position, the net force is zero because the object is motionless, and hence the acceleration is zero. At this point, the velocity is maximum and not zero. This justifies option A
When the object is swinging and it is neither at the middle nor at the extremes, both acceleration and velocity are not zero. This justifies option D
It is not possible for both the acceleration and the velocity of the swinging object to be simultaneously zero. Option C is wrong
An object can have zero acceleration and non-zero velocity or have non-zero velocity and acceleration simultaneously at some point during its motion. However, an object can also have zero velocity and zero acceleration at all times during its motion.
Explanation:Both the statements "The object can have zero acceleration and, simultaneously, nonzero velocity" and "The object can have nonzero velocity and nonzero acceleration simultaneously" are true at some time during the course of motion.
For example, when an object is thrown upwards, at the highest point of its trajectory, its velocity becomes zero, but it still experiences the acceleration due to gravity, which is nonzero. This satisfies the first statement. Additionally, when an object moves in a circular path at a constant speed, it has a nonzero velocity and nonzero acceleration directed towards the center of the circle. This satisfies the second statement.
The statement "The object can have zero velocity and, simultaneously, zero acceleration." is true at all times during the course of motion. For instance, when an object is at rest, both its velocity and acceleration are zero.
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A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the bicycle rests on. Assume that the rear tire is rotating at ω = 32k rad/s. What are the angular velocities of the two cylinders? Consider r1 = 460 mm and r2 = 46 mm.
Answer:
ω2 = 216.47 rad/s
Explanation:
given data
radius r1 = 460 mm
radius r2 = 46 mm
ω = 32k rad/s
solution
we know here that power generated by roller that is
power = T. ω ..............1
power = F × r × ω
and this force of roller on cylinder is equal and opposite force apply by roller
so power transfer equal in every cylinder so
( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2
so
ω2 = [tex]\frac{460\times 32}{34\times 2}[/tex]
ω2 = 216.47
A charged particle is accelerated from rest through a potential difference of magnitude |ΔV|. After exiting the potential difference at an emission point, the particle enters a region of uniform magnetic field. The magnetic field is perpendicular to the particle's velocity, and the particle travels along a complete circular path. The particle's mass is 2.10 ✕ 10−16 kg, its charge is 26.0 nC, and the magnetic field magnitude is 0.600 T. The particle's circular path, as it returns to the emission point, encloses a magnetic flux of 15.0 µWb.
a. What is the speed in (m/s) of the particle when it is in the region of the magnetic field?
b. What is the magnitude of the potential difference through which the particle was accelerated?
Answer:
a
The speed of the particle is [tex]v = 209485.71 m/s[/tex]
b
The potential difference is [tex]\Delta V = 177.2 \ V[/tex]
Explanation:
From the question we are told that
The mass of the particle is [tex]m = 2.10 *10^{-16} kg[/tex]
The charge on the particle is [tex]q = 26.0 nC = 26.0 *10^{-9} C[/tex]
The magnitude of the magnetic field is [tex]B = 0.600 T[/tex]
The magnetic flux is [tex]\O = 15.0 \mu Wb = 15.0 *10^{-6} Wb[/tex]
The magnetic flux is mathematically represented as
[tex]\O = B *A[/tex]
Where A is the the area mathematically represented as
[tex]A = \pi r^2[/tex]
Substituting this into the equation w have
[tex]\O = B (\pi r^2 )[/tex]
Making r the subject of the formula
[tex]r = \sqrt{\frac{\O}{B \pi} }[/tex]
Substituting value
[tex]r = \sqrt{\frac{15 *10^{-6}}{3.142 * 0.6} }[/tex]
[tex]r = 2.82 *10^{-3}m[/tex]
For the particle to form a circular path the magnetic force the partial experience inside the magnetic must be equal to the centripetal force of the particle and this is mathematically represented as
[tex]F_q = F_c[/tex]
Where [tex]F_q = q B v[/tex]
and [tex]F_c = \frac{mv^2 }{r}[/tex]
Substituting this into the equation above
[tex]qBv = \frac{mv^2}{r}[/tex]
making v the subject
[tex]v = \frac{r q B}{m}[/tex]
substituting values
[tex]v = \frac{2.82 * 10^{-3} * 26 *10^{-9} 0.6}{2.10*10^{-16}}[/tex]
[tex]v = 209485.71 m/s[/tex]
The potential energy of the particle before entering the magnetic field is equal to the kinetic energy in the magnetic field
This is mathematically represented as
[tex]PE = KE[/tex]
Where [tex]PE = q * \Delta V[/tex]
and [tex]KE = \frac{1}{2} mv^2[/tex]
Substituting into the equation above
[tex]q \Delta V = \frac{1}{2} mv^2[/tex]
Making the potential difference the subject
[tex]\Delta V = \frac{mv^2}{2 q}[/tex]
[tex]\Delta V = \frac{2.16 * 10^ {-16} * (209485.71)^2 }{2 * 26 *10^{-9}}[/tex]
[tex]\Delta V = 177.2 \ V[/tex]
Final answer:
The speed of the particle in the magnetic field is found using the force equation for a moving charge in a magnetic field and the work-energy principle is used to determine the magnitude of the potential difference.
Explanation:
To answer the student's question:
The speed of the particle in the magnetic field can be calculated using the formula for the magnetic force acting on a moving charge, which is F = qvB. Since this force provides the centripetal force required for circular motion, F = mv2/r, where m is the mass of the particle, q is the charge of the particle, v is the speed of the particle, B is the magnetic field, and r is the radius of the circular path. Combining these equations, we can solve for the speed: v = qBr/m. To determine the radius, we can use the relationship between magnetic flux (ΦB), magnetic field strength (B), and area of the circular path (A): ΦB = BA or r = sqrt(ΦB / πB). The mass, charge, and magnetic field are given, and we can calculate the radius from the magnetic flux to then find the speed.
The magnitude of the potential difference the particle was initially accelerated through can be determined using the work-energy principle. The work done by the electric field on the charge is equal to the change in kinetic energy of the particle. Using the formula W = ΔKE = qΔV and knowing that the kinetic energy is KE = 1/2 mv2, we can solve for the potential difference: ΔV = KE/q, where KE is the kinetic energy of the particle right before entering the magnetic field.
1. Consider a head-on collision between two identical billiard balls. Ball 1 is initially in motion toward ball 2, which is initially at rest. After the collision, ball 2 departs with the same velocity that ball 1 originally had. Disregard any friction between the balls and the surface. What happens to ball 1? What happens to ball 2?
Answer:
first ball is at rest after collision and the kinetic energy of the first ball is transferred to the second ball after collision.
Explanation:
mass of each ball = m
initial velocity of first ball = u
initial velocity of second ball = 0 m/s
final velocity of second ball = initial velocity of first ball = u
Use conservation of momentum
let v is the final velocity of the first ball after collision
m x u + m x 0 = m x v + m x u
So, v = 0
It means after the collision the second ball moves with the velocity which is equal to the initial velocity of the first ball and the first ball comes at rest.
As there is no friction between the balls during the collision, the collision of two balls is perfectly elastic and thus the kinetic energy of the system is conserved.
The entire kinetic energy of the first ball is transferred to the second ball after the collision.
The entire kinetic energy of ball 1 will be totally transferred to the second ball after the collision.
According to the law of conservation of momentum, the change in momentum of a body before the collision is equal to the change in momentum after the collision.
The formula for calculating momentum is expressed as:
[tex]\rho = mv \\[/tex]
According to the conservation of momentum
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
Since ball 2 is initially at rest [tex]u_2=0[/tex]
Also if after the collision, ball 2 departs with the same velocity that ball 1 originally had, hence [tex]v=u_1[/tex]
Substituting the given parameters into the formula:
[tex]m_1u_1+m_2(0)=(m_1+m_2)u_1\\m_1u_1=m_1u_1+m_2u_2\\m_2u_2=0[/tex]
This shows that the velocity of the second ball is also zero
Due to the absence of friction between the colliding object, the collision is elastic in nature showing that the kinetic energy of the system is conserved.
The entire kinetic energy of ball 1 will be totally transferred to the second ball after the collision.
Learn more here: https://brainly.com/question/1113396
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.785 of the escape speed from Earth and (b) its initial kinetic energy is 0.785 of the kinetic energy required to escape Earth
Answer:
Explanation:
Given that,
We have a projectile that is shot from the earth
Then, it's escape velocity is
Ve = √(2GM / R)
Ve² = 2GM / R
M is the mass of earth
R is the radius of Earth
Using conservation of energy
Ki + Ui = Kf + Uf
The final kinetic energy is zero since the body is finally at rest
Then,
Ki + Ui = Uf
Kinetic energy can be determine using
K = ½mv²
Potential energy can be determine using
U = -GmM / R
Where m is mass of the body
Then,
Ki + Ui = Uf
½mv² - GmM / R = -GmM / r
r is the maximum height reached
m cancel out
½ v² - GM / R = -GmM / r
A. We want to find the maximum height reached when initial speed (v) is equal to 0.785 escape velocity
So, v = 0.785•Ve
So,
½v² - GM / R = -GM / r
½(0.785•Ve)² - GM / R = -GM / r
0.308•Ve² - GM / R = -GM / r
0.308 × 2GM / R - GM / R = -GM / r
Divide through by GM
0.616 / R - 1 / R = -1 / r
-0.384 / R = -1 / r
Cross multiply
-0.384r = -R
r = -R / -0.384
r = 2.6 R
B. When it initial kinetic energy is is 0.785 of the kinetic energy required to escape Earth
Ki = 0.785 Ke
Ki = 0.785 ½mVe²
Ki = 0.785 × ½m× 2GM / R
Ki = 0.785•G•m•M / R
So,
Ki + Ui = Uf
0.785•G•m•M / R - GmM / R = -GmM / r
Let G•m•M cancels out
0.785 / R - 1 / R = -1 / r
-0.215 / R = -1 / r
Cross multiply.
-0.215r = -R
r = -R / -0.215
r = 4.65 R
A plate of glass with parallel faces having a refractive index of 1.57 is resting on the surface of water in a tank. A ray of light coming from above in air makes an angle of incidence 37.5 ∘ with the normal to the top surface of the glass.
θ₃ is 27.12°
Explanation:
Using Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
where,
n₁ = refractive index of material with incident light
n₂ = refractive index of material with refracted light
(a)
1 → air
2 → glas
3 → water
n(water) = n₃ = 1.333
From air to glass,
n₁ sin θ₁ = n₂sin θ₂
The angle of refraction 2 is the angle of incidence when the light comes from glass to water.
From glass to water:
n₁ sin θ₁ = n₃ sin θ₃
1 X sin 37.5° = 1.333 sin θ₃
sin θ₃ = sin (37.5)/ 1.333
sin θ₃ = 0.456
θ₃ = 27.12°
Therefore, θ₃ is 27.12°
An electron is accelerated through a potential difference of 2.8 kV and directed into a region between two parallel plates separated by 17 mm with a potential difference of 100 V between them. The electron is moving perpendicular to the electric field when it enters the region between the plates. What magnetic field is necessary perpendicular to both the electron path and the electric field so that the electron travels in a straight line
Answer:
B=1.89*10^{-4} T
Explanation:
First we have to calculate the electric force on the electron, and then we have to take into account that this force is equal to the force generated by the magnetic field.
The formula is
[tex]F=q_eE\\\\E=\frac{V}{d}[/tex]
q: charge of the electron = 1.6*10^{-19}C
V: potential
d: separation between plates
[tex]E=\frac{100V}{17*10^{-3}m}=5882.35N/C[/tex]
[tex]F=(1.6*10^{-19}C)(5882.35N/C)=9.4*10^{-16}N[/tex]
This force must equal the Lorentz's force
[tex]F_E=F_B\\\\F_B=qvB\\\\B=\frac{F_B}{q_ev}[/tex]
But before we have to calculate the speed of the electron by using (me=9.1*10^{-31}kg)
[tex]E_e=\frac{1}{2}m_ev^2=q_eV\\\\v=\sqrt{\frac{2q_eV}{m_e}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.8*10^3V)}{9.1*10^{-31}kg}}\\\\v=3.13*10^7\frac{m}{s}[/tex]
Hence, we have
[tex]B=\frac{9.4*10^{-16}N}{(1.6*10^{-19}C)(3.1*10^{7})\frac{m}{s}}=1.89*10^{-4}T[/tex]
hope this helps!!
Answer:
The magnetic field that is necessary is equal to 1.88x10⁻⁴ T
Explanation:
If the electron is accelerated:
[tex]e*V=\frac{1}{2} mv^{2} \\v=\sqrt{\frac{2eV}{m} }[/tex]
Where
e = 1.6x10⁻¹⁹C
V = 2.8 kV = 2800 V
m = 9.1x10⁻³¹kg
Replacing:
[tex]v=\sqrt{\frac{2*1.6x10^{-19}*2800 }{9.1x10^{-31} } } =3.13x10^{7} m/s[/tex]
When the electron is moving in straight line, the magnetic force is balanced with the electric force, thus:
V = E * d
Where V = 100 V
d = 17 mm = 0.017 m
E = V/d = 100/0.017 = 5882.35 N/C
The magnetic field that is necessary is equal to:
B = E/v = 5882.35/3.13x10⁷ = 1.88x10⁻⁴ T
Electromagnetic radiation is more common than you think. Radio and TV stations emit radio waves when they broadcast their programs; microwaves cook your food in a microwave oven; dentists use X rays to check your teeth. Even though they have different names and different applications, these types of radiation are really all the same thing: electromagnetic (EM) waves, that is, energy that travels in the form of oscillating electric and magnetic fields.
Answer:
Explanation:
1. They have different wavelengths - Because These radiations form a spectra that differs by the size of the wavelength.
2. They have different Frequencies (f) that is frequency = 1/ wavelength
(f = 1/wavelength)
3. They propagate at different speed though a non vacuum media (non vacuum media affect the speed based on the wavelength)
An electron in an electron gun is accelerated from rest by a potential of 25 kV applied over a distance of 1 cm.
The final velocity of the electrons is _____.
The mass of the electron is 9.1x10^(-31) kg and its charge is 1.6x10^(-19) C.
Answer:
[tex]9.38\times 10^7 m/s[/tex]
Explanation:
We are given that
Potential ,V=25 kV=[tex]25\times 10^3 V[/tex]
Distance,r =1 cm=[tex]\frac{1}{100}=0.01 m[/tex]
1 m=100 cm
Mass of electron, m=[tex]9.1\times 10^{-31} kg[/tex]
Charge, q=[tex]1.6\times 10^{-19} C[/tex]
We have to find the final velocity of the electron.
Speed of electron,[tex]v=\sqrt{\frac{2qV}{m}}[/tex]
Using the formula
[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 25\times 10^3}{9.1\times 10^{-31}}[/tex]
v=[tex]9.38\times 10^7 m/s[/tex]
Hence, the final velocity of the electron=[tex]9.38\times 10^7 m/s[/tex]
Consider the standing wave pattern below created by a string fixed at both ends. The string is under a tension of 0.98 N and has a mass of 2.0 grams. What is the frequency of the standing wave?
Answer:
11.07Hz
Explanation:
Check the attachment for diagram of the standing wave in question.
Formula for calculating the fundamental frequency Fo in strings is V/2L where;
V is the velocity of the wave in string
L is the length of the string which is expressed as a function of its wavelength.
The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)
Therefore L = 1.5λ
If L = 3.0m
1.5λ = 3.0m
λ = 3/1.5
λ = 2m
Also;
V = √T/m where;
T is the tension = 0.98N
m is the mass per unit length = 2.0g = 0.002kg
V = √0.98/0.002
V = √490
V = 22.14m/s
Fo = V/2L (for string)
Fo = 22.14/2(3)
Fo = 22.14/6
Fo = 3.69Hz
Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo
Frequency of the wave = 3×3.69
Frequency of the wave = 11.07Hz
What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? View Available Hint(s) What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude.
Answer:
It has no effect on the amplitude.
Explanation:
When the sandbag is dropped, then the cart is at its maximum speed. Dropping the sand bag does not affect the speed instantly, this is because the energy remains within the system after the bag as been dropped. The cart will always return to its equilibrium point with the same amount of kinetic energy, as a result the same maximum speed is maintained.
Answer:
It decreases the amplitude.
Explanation:
If it is located at the end, its kinetic energy will be equal to zero, so the system as a whole would only have potential energy, which is defined as:
Ep = (1/2) * k * A²
Where A is the amplitude of the oscillation.
According to the energy principle, said energy must remain conserved, therefore, the total energy is equal to:
Etot = (1/2) * k * A²
Since the system is in equilibrium, according to the expression of conservation of energy, the energy in the position located at the end would be equal to the energy in equilibrium. As the sandbag falls, its kinetic energy decreases, which is why the total energy of the system also decreases.
According to the total energy expression, it is directly proportional to the square of the oscillation amplitude. If the total energy is decreased, the amplitude of the oscillation would also decrease.
A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reach a target 22 m above the marble's position on the compressed spring. (a) What is the change ΔUg in the gravitational potential energy of the marble-Earth system during the 22 m ascent? (b) What is the change ΔUs in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?
Answer:
a) [tex]\Delta U_{g} = 12.945\,J[/tex], b) [tex]\Delta U_{k} = 12.945\,J[/tex], c) [tex]k = 2930.059\,\frac{N}{m}[/tex]
Explanation:
a) The change in the gravitational potential energy of the marble-Earth system is:
[tex]\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)[/tex]
[tex]\Delta U_{g} = 12.945\,J[/tex]
b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:
[tex]\Delta U_{k} = 12.945\,J[/tex]
c) The spring constant of the gun is:
[tex]\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}[/tex]
[tex]k = \frac{2\cdot \Delta U_{k}}{x^{2}}[/tex]
[tex]k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}[/tex]
[tex]k = 2930.059\,\frac{N}{m}[/tex]
A mass M is attached to an ideal massless spring. When the system is set in motion, it oscillates with a frequency f. What is the new oscillation frequency if the mass is doubled to 2M? g
Answer:
[tex]\frac {f_i}{\sqrt 2}[/tex]
Explanation:
Frequency is given by [tex]\frac {1}{2\pi}\sqrt{{\frac {k}{m}}[/tex]
Let the initial frequency be denoted by [tex]f_i[/tex]
[tex]f_i=\frac {1}{2\pi}\sqrt{{\frac {k}{M}}[/tex]
When M is 2M then
[tex]f_f=\frac {1}{2\pi}\sqrt{{\frac {k}{2M}}=\frac {f_i}{\sqrt 2}[/tex]
Therefore, the final frequency is
[tex]\frac {f_i}{\sqrt 2}[/tex]
A fixed 16.1-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.23 s , the field is changed to 0.19 T pointing down. Part APart complete What is the average induced emf in the coil?
Answer:
Average induced emf will be equal to 0.00156 volt
Explanation:
We have given diameter of the wire d = 16.1 cm
So radius [tex]r=\frac{16.1}{2}=8.05cm=0.08m[/tex]
Cross sectional area of the wire [tex]A=\pi r^2=3.14\times 0.08^2=0.02m^2[/tex]
Magnetic field is changing from 0.53 T to 0.19 T
So change in magnetic field dB = 0.19-0.53 = -0.34 T
Time taken to change in magnetic field dt = 0.23 sec
Induced emf is given by [tex]e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}=-0.02\times \frac{0.34}{0.23}=0.00156volt[/tex]
So average induced emf will be equal to 0.00156 volt
A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated at 4.0 W (at 12.0 V), which are connected in parallel.
What is the battery's internal resistance?
0.46Ω
Explanation:The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;
E = V + Ir --------------------(a)
Where;
I = current flowing through the circuit
But;
V = I x Rₓ ---------------------(b)
Where;
Rₓ = effective or total resistance in the circuit.
First, let's calculate the effective resistance in the circuit:
The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.
Let;
R₁ = resistance in the first bulb
R₂ = resistance in the second bulb
Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;
P = [tex]\frac{V^{2} }{R}[/tex]
=> R = [tex]\frac{V^{2} }{P}[/tex] -------------------(ii)
Where;
P = Power of the bulb
V = voltage across the bulb
R = resistance of the bulb
To get R₁, equation (ii) can be written as;
R₁ = [tex]\frac{V^{2} }{P}[/tex] --------------------------------(iii)
Where;
V = 12.0V
P = 4.0W
Substitute these values into equation (iii) as follows;
R₁ = [tex]\frac{12.0^{2} }{4}[/tex]
R₁ = [tex]\frac{144}{4}[/tex]
R₁ = 36Ω
Following the same approach, to get R₂, equation (ii) can be written as;
R₂ = [tex]\frac{V^{2} }{P}[/tex] --------------------------------(iv)
Where;
V = 12.0V
P = 4.0W
Substitute these values into equation (iv) as follows;
R₂ = [tex]\frac{12.0^{2} }{4}[/tex]
R₂ = [tex]\frac{144}{4}[/tex]
R₂ = 36Ω
Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;
[tex]\frac{1}{R_{X} }[/tex] = [tex]\frac{1}{R_1}[/tex] + [tex]\frac{1}{R_2}[/tex] -----------------(v)
Substitute the values of R₁ and R₂ into equation (v) as follows;
[tex]\frac{1}{R_X}[/tex] = [tex]\frac{1}{36}[/tex] + [tex]\frac{1}{36}[/tex]
[tex]\frac{1}{R_X}[/tex] = [tex]\frac{2}{36}[/tex]
Rₓ = [tex]\frac{36}{2}[/tex]
Rₓ = 18Ω
The effective resistance (Rₓ) is therefore, 18Ω
Now calculate the current I, flowing in the circuit:
Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;
11.7 = I x 18
I = [tex]\frac{11.7}{18}[/tex]
I = 0.65A
Now calculate the battery's internal resistance:
Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;
12.0 = 11.7 + 0.65r
0.65r = 12.0 - 11.7
0.65r = 0.3
r = [tex]\frac{0.3}{0.65}[/tex]
r = 0.46Ω
Therefore, the internal resistance of the battery is 0.46Ω
Answer:
[tex]R_i_n_t=0.45 \Omega[/tex]
Explanation:
Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions that occur within a battery. When a charge is applied to a battery, the internal resistance can be calculated using the following equation:
[tex]R_i_n_t=(\frac{V_N_L}{V_F_L} -1)R_L[/tex]
Where:
[tex]V_F_L=Load\hspace{3}voltage=11.7V\\V_N_L= O pen\hspace{3}circuit\hspace{3}voltage=12V\\R_L=Load\hspace{3}resistance[/tex]
As you can see, we don't know the exactly value of the [tex]R_L[/tex]. However we can calculated that value using the next simple operations:
The problem tell us that the power of each lightbulb is 4.0 W at 12.0 V, hence let's calculated the power at 11.7V using Cross-multiplication:
[tex]\frac{12}{11.7} =\frac{4}{P}[/tex]
Solving for [tex]P[/tex] :
[tex]P=\frac{11.7*4}{12} =3.9W[/tex]
Now, the electric power is given by:
[tex]P=\frac{V^2}{R_b}[/tex]
Where:
[tex]R_b=Resistance\hspace{3}of\hspace{3}each\hspace{3}lightbulb[/tex]
So:
[tex]R_b=\frac{V^2}{P} =\frac{11.7^2}{3.9} =35.1\Omega[/tex]
Now, because of the lightbulbs are connected in parallel the equivalent resistance is given by:
[tex]\frac{1}{R_L} =\frac{1}{R_b} +\frac{1}{R_b} =\frac{2}{R_b} \\\\ R_L= \frac{R_b}{2} =\frac{35.1}{2}=17.55\Omega[/tex]
Finally, now we have all the data, let's replace it into the internal resistance equation:
[tex]R_i_n_t=(\frac{12}{11.7} -1)17.55=0.45\Omega[/tex]
Starting from rest, a 0.0367 kg steel ball sinks into a vat of corn syrup. The thick syrup exerts a viscous drag force that is proportional to the ball's velocity. → F drag = − C → v where C = 0.270 N ⋅ s/m is a constant related to the size and composition of the ball as well as the viscosity of the syrup. Find the rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity.
Answer:
0.48 W
Explanation:
Given that:
the mass of the steel ball = 0.0367 kg
C = 0.270 N
g (acceleration due to gravity) = 9.8
Now;
At Terminal Velocity Weight is balance by drag force
mg =Cv
Making v the subject of the formula:we have:
[tex]v = \frac {mg}{C}[/tex]
[tex]v = \frac {0.0367*9.8}{0.270}[/tex]
v = 1.332 m/s
Thus, the Rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity is:
P=Fv
P = mgv
[tex]P = (0.0367*9.8)*1.332[/tex]
P=0.48 W
Final answer:
To find the rate at which gravitational energy is converted to thermal energy at terminal velocity, equate the drag force to the gravitational force on the ball and solve for power as the product of force and velocity.
Explanation:
The question asks for the rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity in a vat of corn syrup. To find this rate, we must recognize that at terminal velocity, all the gravitational force acting on the ball (minus any buoyant force, which is ignored here) is converted into thermal energy due to viscous drag. The viscous drag force [tex]F_{drag}[/tex] is given by [tex]F_{drag[/tex] = -Cv, where C is the drag coefficient, and v is the velocity of the ball. The gravitational force acting on the ball can be calculated using [tex]F_{gravity[/tex] = mg, where m is the mass of the ball, and g is the acceleration due to gravity. At terminal velocity, [tex]F_{drag[/tex] = [tex]F_{gravity[/tex], hence the power (rate of energy conversion) P can be found as P = Fv = mgv. Assuming standard gravity (9.81 m/s2), the mass of the ball (0.0367 kg), and solving for v using [tex]F_{drag[/tex], one can determine the rate of gravitational energy conversion to thermal energy.
A loop of wire lies flat on a horizontal surface. A bar magnet is held above the center of a loop with south pole pointing down. Magnet is released.
As magnet approaches the loop of wire, the induced current in this wire is:
a) Clockwise
b) Counterclockwise
c) Zero
Answer:
A
Explanation:
Current, magnetic field and motion are mutually dependent and perpendicular to one another
Answer:
b) clockwise
Explanation:
The right hand rule states that to determine the direction of the magnetic force on a positive moving charge, point the thumb of the right hand in the direction of the potential v, the fingers in the direction of the magnetic field, B, and a perpendicular to the palm points in the direction of the force, F.
Since the south pole of the bar magnet is pointing downwards, the induced current developed in the wire repels it and moves in the clockwise direction.