Please ignore that i attempted this question
A firework shell is launched vertically upward from the ground with an initial speed of 44m/s. when the shell is 65 m high on the way up it explodes into two wequal mass halves, one half is observed to continue to rise straight up to a heigh of 120 m. How high does the other half go?
Answer:
[tex]h = 83.093\,m[/tex]
Explanation:
The speed of the firework shell just before the explosion is:
[tex]v = \sqrt{(44\,\frac{m}{s})^{2}-2\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (65\,m)}[/tex]
[tex]v \approx 25.712\,\frac{m}{s}[/tex]
After the explosion, the initial speed of one of the mass halves is:
[tex]v_{f}^{2} = v_{o}^{2} -2\cdot g \cdot s[/tex]
[tex]v_{o}^{2} = v_{f}^{2} + 2\cdot g \cdot s[/tex]
[tex]v_{o} = \sqrt{v_{f}^{2}+2\cdot g \cdot s}[/tex]
[tex]v_{o} = \sqrt{\left(0\,\frac{m}{s}\right)^{2}+ 2 \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (120\,m-65\,m)}[/tex]
[tex]v_{o} \approx 32.845\,\frac{m}{s}[/tex]
The initial speed of the other mass half is determined from the Principle of Momentum Conservation:
[tex]m \cdot (25.712\,\frac{m}{s} ) = 0.5\cdot m \cdot (32.845\,\frac{m}{s} ) + 0.5\cdot m \cdot v[/tex]
[tex]25.842\,\frac{m}{s} = 16.423\,\frac{m}{s} + 0.5\cdot v[/tex]
[tex]v = 18.838\,\frac{m}{s}[/tex]
The height reached by this half is:
[tex]h = h_{o} -\frac{v_{f}^{2}-v_{o}^{2}}{2\cdot g}[/tex]
[tex]h = 65\,m - \frac{(0\,\frac{m}{s} )^{2}- (18.838\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}[/tex]
[tex]h = 83.093\,m[/tex]
Answer:
The other half goes 17.4m high
Explanation:
Pls see calculation in the attached file
Two children are playing with a roll of paper towels. One child holds the roll between the index fingers of her hands so that it is free to rotate, and the second child pulls at constant speed on the free end of the paper towels. As the child pulls the paper towels, the radius of the roll of remaining towels decreases.
a)How does the torque on the roll change with time?
increases
decreases
stays the same
b)How does the angular speed of the roll change in time?
increases
decreases
stays the same
Answer:
a )Decreases
b ) Increases
c.) It's likely to break
Explanation:
a )The torque is as a result of friction, and also the magnitude is dependent on the normal force that exist between the surface in contact. The torque decrease because there is decrease in frictional force as the weight of the roll decrease
b.)The angular velocity will increase as radius o f the roll decrease because it is operating at constant speed.
c)if the child suddenly hit the roll, the papper will break, because it's angular acceleration might not be able to move along with the papper. Situation like this occur when there is large radius
A 6.0 m wire with a mass of 50 g, is under tension. A transverse wave, for which the frequency is 810 Hz, the wavelength is 0.40 m, and the amplitude is 4.0 mm, is propagating on the wire. a) How long will it take for a crest of this wave to travel the length of the wire? b) What is the tension in the wire?
Answer:
a) t = 0.0185 s = 18.5 ms
b) T = 874.8 N
Explanation:
a)
First we find the seed of wave:
v = fλ
where,
v = speed of wave
f = frequency = 810 Hz
λ = wavelength = 0.4 m
Therefore,
v = (810 Hz)(0.4 m)
v = 324 m/s
Now,
v = L/t
where,
L = length of wire = 6 m
t = time taken by wave to travel length of wire
Therefore,
324 m/s = 6 m/t
t = (6 m)/(324 m/s)
t = 0.0185 s = 18.5 ms
b)
From the formula of fundamental frquency, we know that:
Fundamental Frequency = v/2L = (1/2L)(√T/μ)
v = √(T/μ)
where,
T = tension in string
μ = linear mass density of wire = m/L = 0.05 kg/6 m = 8.33 x 10⁻³ k gm⁻¹
Therefore,
324 m/s = √(T/8.33 x 10⁻³ k gm⁻¹)
(324 m/s)² = T/8.33 x 10⁻³ k gm⁻¹
T = 874.8 N
Why do you suffer more pain than me when I slap your face?
A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is given an additional 6-in displacement in the positive direction and then released. The mass is in a medium that exerts a viscous resistance of 6 lb when the mass has a velocity of 3 ft/s. Under the assumptions discussed in this section, formulate the initial value problem that governs the motion of the mass.
Answer:
[tex]\frac{1}{8} y'' + 2y' + 24y=0[/tex]
Explanation:
The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by
[tex]my'' + \zeta y' + ky=0[/tex]
Where m is the mass, ζ is the damping constant, and k is the spring constant.
The spring constant k can be found by
[tex]w - kL=0[/tex]
[tex]mg - kL=0[/tex]
[tex]4 - k\frac{1}{6}=0[/tex]
[tex]k = 4\times 6 =24[/tex]
The damping constant can be found by
[tex]F = -\zeta y'[/tex]
[tex]6 = 3\zeta[/tex]
[tex]\zeta = \frac{6}{3} = 2[/tex]
Finally, the mass m can be found by
[tex]w = 4[/tex]
[tex]mg=4[/tex]
[tex]m = \frac{4}{g}[/tex]
Where g is approximately 32 ft/s²
[tex]m = \frac{4}{32} = \frac{1}{8}[/tex]
Therefore, the required differential equation is
[tex]my'' + \zeta y' + ky=0[/tex]
[tex]\frac{1}{8} y'' + 2y' + 24y=0[/tex]
The initial position is
[tex]y(0) = \frac{1}{2}[/tex]
The initial velocity is
[tex]y'(0) = 0[/tex]
We formulated the initial value problem for a damped spring-mass system:
1/8 * d²x/dt² + 2 * dx/dt + 24 * x = 0
with initial conditions x(0) = 0.5 ft and dx/dt(0) = 0 ft/s.
Let's break down the problem with the given data:
The mass weighs 4 lb.The spring stretches 2 inches under this weight.Additional displacement given is 6 inches in the positive direction.Viscous resistance is 6 lb when the velocity is 3 ft/s.First, find the spring constant k:
The weight of the mass stretches the spring by 2 inches (0.1667 feet).
The force exerted by the weight = 4 lb = mg
The spring force F = kx
So,
k = F/x k= 4 lb / 0.1667 ft k ≈ 24 lb/ftThe general form of the second-order differential equation governing the motion of the spring-mass-damper system is:
m*d²x/dt² + c*dx/dt + k*x = 0The viscous resistance given is 6 lb at 3 ft/s. Therefore, the damping coefficient c:
c = 6 lb / 3 ft/s c = 2 lb·s/ftThe initial conditions are displacement 6 inches (0.5 feet) and initial velocity 0:
x(0) = 0.5 ftCombining these elements, the initial value problem is:
1/8 * d²x/dt² + 2 * dx/dt + 24 * x = 0g A rope, under a tension of 233 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by . where at one end of the rope, is in meters, and is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope
Answer with Explanation:
We are given that
Tension=T=233 N
The displacement of the rope is given by
[tex]y=(0.320 m)sin(\frac{\pi x}{3})sin(10\pi)t[/tex]
a.By comparing with
[tex]y=Asin(kx)sin(\omega t)[/tex]
We get
A=0.32
k=[tex]\frac{\pi}{3}[/tex]
[tex]\omega=10\pi[/tex]
[tex]k=\frac{2\pi}{\lambda}[/tex]
[tex]\frac{\pi}{3}=\frac{2\pi}{\lambda}[/tex]
[tex]\lambda=3\times 2=6m[/tex]
n=2
[tex]n\lambda=2L[/tex]
[tex]2\times 6=2L[/tex]
[tex]L=6[/tex]m
b.[tex]\omega=2\pi f[/tex]
[tex]2\pi f=10\pi[/tex]
[tex]f=\frac{10}{2}=5Hz[/tex]
Speed,[tex]v=f\lambda=5\times 6=30m/s[/tex]
c.Let
Mass of the rope=m
[tex]\mu=\frac{m}{L}=\frac{m}{6}[/tex]
[tex]v^2=\frac{T}{\mu}[/tex]
[tex](30)^2=\frac{233}{\frac{m}{6}}[/tex]
[tex]900\times \frac{m}{6}=233[/tex]
[tex]m=\frac{233\times 6}{900}=1.553 kg[/tex]
When a golfer tees off, the head of her golf club which has a mass of 152 g is traveling 44.8 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 27.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.
Answer:
56.5 m/s²
Explanation:
From the law of conservation of momentum,
mu+m'u' = mv+m'v'........................ Equation 1
Where m = mass of the golf club, u = initial velocity of the golf club, m' = mass of the golf ball, u' = initial velocity of the golf ball, v = final velocity of the golf club, v' = final velocity of the golf ball.
From the question,
The golf ball is at rest, Hence u' = 0 m/s
mu = mv+m'v'
Make v' the subject of the equation
v' = (mu-mv)/m'........................... Equation 2
Given: m = 152 g = 0.152 kg, u = 44.8 m/s, v = 27.7 m/s, m' = 46 g = 0.046 kg.
Substitute into equation 2
v' = (0.152×44.8+0.152×27.7)/0.046
v' = (6.8096-4.2104)/0.046
v' = 2.5992/0.046
v' = 56.5 m/s²
A howler monkey is the loudest land animal and under some circumstances, can be heard up to a distance of 5.0km. Assume the acoustic output of a howler to be uniform in all directions and that the threshold of hearing is 1.0*10^-12 W/m^2. The acoustic power emitted by the howler is clostest to:
A) 0.31mW
B) 1.1mW
C) 3.2mW
D) 11mW
Answer:
Power emitted will be 0.314 mW
So option (A) will be correct option
Explanation:
We have given threshold hearing [tex]I=10^{-12}W/m^2[/tex]
Distance is given r = 5 km =5000 m
We have to find the power emitted
Power emitted is equal to
[tex]P=I\times A[/tex]
[tex]=10^{-12}\times 4\pi r^2[/tex]
[tex]=10^{-12}\times 4\times 3.14\times (5000)^2[/tex]
=[tex]314\times 10^{-6}watt=0.314mW[/tex]
So power emitted will be 0.314 mW
So option (A) will be correct option.
Which of the following statements are true concerning the reflection of light?
Check all that apply.
a. The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane mirror.
b. The reflection of light from a smooth surface is called specular reflection.
c. The reflection of light from a rough surface is called diffuse reflection.
d. For diffuse reflection, the angle of incidence is greater than the angle of reflection.
e. For specular reflection, the angle of incidence is less than the angle of reflection.
Answer:
b. The reflection of light from a smooth surface is called specular reflection.
c. The reflection of light from a rough surface is called diffuse reflection.
Explanation:
a. The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane mirror.
This is wrong: Based on law of reflection "The angle of incidence is equal to the angle of reflection when light strikes any plane surface" examples plane mirrors, still waters, plane tables, etc
b. The reflection of light from a smooth surface is called specular reflection.
This is correct
c. The reflection of light from a rough surface is called diffuse reflection.
This is correct
d. For diffuse reflection, the angle of incidence is greater than the angle of reflection.
This is wrong: the angle of incident is equal to angle of reflection. The only difference between this type of reflection and specular reflection, is that the normal for diffuse reflection is not parallel to each due to the rough surface in which the light incidents.
For specular reflection, the angle of incidence is less than the angle of reflection.
This is wrong: the angle of incident is equal to angle of reflection
The correct statements are statement 2 and statement 3.The true statements are: the reflection of light from a smooth surface is called specular reflection, and the reflection of light from a rough surface is called diffuse reflection. The others are incorrect as the angle of incidence always equals the angle of reflection.
To answer the question about the reflection of light, let's analyze each statement:
The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane mirror.In summary, the true statements concerning the reflection of light are:
The reflection of light from a smooth surface is called specular reflection.The reflection of light from a rough surface is called diffuse reflection.
A person stands in an elevator
weighing a Cheeseburger with a
kitchen scale « lit could happent
The mass of the cheese burger at
10 ilsoka The scale reads l-ION
Determine the magnitude ana
Lirecteon of the netforce on the
Cheeseburger
Answer:
Noennt ied olod l qeu es ocmo a balazuna de cozcna
in
Explanation:
A hemispherical droplet of liquid water, lying on a flat surface, evaporates by molecular diffusion through still air surrounding the droplet. The droplet initially has a radius R. As the liquid water slowly evaporates, the droplet shrinks slowly with time, but the flux of the water vapor is at a nominal steady state. The temperature of the droplet and the surrounding still air are kept constant. The air contains water vapor of fixed concentration at an infinitely long distance from the droplet’s surface. After drawing a picture of the physical process, select a coordinate system that will best describe this diffusion process, list at least five reasonable assumptions for the mass-transfer aspects of the water-evaporation process, and simplify the general differential equation for mass transfer in terms of the flux NA. Finally, specify the simplified differential form of Fick’s flux equation for water vapor (species A), and propose reasonable boundary conditions.
Answer:
Explanation:
question answered below
We described the diffusion process of a slowly evaporating hemispherical water droplet by adopting a spherical coordinate system. We made five assumptions: steady-state conditions, isotropic medium, uniform temperature, constant concentration of water vapor in the surrounding air, and no bulk motion of air. The differential form of Fick's law was derived, and two boundary conditions regarding the concentration of water were proposed.
Explanation:To answer your question regarding the diffusion process of a hemispherical droplet of water slowly evaporating on a flat surface, we will start by selecting a spherical coordinate system. This is chosen because a spherical system best represents a hemispherical droplet with a shrinking radius R.
Now, let's list at least five reasonable assumptions for the mass-transfer aspects of the water-evaporation process:
Steady-state conditions - The rate of evaporation is constant over time.Isotropic medium - The properties of the medium (air) are the same in all directions.Uniform temperature - The temperature of the system is constant throughout the process.Constant concentration of water vapor - At an infinite distance from the droplet surface, the air contains a fixed concentration of water vapor.No bulk motion of air - The air around the droplet is still, causing diffusion to occur solely due to molecular motion.To express the mass transfer in terms of the flux NA, we need to derive from Fick's first law, which states that the molar flux due to diffusion is proportional to the concentration gradient. In this case, it simplifies to: NA = -D(dCA/dx), where D is the diffusion coefficient and dCA/dx is the concentration gradient of species A (water).
As for the boundary conditions, we can consider two conditions:
The concentration of water at the droplet's surface is the saturation concentration.The concentration at infinite distance from the droplet is essentially constant.Learn more about Diffusion of Evaporating Water Droplet here:https://brainly.com/question/33422701
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A 56.0-kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 22.0 m. (a) What is the centripetal acceleration of the child? magnitude 1.76 m/s2 direction (b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? magnitude 647.4 N direction (c) What force does the seat exert on the child at the highest point of the ride? magnitude 450.2 N direction (d) What force does the seat exert on the child when the child is halfway between the top and bottom? (Assume the Ferris wheel is rotating clockwise and the child is moving upward.) magnitude N direction 79.8 ° counter-clockwise from the horizontal
Answer:
a. 3.86 m/s²
b. 332.64 N directed upwards towards the center of the ferris wheel.
c. 764.96 N directed downwards towards the center of the ferris wheel.
d. 589.84 N at 21.5° counterclockwise from the horizontal direction.
Explanation:
a. Since the ferris wheel rotates 4 times per minute, its period, T = 60 s/4 = 15 s.
We now find it angular speed ω = 2π/T = 2π/15 = 0.418 rad/s
We then calculate its centripetal acceleration from a = rω² where r = radius of ferris wheel = 22.0 m.
So, a = 22 m × (0.418 rad/s)² = 3.86 m/s²
b. At the lowest point, the normal force, N and the centripetal force, F both act in opposite directions to the weight, mg of the object. So,
N + F = mg
N = mg - F
N = mg - ma where a is the centripetal acceleration
N = m(g - a)
N = 56 kg(9.8 m/s² - 3.86 m/s²)
N = 56 kg × 5.94 m/s²
N = 332.64 m/s²
The normal force the seat exerts on the child is thus 332.64 N directed upwards towards the center of the ferris wheel.
c. At the highest point, the weight, mg of the object and the centripetal force, F both act in opposite directions to the normal force, N. So,
N = mg + F
N = mg + ma where a is the centripetal acceleration
N = m(g + a)
N = 56 kg(9.8 m/s² + 3.86 m/s²)
N = 56 kg × 13.66 m/s²
N = 764.96 N
The normal force the seat exerts on the child is thus 764.96 N directed downwards towards the center of the ferris wheel.
d. Half way between the top and the bottom of the ferris wheel, the normal force must balance the weight and the centripetal force so the child doesn't fall off. For this to happen, the normal force is thus the resultanf of the centripetal force and the weight of the child. Since these two forces are perpendicular at this instance,
N = √[(mg)² + (ma)²] = m√(g² + a²) = 56 kg√[(9.8 m/s²)² + (3.86 m/s²)²] = 56 kg√[(97.196 (m/s²)² + 14.8996(m/s²)²] = 56kg√110.9396 (m/s²)² = 56 kg × 10.53 m/s² = 589.84 N.
Since the centripetal force acts towards the center of the ferris wheel in the horizontal direction, it is equal to the horizontal component of the normal force, Also, the weight acts downwards and is equal to the vertical component of the normal force.
So, the direction of the normal force is gotten from
tanθ = ma/mg = a/g
θ = tan⁻¹(a/g) = tan⁻¹(3.86 m/s² / -9.8 m/s²) = tan⁻¹(-0.3939) = -21.5°. Since the angle is it shows a counter clockwise direction.
So, the normal force is 589.84 N at 21.5° counterclockwise from the horizontal direction.
(a) The centripetal acceleration of the child is 1.94 m/s².
(b) The force the seat exert on the child at the lowest point of the ride is 657.4 N.
(c) The force the seat exert on the child at the highest point of the ride is 440.2 N.
(d) The force the seat exert on the child when the child is halfway between the top and bottom is 108.64 N.
The given parameters;
mass of the child, m = 56 kgangular speed, ω = 4 rev/mindiameter of the wheel, d = 22 mradius of the wheel, r = 11 mThe centripetal acceleration of the child is calculated as follows;
[tex]a_c = \omega ^2 r\\\\[/tex]
where;
ω is angular speed of the wheel in rad/s
The angular speed of the wheel in rad/s is calculated as follows;
[tex]\omega = 4 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} \\\\\omega = 0.419 \ rad/s[/tex]
The centripetal acceleration of the child is calculated as;
[tex]a_c = \omega^2 r\\\\a_c = (0.419)^2 \times 11\\\\a_c = 1.94 \ m/s^2[/tex]
The force the seat exert on the child at the lowest point of the ride is calculated as;
[tex]T_{bottom} = ma_c + mg\\\\T_{bottom} = m(a_c + g)\\\\T_{bottom} = 56(1.94 + 9.8)\\\\T_{bottom} = 657.4 \ N[/tex]
The force the seat exert on the child at the highest point of the ride is calculated as;
[tex]T_{top} = mg- ma_c\\\\T_{top} = m(g - a_c)\\\\T_{top} = 56(9.8 - 1.94)\\\\T_{top} = 440.2 \ N[/tex]
The force the seat exert on the child when the child is halfway between the top and bottom;
[tex]T = ma_c\\\\T = 56(1.94)\\\\T = 108.64 \ N[/tex]
Learn more here:https://brainly.com/question/15561667
The earth’s magnetic field points toward (magnetic) north. For simplicity, assume that the field has no vertical component (as is the case near the earth’s equator). (a) If you hold a metal rod in your hand and walk toward the east, how should you orient the rod to get the maximum motional emf between its ends? (i) East-west; (ii) north-south; (iii) up-down; (iv) you get the same motional emf with all of these orientations. (b) How should you hold it to get zero emf as you walk toward the east? (i) East-west; (ii) north-south; (iii) up-down; (iv) none of these. (c) In which direction should you travel so that the motional emf across the rod is zero no matter how the rod is oriented? (i) West; (ii) north; (iii) south; (iv) straight up; (v) straight down. ❙
Answer:
A. Up-down
B. East-west & north -south
C. North or south
Explanation:
See attached handwritten document for more details
A 3.0-µF capacitor charged to 40 V and a 5.0-µF capacitor charged to 18 V are connected to each other, with the positive plate of each connected to the negative plate of the other. What is the final charge on the 3.0-µF capacitor?
The final charge on the 3.0-µF capacitor is 90 µC.
Explanation:To find the final charge on the 3.0-µF capacitor, we need to use the concept of charge conservation in a series circuit. In a series circuit, the charge across capacitors connected in series is the same. Therefore, the final charge on the 3.0-µF capacitor will be the same as the charge on the 5.0-µF capacitor.
Using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage, we can calculate the charge on the 5.0-µF capacitor.
Q = (5.0 µF)(18 V) = 90 µC.
Therefore, the final charge on the 3.0-µF capacitor is also 90 µC.
The rotor of a helicopter is gaining angular speed with constant angular acceleration. At tt = 0 it is rotating at 1.25 rad/srad/s. From tt = 0 to tt = 2.00 ss, the rotor rotates through 8.00 radrad. What is the angular acceleration of the rotor?
Answer:
[tex]2.75\ rad/s^2[/tex]
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = 1.25 rad/s
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation = 8 rad
t = Time taken = 2 s
[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 8=1.25\times 2+\frac{1}{2}\times \alpha\times 2^2\\\Rightarrow \alpha=\dfrac{2}{2^2}(8-1.25\times 2)\\\Rightarrow \alpha=2.75\ rad/s^2[/tex]
The angular acceleration of the rotor is [tex]2.75\ rad/s^2[/tex]
A turntable is off and is not spinning. A 0.8 g ant is on the disc and is 9 cm away from the center. The turntable is turned on and 0.8 s later it has an angular speed of 45 rpm. Assume the angular acceleration is constant and determine the quantities for the ant 0.4 s after the turntable has been turned on. Express all quantities using appropriate mks units.
Answer:
I] alpha = wf/t = (45*2pi/60)/1 = 4.712 rad/s
ii] W = alpha*t = 2.356 rad/s
iii] v = Wr =2.356*0.10 = 0.2356 m/s
iv] atan = alpha*r = 0.4712 m/s^2
v] arad= v^2/r = 0.2356^2/0.10 = 0.555 m/s^2
vi] a = sqrt(0.4712^2+0.555^2) = 0.728 m/s^2
vii] Fnet = ma = 1.6e-3*0.728 = 0.00116 N
Explanation:
A charged particle moving along the x-axis enters a uniform magnetic field pointing along the z-axis. Because of an electric field along the y-axis, the charge particle does not change velocity. What is the sign of this particle
Answer:
Explanation:
Let the charge particle have charge equal to +q .
force due to electric field will be along the field that is along y - axis . To balance it force by magnetic force must be along - y axis. ( negative of y axis )
force due to magnetic field = q ( v x B ) , v is velocity and B is magnetic field.
F = q ( v i x B k ) , ( velocity is along x direction and magnetic field is along z axis. )
= (Bqv) - j
= - Bqv j
The force will be along - ve y - direction .
If we take charge as negative or - q
force due to electric field will be along - y axis .
magnetic force = F = -q ( v i x B k )
= + Bqv j
magnetic force will be along + y axis
So it is difficult to find out the nature of charge on the particle from this experiment.
In order to study the long-term effects of weightlessness, astronauts in space must be weighed (or at least "massed"). One way in which this is done is to seat them in a chair of known mass attached to a spring of known force constant and measure the period of the oscillations of this system. The 36.4 kg chair alone oscillates with a period of 1.00 s, and the period with the astronaut sitting in the chair is 2.20 s.
Find the force constant of the spring.
Answer:
Approximately [tex]1.44\times 10^3 \; \rm N \cdot m^{-1}[/tex] assuming that the spring has zero mass.
Explanation:
Without any external force, a piece of mass connected to an ideal spring (like the chair in this question) will undergo simple harmonic oscillation.
On the other hand, the force constant of a spring (i.e., its stiffness) can be found using Hooke's Law. If the spring exerts a restoring force [tex]\mathbf{F}[/tex] when its displacement is [tex]\mathbf{x}[/tex], then its force constant would be:
[tex]\displaystyle k = -\frac{\mathbf{F}}{\mathbf{x}}[/tex].
The goal here is to find the expressions for [tex]F[/tex] and for [tex]x[/tex]. By Hooke's Law, the spring constant would be ratio of these two expressions.
Let [tex]T[/tex] represent the time period of this oscillation. With the chair alone, the period of oscillation is [tex]T = 1.00\; \rm s[/tex].
For a simple harmonic oscillation, the angular frequency [tex]\omega[/tex] can be found from the period:
[tex]\displaystyle \omega = \frac{2\pi}{T}[/tex].
Let [tex]A[/tex] stands for the amplitude of this oscillation. In a simple harmonic oscillation, both [tex]\mathbf{F}[/tex] and [tex]\mathbf{x}[/tex] are proportional to [tex]A[/tex]. Keep in mind that the spring constant [tex]k[/tex] is simply the opposite of the ratio between [tex]\mathbf{F}[/tex] and [tex]\mathbf{x}[/tex]. Therefore, the exact value of [tex]A[/tex] shouldn't really affect the value of the spring constant.
In a simple harmonic motion (one that starts with maximum displacement and zero velocity,) the displacement (from equilibrium position) at time [tex]t[/tex] would be:
[tex]\displaystyle \mathbf{x}(t) = A \cos(\omega \cdot t)[/tex].
The restoring velocity at time [tex]t[/tex] would be:
[tex]\displaystyle \mathbf{v}(t) = \mathbf{x}^\prime(t) = -A\, \omega \sin(\omega\cdot t)[/tex].
The restoring acceleration at time [tex]t[/tex] would be:
[tex]\displaystyle \mathbf{a}(t) = \mathbf{v}^\prime(t) = -A\, \omega^2 \cos(\omega\cdot t)[/tex].
Assume that the spring has zero mass. By Newton's Second Law of motion, the restoring force at time [tex]t[/tex] would be:
[tex]\begin{aligned}& \mathbf{F}(t) \\ &= m(\text{chair}) \cdot \mathbf{a}(t) \\&= -m(\text{chair}) \, A\, \omega^2 \cos(\omega \cdot t)\end{aligned}[/tex].
Apply Hooke's Law to find the spring constant, [tex]k[/tex]:
[tex]\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= -\left(\frac{-m(\text{chair}) \, A\, \omega^2 \cos(\omega \cdot t)}{A\cos(\omega \cdot t)}\right) \\ &= \omega^2 \cdot m(\text{chair}) \end{aligned}[/tex].
Again, [tex]\omega[/tex] stands for the angular frequency of this oscillation, where
[tex]\displaystyle \omega = \frac{2\pi}{T}[/tex].
Before proceeding, note how [tex]A[/tex] was eliminated from the ratio (as expected.) Additionally, [tex]t[/tex] is also eliminated from the ratio. In other words, the spring constant is "constant" at all time. That agrees with the assumption that this spring is indeed ideal. Back to [tex]k[/tex]:
[tex]\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= \cdots \\ &= \omega^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{T}\right)^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{1.00\; \rm s}\right)^2 \times 36.4\; \rm kg\end{aligned}[/tex].
Side note on the unit of [tex]k[/tex]:
[tex]\begin{aligned} & 1\; \rm kg \cdot s^{-2} \\ &= 1\rm \; \left(kg \cdot m \cdot s^{-2}\right) \cdot m^{-1} \\ &= 1\; \rm N \cdot m^{-1}\end{aligned}[/tex].
A certain elastic conducting material is stretched into a circular loop of 12.0 cm radius. It is placed with its plane perpendicular to a uniform 0.800 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 75.0 cm/s . What emf is induced in the loop at that instant
Answer:
The induced emf in the loop is 0.452 volts.
Explanation:
Given that,
Radius of the circular loop, r = 12 cm = 0.012 m
Magnetic field, B = 0.8 T
When released, the radius of the loop starts to shrink at an instantaneous rate of, [tex]\dfrac{dr}{dt}=75\ cm/s=0.75\ m/s[/tex]
The induced emf in the loop is equal to the rate of change of magnetic flux. It is given by :
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA)}{dt}\\\\\epsilon=B\dfrac{-d(\pi r^2)}{dt}\\\\\epsilon=2\pi rB\dfrac{-dr}{dt}\\\\\epsilon=2\pi \times 0.12\times 0.8\times 0.75\\\\\epsilon=0.452\ V[/tex]
So, the induced emf in the loop is 0.452 volts.
Every day a certain amount of water evaporates from Earth’s oceans, lakes, and land surface and forms water vapor and clouds in the atmosphere. Every day a certain amount of rain falls back to Earth. Make the reasonable assumption that, on average, the energy absorbed by the evaporation and lifting of the water is equal to the energy released by its condensation and falling back to earth. The evaporation of one mole of water requires approximately 40.6×103 J of heat. An equal amount of heat is released when a mole of water condenses.
QUESTION: The annual volume of rainfall on Earth is approximately 4.9×105 km3 and the average cloud altitude is 8.8 km above Earth’s surface. How much energy, in joules, is required every day to evaporate and lift the water?
Answer:
0.0000000010 joules
Explanation:
Amount : 1540.3 nanojoules (nJ)
Equals : 0.0 joules (J)
The total amount of energy required everyday to evaporate and lift the water is 1.15 × 10¹⁸ J
Since one mole of water requires 40.6 × 10³ J of heat to evaporate, and the volume of annual rainfall is 4.9 × 10⁵ km³ = 4.9 × 10⁸ m³, we need to find the amount of heat required to evaporate this volume of rainfall.
Amount of heat required to evaporate this volume of rainfall.So, E = nH where
n = number of moles of water in rainfall = m/M where m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = 4.9 × 10⁸ m³ and M = molar mass of water = 18 g/mol = 1.8 × 10⁻² kg/mol and H = energy required to evaporate one mole of water = 40.6 × 10³ J/molSo, E = nH
E = mH/M
E = ρVH/M
Substituting the values of the variables into the equation, we have
E = ρVH/M
E = 1000 kg/m³ × 4.9 × 10⁸ m³ × 40.6 × 10³ J/mol/ 1.8 × 10⁻² kg/mol
E = 198.94 × 10¹⁴ J-kg/mol/ 1.8 × 10⁻² kg/mol
E = 110.52 × 10¹⁶ J
E = 1.1052 × 10¹⁸ J
The amount of energy required to raise this volume to an altitude of 8.8 km
Now, the amount of energy required to raise this volume to an altitude of 8.8 km is E' = mgh where
m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = 4.9 × 10⁸ m³, g = acceleration due to gravity = 9.8 m/s² and h = average cloud altiude = 8.8 km = 8.8 × 10³ mSo, E' = mgh
E' = ρVgh
Substituting the values of the variables into the equation, we have
E' = ρVgh
E' = 1000 kg/m³ × 4.9 × 10⁸ m³ × 9.8 m/s² × 8.8 × 10³ m
E' = 422.576 × 10¹⁴ J
E' = 4.22576 × 10¹⁶ J
The total amount of energy requiredSo, the total energy required to evaporate and raise the volume of water is E" = E + E'
E" = 110.52 × 10¹⁶ J + 4.22576 × 10¹⁶ J
E" = 114.74576 × 10¹⁶ J
E" = 1.1474576 × 10¹⁸ J
E" ≅ 1.15 × 10¹⁸ J
The total amount of energy required everyday to evaporate and lift the water is 1.15 × 10¹⁸ J
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Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon return, the people on earth will have advanced exactly 110 years into the future. According to special relativity, how fast must you travel
Answer:
0.999958c
Explanation:
Remember that the trip involves traveling for six months at a constant velocity thus returning home at same speed . The time interval of this trip will therefore be one year.
Accurate time interval to be measured by the clock on the spaceship Δt0 = 1.0 years
Time interval as advanced on earth observed from the spacecraft Δt = 110 years
the formula for time dilation is
Δt = Δt0 /√(1-v2/c2)
or v = c*√1-(Δt0/Δt)2
=c*√[1-(1year/110years)2]
= 0.999958c
Speculate about some worldwide changes likely to follow the advent of successful fusion reactors. Compare the advantages and disadvantages of electricity coming from a large central power station versus a network of many smaller solar-based stations owned and operated by individuals.
Answer:
Nuclear reaction is the disintegration or integration of nucleus or nuclei of an atom, to produce a different nuclei or nucleus, thereby given out a large amount of heat energy. The energy been extracted from this reaction are very much.
At a successful fusion reaction, the world will have more better opportunity to preserve it's natural resources, as they will be enough energy to produce electricity, cooking, industrial power, and many more. It will also reduce spending, as energy from fusion reaction are not too expensive in obtaining, when compared to energy from natural resources.
ADVANTAGES OF USING CENTRAL POWER STATION THAN SMALLER SOLAR BASED STATION.
1) Central power stations are reliable, because of the abundance of natural resources been used to produce the electricity. Solar based station are not reliable because during winter, when the sun is hardly seen, they will not be enough resource to produce electricity.
2) Central power stations are very efficient because they produce a stable electricity power, while the smaller solar stations are not efficient, because the amount of power produced is dependent on the amount of solar energy it has stored.
DISADVANTAGES OF USING CENTRAL POWER STATION THAN SMALLER SOLAR-BASED STATION.
1) Central power stations causes pollution, and green house effect, because the natural gas been used to produce the electricity causes pollution and green house effect during venting. While smaller solar-based station are pollution free and does not cause green house effect, because it only need sun to store it's charge.
2) Central power station are more dangerous for people to work, as risk of being electrocuted is high. Smaller-based solar stations has less risk, as the power been stored from the sun requires less complex wiring.
3) Central power stations are more expensive to produce and requires much land space. While smaller solar-based station are less expensive, and requires little land space.
Final answer:
Successful fusion reactors could dramatically reshape global politics and economies by reducing reliance on fossil fuels. Centralized fusion power offers consistency, whereas decentralized solar power fosters resilience and local control but is intermittent. Fusion faces technological challenges, while solar technology is readily expanding.
Explanation:
Worldwide Changes with Successful Fusion Reactors
The advent of successful fusion reactors would be a transformative event worldwide. Fusion promises a cleaner, nearly inexhaustible energy source compared to fossil fuels, which could significantly alter the current geopolitical landscape where oil and natural gas are dominant factors in international politics and economics. A shift toward fusion energy could lessen global dependence on fossil fuels, potentially reducing the geopolitical power of oil-rich nations and creating a new paradigm in the world economy. With hydrogen from water as fuel, countries may realign alliances and shift focus towards technological advancements and harnessing fusion energy efficiently.
Central vs. Distributed Power Generation
Electricity from a large central power station, such as a fusion reactor, offers economies of scale and a consistent power supply but requires a significant up-front investment and complex infrastructure. Meanwhile, smaller solar-based stations, which are often owned and operated by individuals, offer distributed generation, empowering local communities and enhancing energy resilience. However, solar energy faces challenges with intermittency and requires storage solutions to provide a reliable power supply.
The disadavantages of fusion, despite its potential, include the complexities and high costs associated with maintaining such a power source, as well as addressing issues related to radioactivity. On the other hand, the technology for solar power is already available today, with deployment expanding rapidly thanks to decreasing costs and improving battery storage technologies.
Challenges in Fusion Technology
Developing fusion reactors for electricity generation requires overcoming significant technical hurdles, such as achieving the necessary high temperatures and creating materials that can contain the fusion reaction without melting. Scientists are optimistic, yet the timeline for fusion reactors to become a commercial reality remains uncertain.
A very long uniform line of charge has charge per unit length λ1 = 4.80 μC/m and lies along the x-axis. A second long uniform line of charge has charge per unit length λ2 = -2.26 μC/m and is parallel to the x-axis at y1 = 0.400 m . (Hint: You will need to know or derive using Gauss law the electric field by a uniform line charge)
Answer:
a) E=228391.8 N/C
b) E=-59345.91N/C
Explanation:
You can use Gauss law to find the net electric field produced by both line of charges.
[tex]\int \vec{E_1}\cdot d\vec{r}=\frac{\lambda_1}{\epsilon_o}\\\\E_1(2\pi r)=\frac{\lambda_1}{\epsilon_o}\\\\E_1=\frac{\lambda_1}{2\pi \epsilon_o r_1}\\\\\int \vec{E_2}\cdot d\vec{r}=\frac{\lambda_2}{\epsilon_o}\\\\E_2=\frac{\lambda_2}{2\pi \epsilon_o r_2}[/tex]
Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.
The net electric field at point r will be:
[tex]E=E_1+E_2=\frac{1}{2\pi \epsilon_o}(\frac{\lambda_1}{r_1}+\frac{\lambda_2}{r_2})[/tex]
a) for y=0.200m, r1=0.200m and r2=0.200m:
[tex]E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C[/tex]
b) for y=0.600m, r1=0.600m, r2=0.200m:
[tex]E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C[/tex]
a) The electric field strength for case 1 will be 228391.8 N/C
b) The electric field strength for case 2 will be 59345.91 N/C
What is gauss law?The total electric flux out of a closed surface is equal to the charge contained divided by the permittivity,
According to Gauss Law. The electric flux in a given area is calculated by multiplying the electric field by the area of the surface projected in a plane perpendicular to the field.
The given data in the problem is;
λ₁ is the charge per unit length for charge 1 = 4.80 μC/m
λ₂ is the charge per unit length for charge 2 =-2.26 μC/m
y₁ is the distance from charge 1 =0.400 m.
From the gauss law, the net electric field is produced by both lines of charges will be .;
[tex]\rm E= \frac{\lambda_}{2 \pi \epsilon_0 r_2} \\\\ \rm E_1= \frac{\lambda_1}{2 \pi \epsilon_0 r_1} \\\\ \rm E_2= \frac{\lambda_2}{2 \pi \epsilon_0 r_2}[/tex]
The total electric field will be the sum of the electric field for the charge 1 and2;
[tex]\rm E=E_1+E_2 \\\\ \rm E=\frac{\lambda_1}{2 \pi \epsilon_0 r_1}+\frac{\lambda_2}{2 \pi \epsilon_0 r_2} \\\\[/tex]
The net electric field for y=0.200m, r₁=0.200m, and r₂=0.200 m will be;
[tex]\rm E=\frac{4.80 \times 10^{-6}}{2 \times 3.14 8.85 \times 10^{-12} \times 0.200}-\frac{2.26 \times 10^{-6}}{2 \times 3.14 \times 8.85 \times 10^{-12}} \\\\\rm E=228391.8 \ N/C[/tex]
The net electric field for y=0.600m, r₁=0.600m, r₂=0.200 m will be;
[tex]\rm E=\frac{4.80 \times 10^{-6}}{2 \times 3.14 8.85 \times 10^{-12} \times 0.600}-\frac{2.26 \times 10^{-6}}{2 \times 3.14 \times 8.85 \times 10^{-12}} \\\\\rm E=-59345.51 \ N/C[/tex]
Hence the electric field strength for case 1 and case 2 will be 228391.8 N/C and 59345.91 N/C respectively.
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A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is low, the penny rides up and down without difficulty. If the frequency is steadily increased, there comes a point at which the penny leaves the surface.A) At what point in the cycle does the penny first lose contact with the piston?midpoint moving uplowest pointhighest pointmidpoint moving downB) What is the maximum frequency for which the penny just barely remains in place for the full cycle?Express your answer with the appropriate units.
Answer:
the penny loses contact at the piston's highest point.
f = 2.5 Hz
Explanation:
Concepts and Principles
1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by
∑F = ma (1)
2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:
a_max = -w^2A (2)
3- The angular frequency w of a wave is related to the frequency f by:
w = 2π f (3)
Given Data
- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)= 0.04 m.
- The frequency of oscillation of the piston is steadily increased.
Required Data
In part (a), we are asked to determine the point at which the penny first loses contact with the piston.
In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.
Solution
(a)
The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.
figure 1 is attached
Apply Newton's second law from Equation (1) in the vertical direction to the penny:
∑F_y -mg= ma
Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So
0 = m(g + a)
a = —g
Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.
(b)
The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):
a = —w^2A
where a = —g at the highest point. So
g = w^2A
Solve for w:
w =√g/A
Substitute for w from Equation (3):
2πf = √g/A
Solve for f :
f = 1/2π√g/A
Substitute numerical values:
f = 1/2π√9.8 m/s^2/0.04
f = 2.5 Hz
The question is based on Simple Harmonic Motion: The penny first loses contact with the piston at its highest point (option c) in the motion. The maximum frequency for which the penny just barely remains in place for the full cycle is approximately 12.5 Hz.
Explanation:The penny will first lose contact with the piston at its highest point, which is option c) highest point in the cycle.
To determine the maximum frequency for which the penny just barely remains in place for the full cycle, we need to consider the condition for the penny to stay in contact with the piston. At the highest point of the motion, the penny experiences an upward gravitational force and a downward centripetal force due to the circular motion.
When the centripetal force becomes greater than or equal to the gravitational force, the penny will lose contact with the piston. This occurs when
mvmax2/r = mg
Solving for the maximum velocity using vmax = 2πfA, where A is the amplitude, and substituting into the equation above, we find that
fmax = g / (4π2A)
Substituting the given values, with a=4.0 cm = 0.04 m and g=9.8 m/s2:
fmax = 9.8 / (4π2(0.04)) ≈ 12.5 Hz
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A cosmic-ray proton in interstellar space has an energy of 19.5 MeV and executes a circular orbit having a radius equal to that of Mars' orbit around the Sun (2.28 1011 m). What is the magnetic field in that region of space?
Answer:
B = (2.80 × 10⁻¹²) T
Explanation:
First of, we convert the 19.5 MeV to Joules
19.5 MeV = 19.5 × 10⁶ × 1.602 × 10⁻¹⁹
= (3.124 × 10⁻¹²) J
The velocity of the cosmic-ray proton can be calculated from the kinetic energy formula
E = (1/2) mv²
m = mass of a proton = (1.67 × 10⁻²⁷) kg
(3.124 × 10⁻¹²) = (1/2)(1.67 × 10⁻²⁷)v²
v = (6.117 × 10⁷) m/s
And since the magnetic force keeps the cosmic ray proton in uniform circular motion,
Magnetic force = Centripetal force keeping the proton in circular motion.
qvB = (mv²/r)
q = charge on a proton = (1.602 × 10⁻¹⁹) C
v = velocity of the proton = (6.117 × 10⁷) m/s
B = magnetic field = ?
r = radius of circular motion = (2.28 × 10¹¹) m
B = (mv/qr)
B = (1.67 × 10⁻²⁷ × 6.117 × 10⁷) ÷ (1.602 × 10⁻¹⁹ × 2.28 × 10¹¹)
B = (2.797 × 10⁻¹²) T
Hope this Helps!!!
Answer:
The magnetic field strength in that region of space is 2.7983 x 10⁻¹² T
Explanation:
Given;
Energy of the cosmic-ray proton, E = 19.5 MeV = 19.5 x 10⁶ x 1.6 x 10⁻¹⁹
E = 3.12 x 10⁻¹² J
Radius of the circular orbit, r = 2.28 x 10¹¹ m
Step 1:
determine the speed of cosmic-ray proton
Kinetic energy of the cosmic-ray proton;
K.E = ¹/₂mv²
[tex]v = \sqrt{\frac{2K.E}{m} }[/tex]
where;
m is mass of proton, m = 1.67 x 10⁻²⁷ kg
[tex]v = \sqrt{\frac{2*3.12*10^{-12}}{1.67*10^{-27}} } \\\\v = 6.1127*10^7 \ m/s[/tex]
Step 2:
determine the magnetic field strength in that region of space
magnetic force = centripetal force
[tex]qvB = \frac{mv^2}{r} \\\\B = \frac{mv}{rq}\\\\[/tex]
Where;
q is charge of proton, q = 1.6 x 10⁻¹⁹ C
[tex]B = \frac{mv}{rq} = \frac{(1.67*10^{-27})(6.1127*10^7)}{(2.28*10^{11})(1.6*10^{-19})}\\\\B =2.7983 *10^{-12} \ T[/tex]
Therefore, the magnetic field strength in that region of space is 2.7983 x 10⁻¹² T
A circuit contains two inductors of 6.0 mH inductance in parallel placed in series with an inductor of 8.0 mH inductance. After one of the 6.0 mH inductors burns out, the repairman wants to replace all three inductors with one inductor of equivalent inductance. Assuming inductors combine in series and parallel the same way resistors do, what inductance should he use
Answer:
Check attachment for circuit diagram
And better understanding
Explanation:
Let find the equivalent inductance.
The 6mH in parallel to 6mH
Parallel connection will give
1 / Leq = 1 / L1 + 1 / L2
L1 = L2 = L = 6mH
1 / Leq = 1 / 6 + 1 / 6
1 / Leq = 2 / 6
1 / Leq = 1 / 3
Then, take reciprocal
Leq = 3 mH
Now the combInation of this parallel connection is connect in series with an 8mH inductor
Series connection is given as
Leq = L1 + L2
So, the equivalent of the parallel connection is 3mH and this will be connect in series with 8mH
Then, final inductance is
Leq = 3 + 8
Leq = 11 mH
Therefore, the repairman need to replace the element with one inductor of inductance 11 mH.
This question involves the concepts of inductance, series combination, and parallel combination.
The inductance to be used is found to be "11 mH".
RESULTANT INDUCTANCEFirst, we will find out the resultant inductance of the two inductors connected in a parallel combination. Since it is given that inductors behave like resistors in combination. Therefore, resultant of two inductances connected in parallel combination will be:
[tex]\frac{1}{L}=\frac{1}{L_1}+\frac{1}{L_2}[/tex]
where,
L = resultant inductance of the two inductors in parallel combination = ?L₁ = L₂ = inductance of the two inductors = 6 mH = 6 x 10⁻³ HTherefore,
[tex]\frac{1}{L}=\frac{1}{6\ x\ 10^{-3}\ H}+\frac{1}{6\ x\ 10^{-3}\ H}\\\\[/tex]
L = 3 x 10⁻³ H = 3 mH
Now, we will calculate the total inductance of this combination in series combination with the third inductor as follows:
[tex]L_T=L+L_3=3\ mH+8\ mH\\L_T=11\ mH[/tex]
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A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angular momentum of the particle about each of the following reference points. The coordinates of the reference points have units of (meters, meters). Assume the positive z-axis is directed out of the screen. Randomized Variables m= 2.5 kg v = 2 i m/sPart (a) Determine the angular momentum in kg.m-/s of the particle about (0,1). Part (b) Determine the angular momentum in kgm/s of the particle about (0, -1). Part (c) Determine the angular momentum in kg.m-/s of the particle about (1,0).Part (d) Determine the angular momentum in kg.m/s of the particle about (-1,0). Part (e) Determine the angular momentum in kg. m²/s of the particle about (1,1). Part (f) Determine the angular momentum in kg ·mº/s of the particle about (-1,1). Part (g) Determine the angular momentum in kg.mʻls of the particle about (-1,-1).Part (h) Determine the angular momentum in kg.m-/s of the particle about (1,-1). Part (i) Determine the angular momentum in kg.m/s of the particle about (0,0).
Answer:
please read the answer below
Explanation:
The angular momentum is given by
[tex]|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)[/tex]
By taking into account the angles between the vectors r and v in each case we obtain:
a)
v=(2,0)
r=(0,1)
angle = 90°
[tex]L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}[/tex]
b)
r=(0,-1)
angle = 90°
[tex]L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}[/tex]
c)
r=(1,0)
angle = 0°
r and v are parallel
L = 0kgm/s
d)
r=(-1,0)
angle = 180°
r and v are parallel
L = 0kgm/s
e)
r=(1,1)
angle = 45°
[tex]L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}[/tex]
f)
r=(-1,1)
angle = 45°
the same as e):
L = 5kgm/s
g)
r=(-1,-1)
angle = 135°
[tex]L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}[/tex]
h)
r=(1,-1)
angle = 135°
the same as g):
L = 5kgm/s
hope this helps!!
The determinant method allows to find the z component of the angular momentum for the particle moving in the x axis are:
a) L_z = - 5.0 kg m / s
b) L_z = 5 kg m/s
c) L_z = 0
d) L_z = 0
e) L_z = -5.0 kg m / s
f) L_z = 5.0 kg / ms
g) L_z = 5.0 kg m / s
The angular momentum is defined by the vector product of the position and the linear momentum, therefore it is a vector quantity.
L = r x p
Where the bold letters represent vectors, L the angular momentum, r the position and p the linear moment
Momentum is defined by.
p = m v
where m is the mass.
The modulus of angular momentum is:
L = m r v sin θ
One way to find the angular momentum in various dimensions is to solve for the determinant method.
[tex]L = m \ \left[\begin{array}{ccc}i&j&k\\x&y&z\\v_x&v_y&v_z\end{array}\right][/tex]
The component z of the angular momentum is
[tex]L_z = m ( x v_y - y v_x)[/tex]
Indicate that the mass of the body is m = 2.5 kg and it moves in the direction x, v = 2 i m/s, let's calculate the angular momentum in direction z for each case:
The velocity is v = 2i
a) point r = (0,1)
[tex]L_z[/tex] = - 2,5 1 2
L_z = - 5.0 kg m / s
b) point r = (0, -1)
L_z = -2.5 (-1) 2
L_z = 5 kg m/s
c) point r = (1,0)
L_z = 2,5 0 2
L_z = 0
velocity is parallel to position
d) point r = (-1,0)
L_z = 0
Velocity is parallel to position
e) point r = (1,1)
L_z= 2.5 (1 0 - 2 1)
L_z = -5.0 kg m / s
f) point r = (- 1, -1)
L_z = 2.5 (1 0 - 2 (-1))
L_z = 5.0 kg / ms
g) point r = (1, -1)
L_z = 2,5 (1 0 - 2 (-1))
L_z = 5.0 kg m / s
In conclusion using the determinant method we can find the z component of the angular momentum for the particle moving in the x axis are:
a) L_z = - 5.0 kg m / s
b) L_z = 5 kg m/s
c) L_z = 0
d) L_z = 0
e) L_z = -5.0 kg m / s
f) L_z = 5.0 kg / ms
g) L_z = 5.0 kg m / s
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The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional boxer exerts a force of 1.8 × 103 N with an effective perpendicular lever arm of 2.85 cm, producing an angular acceleration of the forearm of 140 rad/s2.
What is the moment of inertia of the boxer's forearm?
Answer:
0.366kgm²
Explanation:
F = 1.8*10³N
r = 2.85cm = 0.0285m
α = 140rad/s²
Torque = applied force * distance
τ = r * F
τ = 0.0285 * 1.8*10³
τ = 51.3N.m
but τ = I * α
I = τ / α
I = 51.3 / 140
I = 0.366kgm²
Homework 7 Adaptive Follow-Up Problem 25.68 Item 4 Constants A cylindrical copper cable 3.30 km long is connected across a 220.0-V potential difference. Part A What should be its diameter so that it produces heat at a rate of 50.0 W ? Express your answer with the appropriate units. TemplatesSymbols undoredoresetkeyboard shortcutshelp D = nothing nothing Request Answer Part B What is the electric field inside the cable under these conditions? Express your answer with the appropriate units. TemplatesSymbols undoredoresetkeyboard shortcutshelp E = nothing nothing Request Answer Provide Feedback
First, we calculate the current through the cable using given Power and Voltage. Then, we use this current to find the resistance and subsequently, the diameter of the cable. For Part B, the electric field inside a cable is zero in steady state conditions.
Explanation:To solve Part A of the question, we will use the formula Power = I2R, where I is current and R is resistance. First, we find the Current from the given power and voltage, Power = Voltage * Current. Second, we use this current to calculate the resistance, R = Power / I2. Lastly, we find the diameter using the formula for the Resistance of a cylindrical conductor, R = rho * L / A, where rho is resistivity, L is length and A is area. For Part B, the electric field inside a conductor under steady state conditions is zero.
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The diameter of the copper wire can be determined by solving an equation using its cross-sectional area. The electric field inside the copper wire is calculated by dividing the potential difference by the distance.
Explanation:The question pertains to calculating the diameter of a cylindrical copper cable and the electric field inside it given certain conditions. This involves the use of formulas related to the principles of electromagnetism, and specifically electric current and electric field.
In order to determine the diameter of the cable (Part A), we would first need to deduce the cross-sectional area using data provided. The formula AD² = 4 could be used, where A represents the cross-sectional area and D the diameter of the cable. Substituting the value found for area, we can solve for the diameter.
For Part B, the electric field inside the cable can be calculated using the formula E = V/d, where V is the potential difference and d is the distance. Given that the potential difference is 220V and the length or distance is 3.3 km, the desired electric field can be computed.
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A new concrete mix is being designed to provide adequate compressive strength for concrete blocks. The specification for a particular application calls for the blocks to have a mean compressive strength µ greater than 1350 kPa. A sample of 100 blocks is produced and tested. Their mean compressive strength is 1356 kPa and their standard deviation is 70 kP
Complete question:
new concrete mix is being designed to provide adequate compressive strength for concrete blocks. The specification for a particular application calls for the blocks to have a mean compressive strength µ greater than 1350 kPa. A sample of 100 blocks is produced and tested. Their mean compressive strength is 1356 kPa and their standard deviation is 70 kP
a) Find the p value.
b) Do you believe it is plausible that the blocks do not meet the specification, or are you convinced that they do? Expain your reasoning.
Answer:
a) p-value= 0.1949
b) It is possible the blocks do not meet the specifications
Explanation:
Given:
n = 100
Sample mean, X' = 1356
Standard deviation, s.d = 70
a) To find p- value.
Null hypothesis:
H0: u ≤ 1350
Alternative hypothesis:
H1 : u > 1350
The test statistic wll be:
[tex] z= \frac{X'-u_o}{s.d/ \sqrt{n}}[/tex]
[tex] = \frac{1356-1350}{70/\sqrt{100}}[/tex]
=0.86
The p value will be:
= P(z>0.86)
= 1-P(z≤0.86)
Using the normal distribution table, we now have:
1 - 0.8051
= 0.1949
P value = 0.1949
b) Since our p-value is 0.1949, we do not reject the null hypothesis, because the p-value, 0.1949, is not small. This means that it is possible the blocks do not meet the specifications.
The probability that the compressive strength is greater than 1350 kPa is 80.51%
Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\Where\ x=raw\ score,\mu=mean, \sigma=standard\ deviation, n=sample\ size[/tex]
Given that n = 100, μ = 1356, σ = 70. For x > 1350 kPa:
[tex]z=\frac{1350-1356}{70/\sqrt{100} } =-0.86[/tex]
From the normal distribution table, P(x > 1350) = P(z > -0.86) = 1 - P(z < -0.86) = 1 - 0.1949 = 80.51%
Hence the probability that the compressive strength is greater than 1350 kPa is 80.51%
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