Answer:
a
The number of radians turned by the wheel in 2s is [tex]\theta= 8\ radians[/tex]
b
The angular acceleration is [tex]\alpha =14 rad/s^2[/tex]
Explanation:
The angular velocity is given as
[tex]w(t) = (2.00 \ rda/s^2)t + (1.00 rad /s^4)t^3[/tex]
Now generally the integral of angular velocity gives angular displacement
So integrating the equation of angular velocity through the limit 0 to 2 will gives us the angular displacement for 2 sec
This is mathematically evaluated as
[tex]\theta(t ) = \int\limits^2_0 {2t + t^3} \, dt[/tex]
[tex]= [\frac{2t^2}{2} + \frac{t^4}{4}] \left\{ 2} \atop {0}} \right.[/tex]
[tex]= [\frac{2(2^2)}{2} + \frac{2^4}{4}] - 0[/tex]
[tex]= 4 +4[/tex]
[tex]\theta= 8\ radians[/tex]
Now generally the derivative of angular velocity gives angular acceleration
So the value of the derivative of angular velocity equation at t= 2 gives us the angular acceleration
This is mathematically evaluated as
[tex]\frac{dw}{dt} = \alpha (t) = 2 + 3t^2[/tex]
so at t=2
[tex]\alpha (2) = 2 +3(2)^2[/tex]
[tex]\alpha =14 rad/s^2[/tex]
The wheel turns through 8 radians in the first 2 seconds of its motion. The angular acceleration of the wheel at the end of the first 2 seconds is 14.00 rad/s².
Explanation:The questions are related to the concepts of angular velocity and acceleration in the realm of Physics. To solve parts (a) and (b) we would use the principles of rotational motion.
(a) The number of radians the wheel turns in first 2 seconds is given by the integral of the angular velocity function from 0 to 2. The equation becomes ∫ₒ² ω(t) dt = ∫ₒ² ((2.00 rad/s²)t + (1.00 rad/s⁴)t ³ dt) = [t² + 0.25t⁴]ₒ² = 4 rad + 4 rad = 8 rad.
(b) The angular acceleration is given by the derivative of the angular velocity with respect to time which is ω'(t) = 2.00 rad/s² + 3(1.00 rad/s⁴)t². Evaluating at t = 2s gives an angular acceleration of 2.00 rad/s² + 3(1.00 rad/s⁴)(2s)² = 2.00 rad/s² + 12.00 rad/s² = 14.00 rad/s².
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Suppose we assume that Einstein and Zweistein should compute the same total force on the sphere. Would Zweistein compute a larger or smaller value of the electric field at the sphere than Einstein
Answer:
Forces observed by both is same as both
The frames are non-acceleration
Explanation:
Zweistein = Einstein
[tex](F_B + E_B)_Z[/tex] = [tex](F_B + E_B)_E[/tex]
as [tex](F_B + E_B)_E[/tex] tend to zero (0)
[tex](E_B)_Z[/tex] < [tex](F_B)_Z[/tex] as some quantity is subtracted from the force.
According to Einstein's Theory of Relativity, the laws of physics are the same for all observers, regardless of their motion. Applying this to the calculation of electric field strength around a sphere, both Einstein and Zweistein, assuming they are not accelerating, should compute the same value of the electric field.
Explanation:If we consider the principles of physics consistently at play regardless of the observer, both Einstein and Zweistein must compute the same total force on the sphere. This is following Einstein's postulate stating the laws of physics remain the same in all inertial frames. Therefore, the electric field computed at the sphere by both Einstein and Zweistein should also be the same.
The principle of relativity implies that the laws of physics—including those of electric field strength—are the same for all observers, regardless of their speed or direction as long as they are not accelerating. Einstein's Theory of Relativity extends this to include accelerated motion and gravity.
This principle, combined with the equivalence of inertial and gravitational mass, is enough to show that the force of gravity perceived by an observer in an accelerating spacecraft must be the same as if the observer were stationary on the Earth. So, if Einsteins's Theory of Relativity applies to electric fields in the same way it does for gravitational fields, then the electric field strength calculated by Einstein and Zweistein should be the same.
To clarify, in the context of this question, the 'electric field' refers to the region around a charged particle or object where another charge placed within it would experience a force either of attraction or repulsion.
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A plane electromagnetic wave varies sinusoidally at 86.0 MHz as it travels through vacuum along the positive x-direction. The peak value of the electric field is 2.30 mV/m, and it is directed along the positive y-direction. Determine the average power per unit area (the intensity) this wave carries through space. (Be careful with your units here and make sure to submit your answer in μW/cm2)
Answer:
The intensity is [tex]I = 0.0003053 \mu W/cm^2[/tex]
Explanation:
From the question we are told
The frequency of the electromagnetic wave is [tex]f = 86.0 Hz[/tex]
The peak value of the electric field is [tex]E_o = 2.30 mV/m = \frac{2.30}{1000 } = 2.30 *10^{-3} V/m[/tex]
Generally the intensity of this wave is mathematically represented as
[tex]I = c * \frac{1}{2} * \epsilon_o E^2_o[/tex]
Where c is the speed of light with value [tex]c = 3 *10^8 m/s[/tex]
[tex]\epsilon_o[/tex] is the permittivity of free space with value [tex]\epsilon _o = 8.85 *10^{-12} C^2 /Nm^2[/tex]
Substituting values into equation for intensity
[tex]I = 3.0 *10^8 * 0.5 * 8.85 *10^{-12} * 2.30*10^{-3}[/tex]
[tex]I = 3.053 *10^{-6} W/m^2[/tex]
Converting to [tex]cm^2[/tex] we divide by 10,000
[tex]I = \frac{3.053 *10^{-6}}{10000} W/cm^2[/tex]
[tex]= 3.053 *10^{-10} W/cm^2[/tex]
[tex]= 0.0003053 *10^{-6} W/cm^2[/tex]
[tex]I = 0.0003053 \mu W/cm^2[/tex]
A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each with a radius of 5.00 mm. If water flows through the single pipe at 1.45 m/s, calculate the speed (in m/s) of the water in the narrower pipes.
Answer:
1.24 m/s
Explanation:
Metric unit conversion:
9.25 mm = 0.00925 m
5 mm = 0.005 m
The volume rate that flow through the single pipe is
[tex]\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s[/tex]
This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of
[tex]\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s[/tex]
So the flow speed of each of the narrower pipe is:
[tex]v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}[/tex]
[tex]v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s[/tex]
A camera has a single converging lens with a fixed focal length f. (a) How far should the lens be from the film (or in a present-day digital camera, the CCD chip) in order to focus an object that is infinitely far away (namely the incoming light rays are parallel with the principal axis of the system). (b) How far should the lens be from the film to focus an object at a distance of 2f in front of the lens
Answer:
a) Due to the characteristic that a converging lens focuses light rays from infinity and parallel to its main axis. Therefore, the lens should be placed at a distance "f" from the film, in this way it will form the image of the object placed at infinity in said film.
b) Since the converging lens produces an image of an object placed at a distance of 2f, the lens must be placed at the same distance (2f), so that this object that is placed at a distance of 2f is focused.
Explanation:
To focus an infinitely distant object, the lens should be placed at its focal length from the film or CCD. For an object at 2f, the lens should also be at a distance of 2f to focus the image correctly.
The question pertains to the principles of optics, specifically focusing using converging lenses in cameras.
(a) To focus an object that is infinitely far away, the lens should be placed at its focal length f from the film or CCD. This is because light rays from a distant object enter the lens parallel to its principal axis and converge at the focal point on the opposite side. Therefore, for an object at infinity, the image distance (di) approaches the lens's focal length (f).
(b) To focus an object at a distance of 2f in front of the lens, the lens should also be 2f from the film. This is derived from the lens formula (1/f = 1/do + 1/di) where do is the object distance and di is the image distance. For an object at 2f, solving the equation gives di = 2f.
A 0.700-kg ball is on the end of a rope that is 2.30 m in length. The ball and rope are attached to a pole and the entire apparatus, including the pole, rotates about the pole’s symmetry axis. The rope makes a constant angle of 70.0° with respect to the vertical. What is the tangential speed of the ball?
Answer:
The tangential speed of the ball is 11.213 m/s
Explanation:
The radius is equal:
[tex]r=2.3*sin70=2.161m[/tex] (ball rotates in a circle)
If the system is in equilibrium, the tension is:
[tex]Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}[/tex]
Replacing:
[tex]\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}[/tex]
Replacing:
[tex]v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s[/tex]
Difference Between Photometry and Spectrophotometry ?
Answer:
Spectrophotometry is the quantitative measurement of light spectra reflection and transmission properties of materials as function of the wavelength. Photometry measures the total brightness as seen by the human eye
Explanation:
Please give me brainliest
Show your work and reasoning for below.
After watching a news story about a fire in a high rise apartment building, you and your friend decide to design an emergency escape device from the top of a building.
To avoid engine failure, your friend suggests a gravitational powered elevator. The design has a large, heavy turntable (a horizontal disk that is free to rotate about its center) on the roof with a cable wound around its edge. The free end of the cable goes horizontally to the edge of the building roof, passes over a heavy vertical pulley, and then hangs straight down.
A strong wire cage which can hold 5 people is then attached to the hanging end of the cable. When people enter the cage and release it, the cable unrolls from the turntable lowering the people safely to the ground.
To see if this design is feasible you decide to calculate the acceleration of the fully loaded elevator to make sure it is much less than "g."
Your friend's design has the radius of the turntable disk as 1.5m and its mass is twice that of the fully loaded elevator. The disk which serves as the vertical pulley has 1/4 the radius of the turntable and 1/16 its mass. The moment of inertia of the disk is 1/2 that of a ring.
Answer:
Explanation:
Given that,
His friend design has a turnable disk of radius 1.5m
R = 1.5m
The mass is twice the fully loaded elevator.
Let the mass of the full loaded elevator be M
Then, mass of the turn able
Mt = ½M
Radius of the disk that serves as a vertical pulley is ¼ radius of turntable and 1/16 of the mass.
Mass of pulley is
Mp = 1 / 16 × Mt
Mp = 1 / 16 × M / 2
Mp = M / 32
Also, radius of pulley
Rp = ¼ × R = ¼ × 1.5
Rp = 0.375m
The moment of inertia of the disk of a ring is
I = ½MR²
To calculate the moment of the turntable, we can use the formula
I_t = ½Mt•R²
I_t = ½ × ½M × 1.5²
I_t = 0.5625 M
I_t = 9M / 16
Also, the moment of inertia of the vertical pulley
I_p = ½Mp•Rp
I_p = ½ × (M/16) × 0.375
I_p = 0.01171875 M
I_p = 3M / 256
Let assume that, the tension in the cable between the pulley and the elevator is T1 and Let T2 be the tension between the turntable and the pulley
So, applying newton second law of motion,
For the elevator
Fnet = ma
Mg - T1 = Ma
a = (Mg-T1) / M
For vertical pulley,
The torque is given as
τ_p = I_p × α_p = (T2—T1)•r
τ_p = 3M/256 × α_p = (T2-T1)•r
For turntable
The torque is given as
τ_t = I_t × α_t = T2•r
τ_t = 9M/16 × α_t = T2•r
So, the torque are equal
τ_t = τ_p
9M/16 × α_t = 3M/256 × α_p
M cancel out
9•α_t / 16 = 3•α_p / 256
Cross multiply
9•α_t × 256 = 3•α_p × 16
Divide both sides by 48
48•α_t = α_p
α_t = α_p / 48
Then, from,
τ_t = 9M/16 × α_t = T2•r
T2•r = 9M / 16 × α_p / 48
T2•r = 3Mα_p / 16
Also, from
τ_p = 3M/256 × α_p = (T2-T1)•r
3M/256 × α_p = T2•r - T1•r
T1•r = T2•r - 3M/256 × α_p
T1•r = 3Mα_p / 16 - 3M/256 × α_p
T1•r = 3Mα_p / 16 - 3Mα_p/256
T1•r = 45Mα_p / 256
T1 = 45Mα_p / 256R
Then, from
a = (Mg-T1) / M
a = Mg - (45Mα_p / 256R) / M
a = g - 45α_p / 256
From the final answer, it is show that the acceleration is always less than acceleration due to gravity due to the subtraction of 45α_p / 256 from g
Find A and B
I need help like asap so pleasee help
Answer:
I don't understand?
Explanation:
Could you please elaborate?
A loop of wire is at the edge of a region of space containing a uniform magnetic field B⃗ . The plane of the loop is perpendicular to the magnetic field. Now the loop is pulled out of this region in such a way that the area A of the coil inside the magnetic field region is decreasing at the constant rate c. That is, dAdt=−c, with c>0.Part A
The induced emf in the loop is measuredto be V. What is the magnitude B of the magnetic field that the loop was in?
Part B
For the case of a square loop of sidelength L being pulled out of the magneticfield with constant speed v, what is the rate of change ofarea c = -\frac{da}{dt}?
The question is not clear enough. So i have attached a copy of the correct question.
Answer:
A) B = V/c
B) c = Lv
Explanation:
A) we know that formula for magnetic flux is;
Φ = BA
Where B is magnetic field and A is area
Now,
Let's differentiate with B being a constant;
dΦ/dt = B•dA/dt
From faradays law, the EMF induced is given as;
E = -dΦ/dt
However, we want to express it in terms of V and E.M.F is also known as potential difference or Voltage.
Thus, V = -dΦ/dt
Thus, we can now say that;
-V = B•dA/dt
Now from the question, we are told that dA/dt = - c
Thus;
-V = B•-c
So, V = Bc
Thus, B = V/c
B) according to Faraday's Law or Lorentz Force Law, an electromotive force, emf, will be induced between the two ends of the sidelength:
Thus;
E =LvB or can be written as; V = LvB
Where;
V is EMF
L is length of bar
v is velocity
From the first solution, we saw that;
V = Bc
Thus, equating both of the equations, we have;
Bc = LvB
B will cancel out to give;
c = Lv
The magnitude of the magnetic field B is calculated using Faraday's Law and is equal to V/-c. For a square loop being pulled out from the field, the rate of change of area is proportional to the speed of withdrawal and the length of the side still in the field. Hence, c = Lv.
Explanation:Part A: The magnitude of the magnetic field B can be calculated using Faraday's Law of electromagnetic induction, which states that the induced voltage in a circuit is equal to the negative rate of change of magnetic flux. Mathematically, it is expressed as V = -dB/dt, where V is the induced emf, and dB/dt is the rate of change of magnetic field. Because the area is decreasing at a constant rate, the magnetic field B is V/-c.
Part B: The rate of change of the area, denoted as c = -da/dt, for a square loop of side length L being pulled out of the magnetic field with constant speed v, could be calculated using the formula c = Lv. This is because as the square loop is pulled out with a constant velocity, the rate at which its area decreases is proportional to its speed and the length of the side that is still in the field.
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A square current-carrying loop is placed in a uniform magnetic field B with the plane of the loop parallel to the magnetic field (see the drawing). The dashed line is the axis of rotation. The magnetic field exerts _____________ answer
Answer:
The magnetic field exerts net force and net torque on the loop.
A stainless steel tube with an outside diameter of 54 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 190 MPa, determine the maximum load P that the member can support.
Answer:146.26 kN
Explanation:
Given
Outside diameter of tube [tex]d_o=54\ mm[/tex]
thickness of tube [tex]t=5\ mm[/tex]
therefore inner diameter [tex]d_i=54-2\times 5[/tex]
[tex]d_i=44\ mm[/tex]
Cross-section of tube [tex]A=\frac{\pi }{4}(D_o^2-d_i^2)[/tex]
[tex]A=\frac{\pi }{4}\times 980[/tex]
[tex]A=245\pi mm^2[/tex]
Stress developed must be less than the limited value
thus [tex]\frac{P}{A}\leq \sigma [/tex]
[tex]P\leq \sigma A[/tex]
Maximum value of [tex]P=\sigma \times A[/tex]
[tex]P=190\times 245\times \pi [/tex]
[tex]P=146.26\ kN[/tex]
Final answer:
To calculate the maximum load a stainless steel tube can support, one must first find the cross-sectional area based on its outer diameter and wall thickness. Then, using the limit of the normal stress and the area, the maximum axial load can be computed.
Explanation:
To determine the maximum load P that a stainless steel tube can support when it is used as a compression member, we need to use the formula for axial stress in a cylindrical member, which is σ = P/A, where σ is the normal stress and A is the cross-sectional area. Given that the normal stress must not exceed 190 MPa, we need to first calculate the cross-sectional area. For a tube with an outside diameter (d) of 54 mm and a wall thickness (t) of 5 mm, the internal diameter (di) would be d - 2t = 54 mm - (2 x 5 mm) = 44 mm. The cross-sectional area A can be calculated using the formula for the area of a hollow circle, A = π/4 * (d2 - di2).
A = π/4 * ((55)^2 - (44)^2)
A = 855.298 * 10^-6 m2
Once we have the area, we can calculate the maximum load by rearranging the stress formula to P = σ * A. By plugging in the values for σ and A, we can find the value of P that the member can safely support. Remember to convert diameters to meters when calculating the area in square meters for consistency with the stress units of Pascals (Pa).
P = 190 * 855.298 Pa
P = 1.62 MPa
which is a tool used by scientists to determine the composition of objects?
Answer:
A spectrometer
Explanation:
Who's Faster, Sonic or Flash!?
Answer:
flash
Explanation:
Answer:
Sonic obviously
Explanation:
Sonic's body was made to run, the flashes is just a normal human body.
A beam of light starts at a point 8.0 cm beneath the surface of a liquid and strikes the surface 7.0 cm from the point directly above the light source. Total internal reflection occurs for the beam. What can be said about the index of refraction of the liquid?
Answer:
Explanation:
Given that,
The light starts at a point 8cm beneath the surface
It strikes the surface 7cm directly above the light
The angle of incidence from the given distances is
Using trigonometry
Tan I = opp / adj
tan I= (7.0 cm)/(8.0 cm) = 0.875
I = arctan(0.875)
I = 41.1859
According to Snell's law
n(water) sin I = n(air) sin R
Here R =90° for total reflection to occur
nair = 1
n(water) sin I = n(air) sinR
n(water) = (1)sin 90
n(water) = 1
n(water) = 1/sin I
n(water) =1/ sin(41.1859o)
= 1.52
The refractive index index of the liquid is 1.52
To find the index of refraction of the liquid, we calculate the angle of incidence using the given distances and apply Snell's law. While specific numeric calculations aren't provided, the assumption is the liquid's refractive index is greater than air, as total internal reflection occurs.
A beam of light undergoes total internal reflection when it moves from a medium of higher refractive index to one of lower refractive index, and the angle of incidence is greater than the critical angle. In this case, we are given that a beam of light in a liquid undergoes total internal reflection. To determine the index of refraction of the liquid, we can use the information given about the distances involved to find the angle of incidence and then apply Snell's law.
First, we use the distances given to find the angle of incidence. The light source is 8.0 cm beneath the surface and strikes the surface 7.0 cm from the point directly above it. This forms a right triangle, allowing us to calculate the angle of incidence using trigonometry. The tangent of the angle of incidence (θ) is opposite over adjacent, or θ = tan⁻¹(8.0/7.0). The angle of incidence calculated is therefore tan⁻¹(8/7).
For total internal reflection to occur, the angle of incidence must be greater than the critical angle, which is defined by Snell's law as sin⁻¹(n₂/n₁) where n₂<n₁. Since we know the reflection happens at the interface with air (where n₂ = 1.00 for air), we can find the refractive index of the liquid. However, without exact values for angles given in this scenario, the specific calculation steps cannot be completed, but we understand that the liquid's refractive index is greater than 1, which is typical for most transparent materials compared to air.
A 10.0 μFμF capacitor initially charged to 30.0 μCμC is discharged through a 1.30 kΩkΩ resistor. How long does it take to reduce the capacitor's charge to 15.0 μCμC ? Express your answer with the appropriate units. 0.052s
Answer:
It takes approximately 0.009 seconds for the capacitor to discharge to half its original charge.
Explanation:
Recall that the capacitor discharges with an exponential decay associated with the time constant for the circuit ([tex]\tau_0=RC[/tex]) which in our case is;
[tex]\tau_0=1.3\,k\Omega\,*\,10.0 \mu F=13\,\,10^{-3}\,s[/tex] (13 milliseconds)
Recall as well the expression for the charge in the capacitor (from it initial value [tex]Q_0[/tex], as it discharges via a resistor R:
[tex]Q(t)=Q_0\,e^{-\frac{t}{RC} }[/tex]
So knowing that the capacitor started with a charge of 30 [tex]\mu C[/tex] and reduced after a time "t" to 30 [tex]\mu C[/tex] , and knowing from our first formula what the RC of the circuit is, we can solve for the time elapsed using the charge as function of time equation shown above:
[tex]Q(t)=Q_0\,e^{-\frac{t}{RC} }\\15 \mu C=30 \mu C\,\,e^{-\frac{t}{13\,ms} }\\e^{-\frac{t}{13\,ms} }=\frac{1}{2} \\-\frac{t}{13\,ms}=ln(\frac{1}{2})\\t=-13\,ms\,*\,ln(\frac{1}{2})\\t=9.109\,\, ms[/tex]
In seconds this is approximately 0.009 seconds
Final answer:
The question involves calculating the time it takes for the charge on a capacitor to reduce to a certain value in an RC circuit, using the natural logarithm and the RC time constant formula. This is a Physics problem relevant to High School level.
Explanation:
The subject matter of the student's question is Physics, specifically relating to the discharge of a capacitor through a resistor and the calculation of time constants in RC circuits. The grade level would be High School, as it involves concepts typically taught in a high school physics course.
How to Calculate the Discharge Time
To calculate the time it takes for a capacitor to reduce its charge to a certain value, we use the formula V = V0 * e^(-t/RC), where V0 is the initial voltage, V is the final voltage, R is the resistance, C is the capacitance, t is the time, and e is the base of the natural logarithm. However, since we're dealing with the capacitor's charge rather than voltage, we can adapt the formula to Q = Q0 * e^(-t/RC). By substituting the given values and solving for t, we can find the time it takes for the capacitor's charge to decrease from 30.0 μC to 15.0 μC.
To solve for t, we would take the natural logarithm of both sides after dividing by Q0, resulting in t = -RC * ln(Q/Q0). Inserting the given values (R = 1.30 kΩ and C = 10.0 μF), and calculating the natural logarithm of (Q/Q0 = 15.0 μC / 30.0 μC), we find the time required for the charge to drop to half its initial value.
A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an angular displacement of 13 rad. What is its displacement θ at a time of t = 5.0 s?
Answer:
52 rad
Explanation:
Using
Ф = ω't +1/2αt²................... Equation 1
Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.
Since the object states from rest, ω' = 0 rad/s.
Therefore,
Ф = 1/2αt²................ Equation 2
make α the subject of the equation
α = 2Ф/t².................. Equation 3
Given: Ф = 13 rad, t = 2.5 s
Substitute into equation 3
α = 2(13)/2.5²
α = 26/2.5
α = 4.16 rad/s².
using equation 2,
Ф = 1/2αt²
Given: t = 5 s, α = 4.16 rad/s²
Substitute into equation 2
Ф = 1/2(4.16)(5²)
Ф = 52 rad.
Answer:
[tex]\theta = 52 rads[/tex]
Explanation:
The rotational kinetic equation is given by the formula:
[tex]\theta = w_{0} t + 0.5 \alpha t^{2}[/tex]..............(1)
The object is starting from rest, angular speed, w = 0 rad/s
At t = 2.5 sec, angular displacement, θ = 13 rad
Inserting these parameters into equation (1)
[tex]13 = (0 * 2.5) + 0.5 \alpha 2.5^{2}\\\alpha = \frac{13}{3.125} \\\alpha = 4.16 rad/s^{2}[/tex]
At time, t = 5.0 sec, we substitute the value of [tex]\alpha[/tex] into the kinematic equation in (1)
[tex]\theta = (0*5) + 0.5 *4.16* 5^{2}\\\theta = 52 rad[/tex]
When light propagates through two adjacent materials that have different optical properties, some interesting phenomena occur at the interface separating the two materials. For example, consider a ray of light that travels from air into the water of a lake. As the ray strikes the air-water interface (the surface of the lake), it is partly reflected back into the air and partly refracted or transmitted into the water. This explains why on the surface of a lake sometimes you see the reflection of the surrounding landscape and other times the underwater vegetation.
These effects on light propagation occur because light travels at different speeds depending on the medium. The index of refraction of a material, denoted by n , gives an indication of the speed of light in the material. It is defined as the ratio of the speed of light c in vacuum to the speed v in the material, orn=cv
What is the minimum value that the index of refraction can have?
Answer:
First, the different indices of refraction must be taken into account (in different media): for example, the refractive index of light in a vacuum is 1 (since vacuum = c). The value of the refractive index of the medium is a measure of its "optical density": Light spreads at maximum speed in a vacuum but slower in others transparent media; therefore in all of them n> 1. Examples of typical values of are those of air (1,0003), water (1.33), glass (1.46 - 1.66) or diamond (2.42).
The refractive index has a maximum value and a minimum value, which we can calculate the minimum value by means of the following explanation:
The limit or minimum angle, α lim, is defined as the angle of refraction from which the refracted ray disappears and all the light is reflected. As in the maximum value of angle of refraction, from which everything is reflected, is βmax = 90º, we can know the limit angle (the minimum angle that we would have to have to know the minimum index of refraction) by Snell's law:
βmax = 90º ⇒ n 1x sin α (lim) = n 2 ⇒ sin α lim = n 2 / n 1
Explanation:
When a light ray strikes the separation surface between two media different, the incident beam is divided into three: the most intense penetrates the second half forming the refracted ray, another is reflected on the surface and the third is breaks down into numerous weak beams emerging from the point of incidence in all directions, forming a set of stray light beams.
The minimum value of the index of refraction is 1, which occurs in a vacuum where the speed of light is at its maximum and not slowed by any material.
The minimum value that the index of refraction can have is for a vacuum, where light travels at its fastest and the speed of light is not slowed down by any medium. According to the definition of the index of refraction, n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material. Since the speed of light in a vacuum is the highest possible speed for light and cannot be exceeded, the index of refraction n has a minimum value of 1 in a vacuum. That is because when we plug in the values into the equation, with both c and v being equal (since v is the speed of light in a vacuum), the ratio n = c/v equals 1. Therefore, no material can have an index of refraction less than 1.
A rocket is attached to a toy car that is confined to move in the x-direction ONLY. At time to = 0 s, the car is not moving but the rocket is lit, so the toy car accelerates in the +x-direction at 5.35 m/s2. At t; = 3.60 s, the rocket's fuel is used up, and the toy car begins to slow down at a rate of 1.95 m/s2 because of friction. A very particular physics professor wants the average velocity for the entire trip of the toy car to be +6.50 m/s. In order to make this happen, the physics professor plans to push the car (immediately after it comes to rest by friction) with a constant velocity for 4.50 sec. What displacement must the physics professor give the car (immediately after it comes to rest by friction) in order for its average velocity to be +6.50 m/s for its entire trip (measured from the time the rocket is lit to the time the physics professor stops pushing the car)?
Answer:
What displacement must the physics professor give the car
= 12.91 METERS
Explanation:
Check the attached file for explanation
Final answer:
To find the displacement that the physics professor must give the car, we can break down the different stages of the car's motion and use the equations of motion. By adding the displacements from each stage, we can find the total displacement of the car.
Explanation:
To find the displacement that the physics professor must give the car in order for its average velocity to be +6.50 m/s for the entire trip, we can break down the different stages of the car's motion and use the equations of motion.
1. From time t=0 to t=3.60s: The car is accelerated in the +x direction at 5.35 m/s². We can use the equation v = u + at to find the final velocity v1 at t=3.60s, where u is the initial velocity, a is the acceleration, and t is the time.
2. From time t=3.60s to t=8.10s: The car slows down at a rate of 1.95 m/s² due to friction. We can use the equation v = u + at to find the initial velocity u2 at t=3.60s, where v is the final velocity, a is the acceleration, and t is the time.
3. From time t=8.10s to t=12.60s: The physics professor pushes the car with a constant velocity for 4.50s. The average velocity during this time is +6.50 m/s. We can calculate the displacement during this time using the equation d = vt, where v is the velocity and t is the time.
By adding the displacements from each stage, we can find the total displacement of the car, which is the answer to the question.
An object moving with a speed of 21 m/s and has a kinetic energy of 140 J, what is the mass of the object.
Answer:2940
Explanation:
Answer:
0.635 kg
Explanation:
Recall that Kinetic energy is defined as:
K.E. = (1/2) mv²,
where
K.E = Kinetic energy = given as 140J
m = mass (we are asked to find this)
v = velocity = given as 21 m/s
Substituting the above values into the equation:
K.E. = (1/2) mv²
140 = (1/2) m (21)² (rearranging)
m = (140)(2) / (21)²
m = 0.635 kg
In empty space, which quantity is always larger for X-ray radiation than for a radio wave? In empty space, which quantity is always larger for X-ray radiation than for a radio wave? Speed. Wavelength. Amplitude. Frequency.
In empty space, the quantity that is always larger for X-ray radiation than for a radio wave is the frequency.
Frequency is the number of complete cycles (oscillations) of a wave that occur in one second. It is measured in Hertz (Hz), where 1 Hz means one cycle per second. X-ray radiation has a higher frequency than radio waves.
On the electromagnetic spectrum, X-rays have much higher frequencies compared to radio waves. X-rays have frequencies in the range of [tex]10^16[/tex] to [tex]10^19[/tex] Hz, while radio waves typically have frequencies in the range of [tex]10^3[/tex] to [tex]10^9[/tex] Hz. Therefore, X-ray radiation has significantly higher frequencies than radio waves.
Wavelength and amplitude are other properties of waves, but they are not always larger for X-ray radiation than for a radio wave. The wavelength of X-rays is much shorter than that of radio waves, and the amplitude can vary depending on the specific wave source and conditions.
Hence, In empty space, the quantity that is always larger for X-ray radiation than for a radio wave is the frequency.
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What force is responsible for the
creation of the galaxies and stars?
Answer:
Gravity
Explanation:
I know because im a verified expert
Answer:
gravity
Explanation:
1. gravity is the force that pulls us to the surface of the Earth, keeps the planets in orbit around the Sun and causes the formation of planets, stars and galaxies.
2. electromagnetism is the force responsible for the way matter generates and responds to electricity and magnetism.
A distance-time graph indicates that an object moves 20 m in 10 s and then remains at rest for 20s. What is the average speed of the object?
Answer:
v = 2 m/s
Explanation:
The object moves 20 m in 10 s initially. Then it remains at rest for 20 s. It is required to find the average speed of the object.
Fro 10 s it covers 20 m. For that time, speed is :
[tex]v=\dfrac{d}{t}\\\\v=\dfrac{20\ m}{10\ s}\\\\v=2\ m/s[/tex]
For 20 s, it was at rest means distance covered is 0. So, speed is 0.
The average speed of the object is equal to the sum of speed for 10 s and for 20 s i.e.
v = 2 m/s + 0 = 2 m/s
So, the average speed of the object is 2 m/s.
One can estimate the oil reservoir pressure underground from the height the oil rises. Of the gusher rises to 200 ft above ground and the diameter of the bore pipe that goes down to the oil reservoir underground is 9 inches, calculate the oil pressure in a reservoir that is 1500 m below ground. The oil density is 0.85 g/cm3 and viscosity is 200 cP
Answer:
12,608 kPa
Explanation:
First we need to convert the density to standard units, that is kg/m^3
0.85 g/cm^3 = 0.85 × [tex]\frac{100^{3} }{1000}[/tex] = 0.85 × 1000 = 850 kg/m^3
Pressure of oil = pressure at surface + rho × g × h
= 101,000 + (850×9.81×1500)
= 12,608 kPa
The units in the part 'rho.g.h' will cancel out against each other and you will be left with the unit Pascals - which is the unit for pressure.
Hope that answers the question, have a great day!
A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.10 cm, and at a .. 29.47 particular instant the conduction current in the wires is 0.276 A. (a) What is the displacement current density jD in the air space between the plates? (b) What is the rate at which the electric field between the plates is changing? (c) W
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.10 cm, and at a particular instant the conduction current in the wires is 0.276 A. (a) What is the displacement current density jD in the air space between the plates? (b) What is the rate at which the electric field between the plates is changing? (c) What is the induced magnetic field between the plates at a distance of 2 cm from the axis? (d) What is the induced magnetic field between the plates at a distance of 1 cm from the axis?
Given Information:
Radius of parallel-plate capacitor = 4.10 cm = 0.041 m
Conduction current = Ic = 0.276 A
Required Information:
a) Displacement current density = JD = ?
b) Rate of change of electric field = dE/dt = ?
c) Magnetic field between plates at r = 2 cm = ?
d) Magnetic field between plates at r = 1 cm = ?
Answer:
a) Displacement current density = JD = 52.27 A/m²
b) Rate of change of electric field = dE/dt = 5.904×10¹² V/m.s
c) Magnetic field between plates at r = 2 cm = 6.568×10⁻⁷ Tesla
d) Magnetic field between plates at r = 1 cm = 3.284×10⁻⁷ Tesla
Explanation:
a) What is the displacement current density JD in the air space between the plates?
Displacement current density is given by
JD = Id/A
Where Id is the conduction current and A is the area of capacitor given by
A = πr²
A = π(0.041)²
A = 0.00528 m²
As you can notice in the diagram, conduction current has equal displacement between the capacitor plates therefore, Id = Ic
JD = 0.276/0.00528
JD = 52.27 A/m²
b) What is the rate at which the electric field between the plates is changing?
The rate of change of electric field is given by
dE/dt = JD/ε₀
Where JD is the displacement current density and ε₀ is the permittivity of free space and its value is 8.854×10⁻¹² C²/N.m²
dE/dt = 52.27/8.854×10⁻¹²
dE/dt = 5.904×10¹² V/m.s
c) What is the induced magnetic field between the plates at a distance of 2 cm from the axis?
The induced magnetic field between the plates can be found using Ampere's law
B = (μ₀/2)*JD*r
Where μ₀ is the permeability of free space and its value is 4π×10⁻⁷ T.m/A, JD is the displacement current density and r is the distance from the axis.
B = (4π×10⁻⁷/2)*52.27*0.02
B = 6.568×10⁻⁷ Tesla
d) What is the induced magnetic field between the plates at a distance of 1 cm from the axis?
B = (μ₀/2)*JD*r
B = (4π×10⁻⁷/2)*52.27*0.01
B = 3.284×10⁻⁷ Tesla
This physics-based question involves understanding the properties and behaviors of a parallel-plate, air-filled capacitor. Specifically, it focuses on finding the displacement current density and the rate at which the electric field between the plates is changing as the capacitor is charged.
Explanation:The question revolves around a parallel-plate capacitor, a system of two identical conducting plates separated by a distance. This specific case involves an air-filled capacitor where the plates have a certain radius, and a conduction current is defined.
First, let's address part (a): What is the displacement current density jD in the air space between the plates? The displacement current density, noted as 'jD', can be found by using Maxwell's equation which correlates the time rate of change of electric flux through a surface to the displacement current passing through the same surface.
For (b): What is the rate at which the electric field between the plates is changing? As the current is introducing charge onto the plates, the electric field (which is proportional to the charge on the capacitor's plates) will be changing. This change can be calculated as the derivative of the electric field's magnitude with respect to time.
Without specific numerical values or more information, I'm unable to provide a more detailed analysis or a concrete answer.
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Two identical steel balls, each of mass 67.8 g, are moving in opposite directions at 4.80 m/s.They collide head-on and bounce apart elastically. By squeezing one of the balls in a vise while precise measurements are made of the resulting amount of compression, you find that Hooke's law is a good model of the ball's elastic behavior. A force of 15.9 kN exerted by each jaw of the vise reduces the diameter by 0.130 mm. Model the motion of each ball, while the balls are in contact, as one-half of a cycle of simple harmonic motion. Compute the time interval for which the balls are in contact.
Answer:
Explanation:
To detrmine the time interval at which the balls are in contact.
Given information
The mass of each steal ball 67.8g. The speed of ball towards each other is 4.80 m/s. The exerted force by each jaw 15.9 kN and the forece reduce the diameter by 0.130 mm.
Expression for the effective spring constant ball is shown below.
K = |F|/|x|
Here,
k is a spring constant
F is the force exerted on the ball
x is dispalcement due force
substitute 15.9 kN for F and 0.130 mm in above equation
K = (15.9 kN)(1X10³N) / (0.130 mm)(1x10⁻³m/1mm)122 x 10⁶ N/mThe spring constant is 122 x 10⁶ N/m
A 0.74-m diameter solid sphere can be rotated about an axis through its center by a torque of 10.8 m*N which accelerates it uniformly from rest through a total of 160 revolutions in 14.0 s.
1. What is the mass of the sphere?
Answer:
Mass of the sphere is 19.2 kg
Explanation:
We have given diameter of the sphere d = 0.74 m
So radius r = 0.37 m
Initial angular velocity [tex]\omega _i=0rad/sec[/tex]
Time t = 14 sec
Angular displacement [tex]\Theta =160revolution=160\times 2\pi =1004.8rad[/tex]
From second equation of motion
[tex]\Theta =\omega _it+\frac{1}{2}\alpha t^2[/tex]
So [tex]1004.8=0\times t+\frac{1}{2}\times \alpha \times 14^2[/tex]
[tex]\alpha =10.25rad/sec^2[/tex]
Torque is given [tex]\tau =10.8Nm[/tex]
Torque is equal to [tex]\tau =I\alpha[/tex], here I is moment of inertia and [tex]\alpha[/tex] angular acceleration
So [tex]10.8=10.25\times I[/tex]
[tex]I=1.053kgm^2[/tex]
Moment of inertia of sphere is equal to [tex]I=\frac{2}{5}Mr^2[/tex]
So [tex]1.053=\frac{2}{5}\times M\times 0.37^2[/tex]
M = 19.23 kg
So mass of the sphere is 19.23 kg
An electron is moving at a speed of 2.20 ✕ 104 m/s in a circular path of radius of 4.3 cm inside a solenoid. The magnetic field of the solenoid is perpendicular to the plane of the electron's path. The solenoid has 25 turns per centimeter.(a) Find the strength of the magnetic fieldinside the solenoid.(b) Find the current in the solenoid.
Answer:
a) 2.90*10^-6 T
b) 0.092A
Explanation:
a) The magnitude of the magnetic field is given by the formula for the calculation of B when it makes an electron moves in a circular motion:
[tex]B=\frac{m_ev}{qR}[/tex]
me: mass of the electron = 9.1*10^{-31}kg
q: charge of the electron = 1.6*10^{-19}C
R: radius of the circular path = 4.3cm=0.043m
v: speed of the electron = 2.20*10^4 m/s
By replacing all these values you obtain:
[tex]B=\frac{(9.1*10^{-31}kg)(2.20*10^4 m/s)}{(1.6*10^{-19}C)(0.043m)}=2.90*10^{-6}T=2.9\mu T[/tex]
b) The current in the solenoid is given by:
[tex]I=\frac{B}{\mu_0 N}=\frac{2.90*10^{-6}T}{(4\pi*10^{-7}T/A)(25)}=0.092A=92mA[/tex]
To find the strength of the magnetic field inside the solenoid, we can use the formula for the magnetic field produced by a solenoid. To find the current in the solenoid, we can use the formula for the magnetic force experienced by a charged particle moving in a magnetic field.
Explanation:To find the strength of the magnetic field inside the solenoid, we can use the formula for the magnetic field produced by a solenoid: B = µ0nI, where B is the magnetic field, µ0 is the permeability of free space (4π × 10-7 T·m/A), n is the number of turns per unit length, and I is the current flowing through the solenoid.
Given that the solenoid has 25 turns per centimeter and the radius of the circular path is 4.3 cm, we can find the number of turns per unit length: n = 25 × 100 cm = 2500 turns/m.
The strength of the magnetic field inside the solenoid is then: B = (4π × 10-7 T·m/A) × (2500 turns/m) × (I).
To find the current in the solenoid, we can use the formula for the magnetic force experienced by a charged particle moving in a magnetic field: F = qvB, where F is the magnetic force, q is the charge of the particle, v is its velocity, and B is the magnetic field.
Given that the velocity of the electron is 2.20 × 104 m/s and the radius of the circular path is 4.3 cm, we can find the magnetic force experienced by the electron: F = (1.6 × 10-19 C) × (2.20 × 104 m/s) × (B).
Since the magnetic force is centripetal, it can also be expressed as F = mv2/r, where m is the mass of the electron, v is its velocity, and r is the radius of the circular path.
Combining these two equations, we can find the current in the solenoid: I = (mv2/r) / (vB).
The sound level produced by one singer is 70.8 dB. What would be the sound level produced by a chorus of 37 such singers (all singing at the same intensity at approximately the same distance as the original singer)
Answer:
86.48 dB . Ans
Explanation:
Sound level by one singer = 70.8 dB
= 7.08 B
If I be its intensity in terms of watt/m²
log ( I / 10⁻¹² ) = 7.08 ( given )
I / 10⁻¹² = 10⁷°⁰⁸
I = 10⁻¹² x 10⁷°⁰⁸
= 10⁻⁴°⁹²
= .000012022 W / m²
intensity by 37 singers
= 37 x .000012022 W /m²
= .0004448378
= 4448.378 x 10⁻⁷
intensity level
= log I / 10⁻¹² B
= log 4448.378 x 10⁻⁷ / 10⁻¹²
= log 444837800 B
log 4448378 + 2 B
= 6.6482 +2 B
8.6482 B
intensity level = 8.6482 B
= 86.48 dB . Ans
Answer:
86.48 dB
Explanation:
86.48 dB
Explanation:
sound level intensity, β = 70.8 dB
Io = 10^-12 W/m²
Let the intensity if I at the level of 70.8 dB
Use the formula
7.08 = log I + 12
log I = - 4.92
I = 1.2 x 10^-5 W/m²
There are 37 singers, so the total intensity is I'.
I' = 37 x I
I' = 37 x 1.2 x 10^-5 = 4.45 x 10^-4 W/m²
Let the sound intensity is β' in decibels.
Use the formula
β' = 86.48 dB
An object with mass 3M is launched straight up. When it reaches its maximum height, a small explosion breaks the object into two pieces with masses M and 2M. The explosion provides an impulse to each object in the horizontal direction only. The heavier of the two pieces is observed to land a distance 53 m from the point of the original launch. How far from the original launch position does the lighter piece land
Using conservation of momentum, the lighter piece in the explosion scenario lands 53 meters away from the launch point in the opposite direction in which the heavier piece landed.
The question revolves around a scenario where conservation of momentum applies. An object of mass
3M explodes into two pieces of mass M and 2M. The conservation of momentum dictates that the center of mass of the two pieces and the original object all follow the same horizontal trajectory. If the heavier piece lands 53 meters away from the point of launch, then, due to the center of mass being conserved during the explosion, the lighter piece would have landed at the same point of launch.
Utilizing the conservation law, which states that the total momentum before the explosion must be equal to the total momentum after the explosion, the pieces must move symmetrically about the center of mass of the original object. Given that the larger piece is observed to land 53 meters away from the launch point and has twice the mass of the smaller piece, the smaller piece would have landed 53 meters on the other side of the launch point, making a total separation of 106 meters between the two pieces. Therefore, the lighter piece would have landed 53 meters away in the opposite direction from where the heavier piece landed.
A thin metal cylinder of length L and radius R1is coaxial with a thin metal cylinder of length L and a larger radius R2. The space between the two coaxial cylinders is filled with a material that has resistivity rho. The two cylinders are connected to the terminals of a battery with potential difference ΔV, causing current I to flow radially from the inner cylinder to the outer cylinder. Part A Find an expression for the resistance of this device in terms of its dimensions and the resistivity. Express your answer in terms of some or all of the variables R1, R2, L, and rho. R = nothing
Answer:
The expression for resistance is [tex]R = \frac{\rho}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]
Explanation:
Generally flow of charge at that point is mathematically given as
[tex]J = \frac{I}{2 \pi r L}[/tex]
Where L is length of the cylinder as given the question
The potential difference that is between the cylinders is
[tex]\delta V = -E dr[/tex]
Where is the radius
Where E is the electric field that would be experienced at that point which is mathematically represented as
[tex]E = \rho J[/tex]
Where is the [tex]\rho[/tex] is the resistivity as given the question
considering the formula for potential difference we have
[tex]\delta V = -[\frac{\rho I}{2 \pi r L} ]dr[/tex]
To get V we integrate both sides
[tex]\int\limits^V_0 {\delta V} \, = \int\limits^{R_2}_{R_1} {\frac{\rho I}{2 \pi L r} } \, dr[/tex]
[tex]V = \frac{\rho I}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]
According to Ohm law
[tex]V= IR[/tex]
Now making R the subject we have
[tex]R = \frac{V}{I}[/tex]
Substituting for V
[tex]R = \frac{\rho}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]
The resistance of a coaxial cylindrical configuration with radii R1 and R2, length L, and resistivity rho, is given by the formula R = (rho / (2 * pi * ln(R2/R1))).
The question asks for an expression for the resistance of a coaxial cylindrical configuration, where the space between two metal cylinders of length L and radii R1 and R2 is filled with a material of resistivity rho. To find this, we utilize the formula for the resistance R of a material, which is R = rho * (L/A), where A is the cross-sectional area. However, since the current flows radially through the material between the cylinders, we need to consider the formula for resistance in terms of the radii and length of the cylinders. The resistance can be expressed as R = (rho * L) / (2 * pi * L * ln(R2/R1)), simplifying to R = (rho / (2 * pi * ln(R2/R1))).