The motion of a particle is given by x = A sin3(ωt) where the amplitude is
x/sin3(ωt).
What does a SHM's amplitude mean?The largest deviation a particle makes from its typical position is the amplitude of a SHM. A point on a vibrating body or wave can move as far or as far from its equilibrium position as possible, or at its greatest amplitude. The length of the vibration route is divided in half by this value.
b) The expression for the particle's velocity
=> sin^-1(x)/3t.
c) The expression for the particle's acceleration
=>ωt = a = sin^-1(x)/3.
How do angular velocity and acceleration relate?The rate at which angular velocity changes is referred to as angular acceleration. In an equation, angular acceleration is written as follows: = t, where t is the change in time and is the change in angular velocity.
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Calculate the de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s (about 60 mi/h)
The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.
De Broglie wavelength:The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.
λ=h/p
Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.
Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.
Therefore, λ=h/(mv)
λ=(6.63×10⁻³⁴)/(0.56×26)
λ=4.55×10⁻³⁵ m.
The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.
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In the given figure, weight of stone inside water
is 9N and water displaced by stone is 2N then,
i)What is the actual weight of stone?
ii) Which principle is the
experiment based on?
The actual weight of the stone is 11 N. It is based on the Archimedes principles.
What is Archimedes principle?Archimedes principle states that the up thrust by water on an object is equal to the weight of water displaced.Upthrust by water on an object= actual weight of object - weight inside waterWhat is the actual weight of the object, if its weight inside water is 9N and weight of water displaced is 2N?Actual weight= weight inside water+ weight of water displaced
= 9N + 2N = 11N
Thus, we can conclude that the actual weight of the object is 11N.
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A toroidal solenoid has 600 turns, cross-sectional area 6.90 cm2, and mean radius 4.30 cm.
a) Calculate the coil's self-inductance.
b) If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms , calculate the self-induced emf in the coil.
c) The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a?
(a) The coil's self-inductance is 7.26 mH.
(b) The self-induced emf in the coil is 7.26 V
(c) The direction of the induced emf is from b to a.
Coil's self-inductance
L = N²μA/I
L = (600² x 4π x 10⁻⁷ x 6.9 x 10⁻⁴)/(0.043)
L = 7.26 x 10⁻³ H
L = 7.26 mH
Self-induced emf in the coilemf = N(ΔBA)/t
where;
B is magnetic fieldA is area N is number of turnst is timeB = μNI/L
B1 = (4π x 10⁻⁷ x 600 x 5)/0.043
B1 = 0.0876 T
B2 = (4π x 10⁻⁷ x 600 x 2)/0.043
B2 = 0.035 T
emf = NΔBA/t
emf = (600)(0.0876 - 0.035)(6.9 x 10⁻⁴) / (3 x 10⁻³)
emf = 7.26 V
The direction of the induced emf is always opposite to the direction of the applied current.
Thus, the direction of the induced emf is from b to a.
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By how many newtons does the weight of a 100-kg person decrease when he goes from sea level to mountain top at an altitude of 5000 m? The mean radius of the earth is 6.38 × 106 m.
The weight of a 100-kg person decreases by 4 N when he goes from sea level to mountain top at an altitude of 5000 m.
What determines the weight of a person?The weight of a person is determined by the mass of the body and the acceleration due to gravity.
The acceleration due to gravity, g is dependent on the mass of the earth, M the radius of the earth and the gravitational force constant , G.
Mathematically, the acceleration due to gravity at the mountain top is determined using the formula:
g = GM/r²where:
G = 6.67 × 10⁻¹¹ Nm²/kg²
M = 5.9736 x 10²⁴ kg
r = 638000 + 5000 = 6385000
g = (6.67 × 10⁻¹¹ * 5.9736 x 10²⁴ )(6385000)²
g = 9.77 m/s²
His weight at the mountain top will be:
weight = 100 * 9.77
weight = 977 N
Weight at sea level = 100 * 9.81 = 981 N
Decrease in weight = 981 - 977
Decrease in weight = 4 N
In conclusion, the weight of the man varies according to his distance from the earth.
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The mass of Saturn is 5.68 x 1026 kg, and use an orbital radius of 3.00 x 105 km. (G = 6.67 × 10-11 N ∙ m2/kg2). Find the orbital speed of an ice cube in the rings of Saturn.
The orbital speed of an ice cube in the rings of Saturn is determined as 355,366.5 m/s.
What is orbital speed?
The orbital speed of an astronomical body or object is the speed at which it orbits around the center of mass of the most massive body.
Orbital speed of ice cube in the rings of SaturnThe orbital speed of ice cube in the rings of Saturn is calculated as follows;
v = √GM/r
where;
G is universal gravitation constantM is mass of Saturnr is the distance of the ice cubev = √(6.67 x 10⁻¹¹ x 5.68 x 10²⁶)/(3 x 10⁵)
v = 355,366.5 m/s
Thus, the orbital speed of an ice cube in the rings of Saturn is determined as 355,366.5 m/s.
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A 1300 kg steel beam is supported by two ropes. (Figure
1)
What is the tension in rope 1?
What is the tension in rope 2?
Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.
By Newton's second law,
the net horizontal force acting on the beam is[tex]R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0[/tex]
where [tex]R_1,R_2[/tex] are the magnitudes of the tensions in ropes 1 and 2, respectively;
the net vertical force acting on the beam is[tex]R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0[/tex]
where [tex]m=1300\,\rm kg[/tex] and [tex]g=9.8\frac{\rm m}{\mathrm s^2}[/tex].
Eliminating [tex]R_2[/tex], we have
[tex]\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)[/tex]
[tex]R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2[/tex]
[tex]R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2[/tex]
[tex]-R_1 \sin(50^\circ) = -\dfrac{mg}2[/tex]
[tex]R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}[/tex]
Solve for [tex]R_2[/tex].
[tex]\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0[/tex]
[tex]\dfrac{R_2}2 = -mg\cot(110^\circ)[/tex]
[tex]R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}[/tex]
when a branding iron gets heated in a fire the handle becomes hot, how is the energy transferred to the handle
Conduction is a process through which energy is transferred from the branding iron to the handle.
When two objects with different surface temperatures come into close touch with one another, conduction happens. Up until they reach thermal equilibrium, or the point at which they are at the same temperature, the heat energy moves from the hotter to the cooler object.
The metal tip of the branding iron, which makes contact with the branded surface, becomes extremely hot when it is placed in the fire because of the high fire temperature.
Hence, conduction is a process through which energy is transferred from the branding iron to the handle.
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the temperature at which the velocity of sound in air is twice its velocity at 15°C
With the use of below formula, at 879 °C, velocity will be double the velocity at 15 °C.
What is the relationship between Velocity and sound ?The velocity of sound waves in air is proportional to the square root of Thermodynamic temperature. That is, V = K[tex]\sqrt{T}[/tex]
Given that the temperature at which the velocity of sound in air is twice its velocity at 15°C, Let us make use of the formula;
(v2/v1) = √(T2 / T1)
Where
v2 = final velocityv1 = initial velocityT2 = final absolute temperatureT1 = initial temperature.Recall that absolute temperature = °C + 273.
If v2 = 2 × v1 and temperature in degree Celsius = 15°C, then,
Temperature in Kelvin K = 15 + 273 = 288
Substitute all the parameters into the formula
(2 × v1)/v1 = √(T2/288)
2 = √ (T2 /288)
Square both sides
4 = (T2/288)
T2 = 4 × 288
T2 = 1152K
Temperature in degrees Celsius = 1152 - 273 = 879 °C.
Therefore, at 879 °C, velocity will be double the velocity at 15 °C.
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A canon ball is shot out of a cannon at an angle of 45 degrees. What is the initial velocity of the cannon ball if its initial horizontal velocity is 8 m/s?
Answer:
11.31 [m/s].
Explanation:
1. the required velocity can be calculated according to
[tex]V=\frac{V_{horizontal}}{sin45};[/tex]
2. according to the formula above:
V=8*1.41≈11.3137085 [m/s].
Thanks to the direction finding feature in gmaps application, which most of us use, we can find our way. Here, the maps application offers us alternative routes. Among these suggestions, I want to choose the path that will have the least fuel and do this based on calculations. For example, one of the two directions may be short, but if that short route is also uphill, it will not be an economical route. In my opinion the most important factor is elevation. If we take elevation into account other factors such as friction, where assuming the same asphalt type is often used in the same area for friction, I think the correct result will be achieved. In your opinion, what are the input data required to find the least energy path, what assumptions can be made and what are the necessary formulations and calculations?
In my opinion, I think that the input data that are required to find the least energy path are:
ElevationDistanceWhat is a Map?This refers to the use of a diagram to represent the features of a place that shows its physical landforms to help in navigation.
Hence, we can see that when using maps like gmaps, it is important to consider both elevation and distance to be able to find the path that uses the least energy.
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ATTA-KAY PHYSICS 121 3. A ball A is left to roll down an inclined plane of inclination 30º. Just at the moment a second ball B is pushed up the plane with a velocity of 45ms¹. The balls met at a point where the velocity of B is 1.8 times the velocity of A. Calculate (a) the velocities of A and B when they meet. (b) when the two balls meet. (c) where the two ball meet. (g = 10 ms²)
vr>vs because the rolling ball acquires rotational as well as translational kinetic energy.
The accelerating force acting on the ball as it goes along a smooth plane is mgsin. Its acceleration is therefore equal to gsin. The mgsin acts down the plane as the ball travels down the rough inclined plane, but friction develops that acts up the plane.
Since both balls' potential energy is lost at the same rate, their KEs are actually equal at the base of the planes. However, a ball sliding down a smooth plane has only translational kinetic energy, but a ball rolling down a rough plane contains both translational and rotational kinetic energy at the bottom of the plane. As a result, the ball's translational KE will be lower than its translational Kinetic energy.
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Two spheres with uniform surface charge density, one with a radius of 7.4 cm and the
other with a radius of 5.0 cm, are separated by a center-to-center distance of 38 cm. The
spheres have a combined charge of +55μC and repel one another with a force of 0.62 N.
Assume that the charge of the first sphere is greater than the charge of the second sphere.
Question 1.
What is the surface charge density in the sphere of radius 7.4?
Question 2
What is the surface density on the 2nd sphere?
EXPRESS ANSWER USING TWO SIGNIFICANT FIGURES.
(a) The surface charge density in the sphere of radius 7.4 is 0.0322 C/m³.
(b) The surface density on the 2nd sphere is 3.48 x 10⁻⁴ C/m³.
Total charge of the spheresF = kq₁q₂/r²
Fr² = kq₁q₂
q₁q₂ = Fr²/k
where;
r is distance between the chargesk is Coulomb's constantq₁q₂ = (0.62 x 0.38²) / (9 x 10⁹)
q₁q₂ = 9.95 x 10⁻¹² C
q₂ = 9.95 x 10⁻¹² C/q₁
From the question;
q₁ + q₂ = 55 x 10⁻⁶
q₁ + 9.95 x 10⁻¹² /q₁ = 55 x 10⁻⁶
q₁² + 9.95 x 10⁻¹² = 55 x 10⁻⁶q₁
q₁² - 55 x 10⁻⁶q₁ + 9.95 x 10⁻¹² = 0
solve the quadratic equation using formula method;
q₁ = 5.48 x 10⁻⁵ C
q₂ = 55 x 10⁻⁶ C - 5.48 x 10⁻⁵ = 1.82 x 10⁻⁷ C
Volume of the first sphereV1 = ⁴/₃πr³
V1 = (⁴/₃ π)(0.074)³ = 1.7 x 10⁻³ m³
Surface charge density = (5.48 x 10⁻⁵ C) / (1.7 x 10⁻³ m³) = 0.0322 C/m³
Volume of the second sphereV2 = (⁴/₃ π)(0.05)³ = 5.236 x 10⁻⁴ m³
Surface charge density = ( 1.82 x 10⁻⁷ C) / (5.236 x 10⁻⁴ m³) = 3.48 x 10⁻⁴ C/m³
Thus, the surface charge density in the sphere of radius 7.4 is 0.0322 C/m³.
The surface density on the 2nd sphere is 3.48 x 10⁻⁴ C/m³.
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How big is this restoring force compared with the tensile force stretching the spring?
A. Bigger
B. Not enough info
C. Smaller
D. Same size
The restoring force on the spring is found to have exactly the same magnitude as the stretching force. Option D
What is the restoring force?The restoring force is the force that seeks to restore the spring to its equilibrium position. It has the same magnitude as the stretching force but acts in opposite direction.
Thus, the restoring force on the spring is found to have exactly the same magnitude as the stretching force.
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How much work must be done to stop a 975- kg car traveling at 105 km/h ?
Express your answer to two significant figures and include the appropriate units.
The amount of work done to stop a 975- kg car traveling at 105 km/h is 414,808.34J.
How to calculate work done?The amount of work done by a moving object can be calculated using the following formula:
W (Kinetic energy) = ½ mv²
Where;
m = massv = velocityAccording to this question, a car of 975 kg is traveling at 105 km/h. This speed in m/s is 29.17m/s.
K.E = ½ × 975 × 29.17²
K.E = 414,808.34J
Therefore, the amount of work done to stop a 975- kg car traveling at 105 km/h is 414,808.34J.
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An object of height 8.50 cm is placed 20.0 cm to the left of a converging lens with a focal length of 12.0 cm. Determine the image location in cm, the magnification, and the image height in cm.
The image distance is 33.3 cm while the image height is 14.2 cm.
What is a converging lens?A converging lens will always have a positive focal length hence, we have to find the object distance as follows;
1/f = 1/v + 1/u
1/12 = 1/v + 1/20
1/v = 1/12 - 1/20
1/v = 0.08 - 0.05
v =33.3 cm
Now;
Magnification = 33.3 cm/20.0 cm =1.67
M = Image height/Object height
1.67 = Image height/8.50 cm
Image height = 1.67 * 8.50 cm
Image height = 14.2 cm
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A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale.
Evaluate the magnitude of the force on the left hand pole.
The Force on the left hand pole, F' = 0.167N
What is the force on the left hand pole?Force is an agent which produces a change in the motion or state of an object.
Force is a vector quantity.
The general force is calculated as follows:
F = mg/sinθ
m = 17.1 g = 0.0171 kg
g = 9.81 m/s²
θ = 45°
F = 0.0171 * 9.81/sin45
F = 0.237 N
Force on the left hand pole, F' = Fcosθ
F' = 0.237 * cos 45
F' = 0.167N
In conclusion, the force on the left hand pole is the horizontal component of force.
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An object with a density of 941.0 kg/m3 and a mass of 1039.0 kg is thrown into the ocean. Find the volume that sticks out of the water. (use ρseawater = 1024 kg/m3)
The volume that sticks out of the water is 83 m³.
To find the answer, we need to know about the archimedes principle.
What's archimedes principle?It says that when an object is on a water surface, the amount of force on the object is equal to the weight of water displaced by it.Mathematically, weight of the object= weight of water displacedWhat's the volume of an object remain on the water surface, if the density and mass of the object are 941.0 kg/m³, 1039.0 kg respectively?Let V = volume of the object, v= volume of water displacedV-v = volume that sticks out of the waterWeight of the object = V× density of object × gWeight of water displaced= v× density of water × gAs per archimedes principle, V× density of object × g=v× density of water × gV-v = density of water - density of object= 1024 - 941 = 83 m³
Thus, we can conclude that the volume that sticks out of the water is 83 m³.
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A shot-putter accelerates a 7.5 kg shot from rest to 12 m/s .
If this motion takes 1.6 s , what average power was developed?
Express your answer using two significant figures.
From the calculations, the power developed is 337.5 W.
What is the power developed?First we must obtain the acceleration from;
v = u + at
u = 0 m/s because the motion started rom rest
v = at
a = v/t
a = 12 m/s/ 1.6 s
a= 7.5 m/s^2
The force is obtained from;
F = ma = 7.5 kg * 7.5 m/s^2 = 56.25 N
Now the distance covered is obtained from';
v^2 = u^2 + 2as
v^2 = 2as
s = v^2/2a
s = (12)^2/2 * 7.5
s = 9.6 m
Now;
Work = Fs = 56.25 N * 9.6 m = 540 J
Power expended = 540 J/ 1.6 s = 337.5 W
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Light of intensity I0 is polarized vertically and is incident on an analyzer rotated at an angle from the vertical. Find the angle if the transmitted light has intensity
I = (0.750)I0, I = (0.500)I0, I = (0.250)I0, and I = 0.
(Enter your answers in degrees.)
a. θ = 41. 4°
b. θ = 60°
c. θ = 75. 5°
d. θ = 90°
How to determine the angleFrom the given information, we would be using the Malus' law
It is given as;
I = I0 cos²θ
Where I0 is the intensity of the polarized light after passing through P
a. To find the angle, compare with the given equation
I = (0.750)I0
I = I0 cos θ
then
cos θ = 0. 750
θ = [tex]cos^-^1(0. 750)[/tex]
θ = 41. 4°
b. I = (0.500)I0
cos θ = 0. 500
θ = [tex]cos^-^1(0. 500)[/tex]
θ = 60°
c. I = (0.250)I0
cos θ = 0. 250
θ = [tex]cos^-^1 (0. 250)[/tex]
θ = 75. 5°
d. I = 0
cos θ = 0
θ = [tex]cos^-^1 (0)[/tex]
θ = 90°
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Discuss the aspects of either the Gemini Program or the Soyuz Program.
Answer:
The Gemini program
Explanation:
The Gemini Program was the second human spaceflight program hosted by Nasa in the year 1961. Taking place between mission Mercury and Apollo, the Gemini spacecraft carried two people to space and marked the foundation to the upcoming Apollo mission to Moon. It was a series of missions into the outer orbital which took place between 1965 and 1966. Prior to the Gemini missions, NASA had little to no information about space and space traveling. It was crucial for them to get acquainted with life outside before establishing successful Moon landings. And the series of Gemini missions helped them do just that.
Haley is trying to pull an object upward. The below forces are acting on the object.
Fp= 5500 N
Fg= 6000 N
Which represents the net force?
The net force is represented by ↓ 500N.
What is the net force?The net force is the force that has the same effect in magnitude and direction as two or more forces acting together.
Now we have the forces;
Fp= 5500 NFg= 6000 NThus we can obtain the net force as;
5500 N - 6000 N
= - 500 N
Therefore the net force is represented by ↓ 500N.
Missing parts:
Haley is trying to pull an object upward. The below forces are acting on the object.
Fp = 5500N
Fg = 6000N
Which represents the net force?
← 500N
→ 500N
↑ 500N
↓ 500N
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A hypothetical planet has a mass 2.81 times that of Earth, but the same radius.
What is g near its surface?
The acceleration due to gravity near the surface of the planet is 27.38 m/s².
Acceleration due to gravity near the surface of the planet
g = GM/R²
where;
G is universal gravitation constantM is mass of the planetR is radius of the planetg is acceleration due to gravity = ?g = (6.626 x 10⁻¹¹ x 2.81 x 5.97 x 10²⁴) / (6371 x 10³)²
g = 27.38 m/s²
Thus, the acceleration due to gravity near the surface of the planet is 27.38 m/s².
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What is the x-component of a vector with a magnitude of 115 km at an angle of 22°?
The x-component of a vector are < 106.6, 43.07 >
Depending on the angle we are provided, the x-component of a vector can either be cos or sin. Cos always corresponds to the right triangle's side that contacts the specified angle.
If a vector v with magnitude ||v|| makes an angle θ with the positive x-axis then,
v = ||v|| cos θi + ||v|| sin θj
= < ||v|| cos θ , ||v|| sin θ >
Magnitude p = 115 km
Angle = 22°
p = ||p|| < cos θ, sin θ >
p = 115 < cos 22°, sin 22° >
p = 115 < 0.927, 0.3746 >
p = < 106.6, 43.07 >
Therefore, the x-component of a vector are < 106.6, 43.07 >
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Why is infinity the zero point for electric potential energy true?
A. Electric field strength is zero at the charge
B. Electric field strength is zero at infinity
C. Not enough info
D. Electric field strength is zero somewhere in the middle
Infinity is the zero point for electric potential energy because Electric field strength is zero at infinity.
This conclusion is not "derived" from anything else; the electrostatic potential is zero at infinity because that is how we define it.
Define electric field and electric field strength.
The physical field that contains electrically charged particles and pulls or attracts all other charged particles in the vicinity is known as an electric field. Additionally, it describes the physical environment of a system of charged particles.
The intensity of an electric field at a specific location is expressed quantitatively as "electric field strength." The volt per meter (v/m or v m-1) is the common unit.
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Three ropes A, B and C are tied together in one single knot K.
If the tension in rope A is 65.3 N, then what is the tension in rope B?
The tension in the rope B is determined as 10.9 N.
Vertical angle of cable Btanθ = (6 - 4)/(5 - 0)
tan θ = (2)/(5)
tan θ = 0.4
θ = arc tan(0.4) = 21.8 ⁰
Angle between B and Cθ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰
Apply cosine rule to determine the tension in rope B;
A² = B² + C² - 2BC(cos A)
B = C
A² = B² + B² - (2B²)(cos A)
A² = 2B² - 2B²(cos 43.6)
A² = 0.55B²
B² = A²/0.55
B² = 65.3/0.55
B² = 118.73
B = √(118.73)
B = 10.9 N
Thus, the tension in the rope B is determined as 10.9 N.
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Hello!
This is an example of a force summation in the vertical direction.
We have the tension of rope A upward (+), and the equal vertical components of the tensions of rope B and C downward (-).
These forces sum to zero, since the knot is stationary.
[tex]\Sigma F = T_A - T_{By} - T_{Cy} \\\\0 = T_A - T_{By} - T_{Cy}[/tex]
Ropes 'B' and 'C' form equivalent angles from the vertical. (If you were to draw a line from rope A down). We can use right-triangle trig to determine the angle:
[tex]tan^{-1}(\frac{O}{A}) = \theta[/tex]
The ropes are 5 m long and 2 m tall, which are the opposite and adjacent sides respectively:
[tex]tan^{-1}(\frac{5}{2}) = 68.2^o[/tex]
The vertical components are the adjacent sides from this angle, so, we would use cosine.
[tex]0 = T_A - T_Bcos\theta - T_Ccos\theta[/tex]
Rope 'B' and 'C' have the same tensions since they form the same angle with the vertical and are the same length, so we can call them 'T'.
[tex]0 = T_A - 2Tcos\theta[/tex]
Solving for 'T':
[tex]2Tcos\theta = T_A \\\\T = \frac{T_A}{2cos\theta}\\\\T = \frac{65.3}{2cos(68.2)} = \boxed{87.92 N}[/tex]
Using a schematic diagram, explain the steps of the laser technique.
The steps of the use of the laser technique is explained below:
A laser beam is emitted from the diode in the unit and spread into a laser plane (1). The laser plane, appearing as a line on the sphere (2), is reflected and collected by dual CCD arrays (1). The resulting 2D profile is digitized and as the unit travels along the x-axis of the object, multiple profiles are collected yielding a 3D coordinate point cloud of the surface (3). What is a Laser Technique?This refers to the type of surgery that makes use of special light beams in order to cut open the human body in a surgical procedure.
Hence, we can see that the laser technique is considered safer than conventional surgical methods.
Laser techniques include:
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A closed curve encircles several conductors. The line integral around this curve is (image attached below)
a) What is the net current in the conductors?
b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral?
The net current in the conductors and the value of the line integral
[tex]I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A[/tex]The resultant remains same 3.2 *10^4 TmThis is further explained below.
What is the net current in the conductors?Generally,
To put it another way, the total current In flowing across a surface S (contained by C) is proportional to the line integral of the magnetic B-field (in tesla, T).
[tex]\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \iint_S \mathbf{J} \cdot \mathrm{d}\mathbf{S} = \mu_0I_\mathrm{enc}[/tex]
[tex]I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A[/tex]
B)
In conclusion, It is possible for the line integral to go around the loop in either direction (clockwise or counterclockwise), the vector area dS to point in either of the two normal directions and Ienc, which is the net current passing through the surface S, to be positive in either direction—but both directions can be chosen as positive in this example. The right-hand rule solves these ambiguities.
The resultant remains the same at 3.2 *10^4 Tm
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A 2.30 mH toroidal solenoid has an average radius of 6.20 cm and a cross-sectional area of 2.80 cm2.
a) How many coils does it have? In calculating the flux, assume that B is uniform across a cross section, neglect the variation of B with distance from the toroidal axis.
b) At what rate must the current through it change so that a potential difference of 2.60 V is developed across its ends?
(a) The number of turns of the coil is determined as 1,596 turns.
(b) The rate of change of current is determined as 1,130.43 A/s.
Number of turns of the solenoid
L = N²μA/l
where;
L is inductance N is number of turnsA is areal is average length = 2πrN²μA = LI
N² = LI/μA
N² = (2.3 x 10⁻³ x 2π x 0.062)/(4π x 10⁻⁷ x 2.8 x 10⁻⁴)
N² = 2,546,428.6
N = √2,546,428.6
N ≈ 1,596 turns
Rate of current changeL = (emf)/I
I = (emf)/L
I = (2.6)/(2.3 x 10⁻³)
I = 1,130.43 A/s
Thus, the number of turns of the coil is determined as 1,596 turns.
The rate of change of current is determined as 1,130.43 A/s.
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What is the minimum work needed to push a 950- kg car 700 m up along a 8.5 ∘ incline? Ignore friction.
Express your answer with the appropriate units.
The minimum work needed to pus the cart up the inclined plane is 960000 J.
What is work done on a inclined plane?The work done on a inclined plane is given below as:
Work done = force * distanceDistance = 700 m
The force on an inclined plane, F = mgsinθ
where;
m is mass in kg
g = 9.81 m/s²
θ = 8.5°
Work done = 950 * 9.81 * sin 8.5 * 700
Work done = 960000 J
Therefore, minimum work needed is 960000 J.
In conclusion, the work done is a product of force and distance.
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Earth travels around the Sun each year in sn elliptical path, as opposed to a perfect curcle. This means that the speed if earth and its dustance from the Sun change over rhe course if a year. What does this sayabout the magnitude of the centripetal acceleration if earth over the course of a year
A change in the linear speed of the Earth around the sun will cause a change in the magnitude of the centripetal acceleration.
What is centripetal acceleration ?
Centripetal acceleration is the acceleration of a body moving a circular path.
The relationship between centripetal acceleration and speed;
a = v²/r
where;
v is linear speeda is centripetal accelerationr is radius of the pathSince the centripetal acceleration is directly proportional to square of linear speed, a change in the linear speed of the Earth around the sun will cause a change in the magnitude of the centripetal acceleration.
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