Answer:
Up And Down
Explanation:
In this case, the particles of the medium move parallel to the direction that the pulse moves. This type of wave is a longitudinal wave. Longitudinal waves are always characterized by particle motion being parallel to wave motion.
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Final answer:
In a longitudinal wave on a horizontal spring, the medium oscillates back and forth horizontally, parallel to the direction of the wave's travel.
Explanation:
The question pertains to the behavior of longitudinal waves in a spring. When one student shakes the spring to create a longitudinal wave, the oscillation of the spring occurs in the same direction as the wave's propagation. This means that in a longitudinal wave, the medium—the spring in this case—oscillates parallel to the wave's direction of motion. Therefore, if the spring is held horizontally and the wave travels from left to right, the individual coils of the spring will move left and right along the same horizontal line, compressing and expanding as the wave passes through.
In contrast, with transverse waves, the medium moves perpendicularly to the direction of the wave's travel, such as in the motion of a rope being moved up and down while the wave travels horizontally.
Two narrow slits 0.02 mm apart are illuminated by light from a CuAr laser (λ = 633 nm) onto a screen. a) If the first fringe is 0.2 cm away from the central fringe, what is the screen distance? b) What is the angle of the first dark fringe? c) How many fringes are visible? d) What wavelength would a laser have to provide a fringe that coincides (lines up) with the third order fringe of the CuAr laser?
Answer:
a) 0.063m
b) 2.72°
c) 3151 fringes
d) 1.87*10^-6m
Explanation:
a) To find the screen distance you use the following formula:
[tex]y=\frac{m\lambda D}{d}\\\\D=\frac{dy}{m\lambda}[/tex]
D: screen distance
d: distance between slits
m: order of the fringes
λ: wavelength
By replacing the values of the parameters you obtain:
[tex]D=\frac{(0.02*10^{-3}m)(0.2*10^{-2}m)}{(1)(633*10^{-9}m)}=0.063m[/tex]
b) The condition for dark fringes is given by:
[tex]\lambda(m+\frac{1}{2})=dsin\theta[/tex]
for the first dark fringe the angle is:
[tex]\theta=sin^{-1}(\frac{\lambda(m+\frac{1}{2})}{d})\\\\\theta=sin^{-1}(\frac{(633*10^{-9}m)(1+\frac{1}{2})}{0.02*10^{-3}m})=2.72\°[/tex]
c) the visible number of fringes is given by:
[tex]N=1+2\frac{D}{d}=1+\frac{0.063m}{0.02*10^{-3}}=3151 \ fringes[/tex]
d) the wavelength of a laser in which its first order fringe coincides with the third one of the CuAr laser is:
[tex]y=\frac{(3)(633*10^{-9}m)(0.063m)}{0.02*10^{-3}m}=5.98*10^{-3}m\approx0.59cm\\\\\lambda'=\frac{dy}{mD}=\frac{(0.02*10^{-3}m)(0.59*10^{-2}m)}{(1)(0.063m)}=1.87*10^{-6}m[/tex]
An undamped 2.47 kg2.47 kg horizontal spring oscillator has a spring constant of 32.8 N/m.32.8 N/m. While oscillating, it is found to have a speed of 2.30 m/s2.30 m/s as it passes through its equilibrium position. What is its amplitude AA of oscillation? A=A= mm What is the oscillator's total mechanical energy EtotEtot as it passes through a position that is 0.6520.652 of the amplitude away from the equilibrium position? Etot=Etot= J
Answer:
0.631 m
6.53315 J
Explanation:
m = Mass = 2.47 kg
v = Velocity = 2.30 m/s
k = Spring constant = 32.8 N/m
A = Amplitude
In this system the energy is conserved
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{2.47\times 2.3^2}{32.8}}\\\Rightarrow A=0.631\ m[/tex]
The amplitude is 0.631 m
Mechanical energy is given by
[tex]E=\dfrac{1}{2}mv^2\\\Rightarrow E=\dfrac{1}{2}2.47\times 2.3^2\\\Rightarrow E=6.53315\ J[/tex]
The mechanical energy is 6.53315 J
Upper A 12-pound box sits at rest on a horizontal surface, and there is friction between the box and the surface. One side of the surface is raised slowly to create a ramp. The friction force f opposes the direction of motion and is proportional to the normal force Upper F Subscript Upper N exerted by the surface on the box. The proportionality constant is called the coefficient of friction, mu. When the angle of the ramp, theta, reaches 20degrees, the box begins to slide. Find the value of mu.
Answer:
The coefficient of static friction is : 0.36397
Explanation:
When we have a box on a ramp of angle [tex]\alpha[/tex] , and the box is not sliding because of friction, one analyses the acting forces in a coordinate system system with an axis parallel to the incline.
In such system, the force of gravity acting down the incline is the product of the box's weight times the sine of the angle:
[tex]F_g=W\,*\, sin(\alpha)\\F_g=m*g\,*\, sin(\alpha)[/tex]
Recall as well that component of the box's weight that contributes to the Normal N (component perpendicular to the ramp) is given by:
[tex]N=m*g*cos(\alpha)[/tex]
and the force of static friction (f) is given as the static coefficient of friction ([tex]\mu[/tex]) times the normal N:
[tex]f=\mu *m*g*cos(\alpha)[/tex]
When the box starts to move, we have that the force of static friction equals this component of the gravity force along the ramp:
[tex]f=F_g\\\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)[/tex]
Now we use this last equation to solve for the coefficient of static friction, recalling that the angle at which the box starts moving is 20 degrees:
[tex]\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)\\\mu=\frac{sin(\alpha)}{cos(\alpha)}\\\mu = tan(\alpha)\\\mu=tan(20^o)\\\mu=0.36397[/tex]
The coefficient of static friction (μ) for the box on the 20-degree incline can be found using the tangent of the angle, resulting in μ ≈ 0.3640.
To find the value of the coefficient of static friction (μ) when the box begins to slide at a 20-degree angle, we first need to understand the relationship between the normal force (Fn), the weight of the box (W), and the angle of the incline (θ).
Since the box is sliding at an angle of 20 degrees, we can use the component of the gravitational force perpendicular to the ramp to find Fn. The normal force is given by Fn = Wcos(θ), where W = mg and g is the acceleration due to gravity (9.8 m/s²). In this case, Fn = (12 pounds)(cos(20°)) × 4.44822, converting pounds to Newtons.
At the point of sliding, the static friction force is at its maximum, and it equals the component of the weight of the box parallel to the ramp, which is Wsin(θ). Thus, Fstatic_max = Wsin(θ), and since Fstatic_max = μFn, we can solve for μ.
μ = Fstatic_max / Fn = (Wsin(θ)) / (Wcos(θ)) = tan(θ). Substituting the given values, we get μ = tan(20°) ≈ 0.3640, which is the coefficient of static friction.
The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in the horizontal position θ = 0 where the spring is unstretched. If the bar is observed to momentarily stop in the position θ = 64°, determine the spring constant k. For your computed value of k, what is magnitude of the angular velocity of the bar when θ = 32°.
Answer:
The spring constant = 104.82 N/m
The angular velocity of the bar when θ = 32° is 1.70 rad/s
Explanation:
From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:
[tex]T_1+V_1=T_2+V_2[/tex]
[tex]0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}[/tex]
Also;
[tex]\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}[/tex]
Thus;
[tex]k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}[/tex]
where;
[tex]\delta[/tex] = deflection in the spring
k = spring constant
b = remaining length in the rod
m = mass of the slender bar
g = acceleration due to gravity
[tex]k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}[/tex]
[tex]k = 104.82\ \ N/m[/tex]
Thus; the spring constant = 104.82 N/m
b
The angular velocity can be calculated by also using the conservation of energy;
[tex]T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0[/tex]
[tex]\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0[/tex]
[tex]\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0[/tex]
[tex]0.7344 \omega_3^2 = 2.128[/tex]
[tex]\omega _3 = \sqrt{\frac{2.128}{0.7344} }[/tex]
[tex]\omega _3 =1.70 \ rad/s[/tex]
Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s
You've recently read about a chemical laser that generates a 20.0-cm-diameter, 30.0 MW laser beam. One day, after physics class, you start to wonder if you could use the radiation pressure from this laser beam to launch small payloads into orbit. To see if this might be feasible, you do a quick calculation of the acceleration of a 20.0-cm-diameter, 99.0 kg, perfectly absorbing block.
What speed would such a block have if pushed horizontally 100 m along a frictionless track by such a laser?
A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with closely packed coils made from 12.0 mm of insulated wire 1.1 mmmm in diameter. The permeability of free space is 4π×10−7T⋅m/A4π×10−7T⋅m/A.
How many loops will the inductor contain?
Final answer:
To calculate the number of loops for an inductor in a resonant circuit with a given capacitor and frequency, the resonant frequency formula can be used in conjunction with the inductance formula for a solenoid. These formulas involve calculations that may require additional information not provided in the question, such as the length of the solenoid.
Explanation:
Calculating the Required Number of Loops for the Inductor
To find the number of loops needed for a solenoid inductor to resonate with a 286 nF capacitor at 18.0 kHz, we must use the formula for the resonant frequency (f) of an LC circuit:
f = 1 / (2π√(LC))
Where:
C is the capacitance (286 nF or 286 × 10-9 F).
L is the inductance we need to calculate.
f is the resonant frequency (18.0 kHz or 18,000 Hz).
The inductance (L) of a solenoid inductor is given by:
L = (μ0× N2× A) / l
Where:
N is the number of loops.
A is the cross-sectional area of the coil (π × radius2).
l is the length of the solenoid.
μ0 is the permeability of free space (4π × 10-7 T·m/A).
By rearranging the resonant frequency formula to solve for L and then equating it to the inductance formula of a solenoid, we can determine the number of loops required for the solenoid to meet the resonance condition with the given capacitor at the specified frequency.
A trooper is moving due south along the freeway at a speed of 30.1 m/s when a red car passes the trooper. The red car moves with constant velocity of 45.4 m/s southward. At the instant the trooper's car is passed, the trooper begins to speed up at a constant rate of 2.0 m/s2. What is the maximum distance ahead of the trooper that is reached by the red car
Answer:
The correct answer to the following question will be "41.87 m".
Explanation:
The given values are:
The speed of trooper = [tex]30.1 \ m/s[/tex]
The velocity of red car = [tex]45.4 \ m/s[/tex]
Now,
A red car goes as far as possible until the speed or velocity of the troops is the same as that of of the red car at
[tex]t=\frac{45-30}{2.7}[/tex] (∵ [tex]time=\frac{distance}{time}[/tex])
[tex]t=5.56 \ sec[/tex]
then,
The distance covered by trooper,
[tex]t1=30\times 5.56+\frac{1}{2}\times 2.7\times (5.56)^2[/tex]
[tex]=208.33 \ m[/tex]
The distance covered by red car,
= [tex]45\times 5.56[/tex]
= [tex]250.2 \ m[/tex]
Maximum distance = [tex]250.2-208.33[/tex]
= [tex]41.87 \ m[/tex]
Final answer:
The trooper's car catches up with the red car after 7.65 seconds. The maximum distance ahead reached by the red car is 347.01 meters.
Explanation:
To find the maximum distance ahead reached by the red car, we need to calculate the time it takes for the trooper's car to catch up with the red car.
First, find the time it takes for the trooper's car to accelerate from 30.1 m/s to 45.4 m/s:
Initial velocity of the trooper's car, v₀ = 30.1 m/sFinal velocity of the trooper's car, v = 45.4 m/sAcceleration of the trooper's car, a = 2.0 m/s²Using the equation v = v₀ + at, we can solve for t:t = (v - v₀) / a = (45.4 - 30.1) / 2.0 = 7.65 seconds
Now, we can find the maximum distance ahead by multiplying the velocity of the red car by the time it took for the trooper's car to catch up:
Distance = velocity x time = 45.4 m/s x 7.65 s = 347.01 meters
Notice that the flux through the cube does not depend on aaa or ccc. Equivalently, if we were to set b=0b=0, so that the electric field becomes E′→=ai^+cj^E′→=ai^+cj^, then the flux through the cube would be zero. Why?
Electric flux through a cube with an external electric field is zero if there are no charges inside the cube, as Gauss's law states that the net flux is proportional to enclosed charge, leading to zero flux when the electric field vectors align perpendicularly with the area vectors of the cube's sides.
Explanation:The electric flux through a cube can be analyzed when an electric field is applied. If the electric field is given by E' = ai + cj, where i and j are unit vectors along the x and y axes respectively, and there is no field component along b, zero flux is observed through most of the cube's surfaces due to the perpendicular directions of the electric field and the area vectors of the surfaces. Specifically, the sides of the cube have area vectors perpendicular to the given field components, resulting in a scalar product of zero. Consequently, the flux through these sides is zero. Gauss's law states that the net flux through a closed surface is proportional to the charge enclosed within the surface. If there are no charges inside the cube, the net flux must be zero regardless of whether the charges are outside the cube or if the cube has an induced electric field due to a changing current, as would be the case with a wire acting as an inductor passing through it.
Two stereo speakers mounted 4.52 m apart on a wall emit identical in-phase sound waves. You are standing at the opposite wall of the room at a point directly between the two speakers. You walk 2.11 m parallel to the wall, to a location where you first notice that the sound intensity drops to zero.
If the wall along which you are walking is 10.7 m from the wall with the speakers, what is the wavelength of the sound waves?
a) 2.1 m
b) 2.9 m
c) 0.9 m
d) 1.7 m
The wavelength of the sound waves is equal to 1.714 m. Therefore, option (d) is correct.
What are sound waves?Sound waves can transmit through liquids, and gases, as well as plasma as longitudinal waves, called compression waves. It requires a medium to travel so it can be traveled through solids in the form of both longitudinal waves and transverse waves.
Given, the distance between two speakers, d₁ = 4.52 m
The distance between two walls, d₂ = 10.2 m
The expression to calculate wavelength: pd =(n + 0.5 λ)
As there is no slit, n =0 therefore, pd = 0.5 λ ...........(1)
Given, the triangle-1, the value of s₁ = 10.7 m
The value of s₂ = (4.52/2) - 2.11 = 0.15 m
The value of s₃ is equal to: [tex]s_3 =\sqrt{(10.7)^2+(0.15)^2}[/tex]
s₃ = 10.7 m
Given, the triangle-2, the value of a₁ = 10.7 m
The value of a₂ = (4.52/2) + 2.11 = 4.37 m
The value of s₃ is equal to: [tex]s_3 =\sqrt{(10.7)^2+(4.37)^2}[/tex]
a₃ = 11.557 m
The constructive interference, pd = a₃ - s₃ = 11.557 - 10.7 = 0.857 m
From the equation (1) we can determine wavelength:
pd = 0.5 λ
0.857 = 0.5 λ
λ = 1.714 m
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Consider a steel guitar string of initial length L=1.00L=1.00 meter and cross-sectional area A=0.500A=0.500 square millimeters. The Young's modulus of the steel is Y=2.0×1011Y=2.0×1011 pascals. How far ( ΔLΔLDelta L) would such a string stretch under a tension of 1500 newtons?
Answer:
The extent to which it would stretch is [tex]\Delta L = 0.015 \ m[/tex]
Explanation:
From the question we are told that
The initial length is [tex]L = 1.00m[/tex]
The area is [tex]A = 0.500 mm^2 = \frac{0.500}{1 *10^6} = 0.500*10^6 \ m^2[/tex]
The Young modulus of the steel is [tex]Y = 2.0*10^{11} Pa[/tex]
The tension is [tex]T =1500 N[/tex]
The Young modulus is mathematically represented as
[tex]Y = \frac{\sigma}{e}[/tex]
Where [tex]\sigma[/tex] is the stress which is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
Substituting values
[tex]\sigma = \frac{1500}{0.500*10^{-6}}[/tex]
[tex]\sigma = 3.0*10^9 N/m^2[/tex]
And e is the strain which is mathematically represented as
[tex]e = \frac{\Delta L}{L }[/tex]
Where [tex]\Delta L[/tex] The extension of the steel string
Substituting these into the equation above
[tex]Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }[/tex]
Substituting values
[tex]2.0 *10^{11} = \frac{3.0*10^9}{\frac{\Delta L}{L} }[/tex]
[tex]\Delta L = \frac{3.0*10^9 * 1}{2.0 *10^{11}}[/tex]
[tex]\Delta L = 0.015 \ m[/tex]
d. What do the results suggest about the ability of touch therapists to select the correct hand by sensing energy fields? A. Since the confidence interval is not entirely above 0.5, there does not appear to be sufficient evidence that touch therapists can select the correct hand by sensing energy fields. B. Since the confidence interval is not entirely below 0.5, there appears to be evidence that touch therapists can select the correct hand by sensing energy fields. C. Since the lower confidence limit is below 0.5, there does not appear to be sufficient evidence that touch therapists can select the correct hand by sensing energy fields. D. Since the upper confidence limit is above 0.5, there appears to be evidence that touch therapists can select the correct hand by sensing energy fields.
Answer:
A) Since the confidence interval is not entirely above 0.5, there does not appear to be sufficient evidence that touch therapists can select the correct hand by sensing energy fields.
Explanation:
See answer, it actually doubles as explaination.
Which of the following is a true statement about recycling?
Recycling allows resources to be reused, instead of dumped in landfills.
Recycled products emit oxygen into the atmosphere.
Recycling causes the earth to cool down, instead of warm up.
Recycling turns non-renewable resources into renewable ones.
Answer:
A. recycling allows resources to be reused, instead of dumped in landfills.
Explanation:
A large turntable with radius 6.00 m rotates about a fixed vertical axis, making one revolution in 8.00 s. The moment of inertia of the turntable about this axis is 1200 kg⋅m². You stand, barefooted, at the rim of the turntable and very slowly walk toward the center, along a radial line painted on the surface of the turntable. Your mass is 73.0 kg. Since the radius of the turntable is large, it is a good approximation to treat yourself as a point mass. Assume that you can maintain your balance by adjusting the positions of your feet. You find that you can reach a point 3.00 m from the center of the turntable before your feet begin to slip.
What is the coefficient of static friction between the bottoms of your feet and the surface of the turntable?
Answer:
0.8024
Explanation:
From the given question; we can say that the angular momentum of the system is conserved if the net torque is is zero.
So; [tex]I_o \omega _o = I_2 \omega_2[/tex]
At the closest distance ; the friction is :
[tex]f_s = \mu_s (mg)[/tex]
According to Newton's Law:
F = ma
F = mrω²
From conservation of momentum:
[tex]I_o \omega _o = I_2 \omega_2[/tex]
[tex]\omega_2= \frac { I_o \omega _o}{ I_2 }[/tex]
[tex]\omega_2=( \frac { I_1+ mr_o^2}{ I_1 + mr^2 })* \omega_o[/tex]
However ; since the static friction is producing the centripetal force :
[tex]\mu_s (mg) = mr \omega_2^2[/tex]
[tex]\mu _s = \frac{ \omega^2_2 *r }{g}[/tex]
The coefficient of static friction between the bottoms of your feet and the surface of the turntable can now be calculated by using the formula :
[tex]\mu _s = \frac{ \omega^2_2 *r }{g}[/tex]
= [tex]( \frac { I_1+ mr_o^2}{ I_1 + mr^2 })^2 * \frac{ \omega^2_o *r }{g}[/tex]
= [tex][\frac{1200+(73(6)^2)}{((1200)+(73)(3)^2)} ]^2*[\frac{(\frac{\pi}{4})^2(3)} {9.8}][/tex]
= 0.8024
The friction force between the bottom of the feet and the surface of the
turntable balance the centrifugal force due to rotation.
The coefficient static friction is approximately 0.802Reasons:
Radius of the turntable, r₀ = 6.00 m
The period of rotation of the turntable, T₀ = 8.00s
Moment of inertia of the turntable, [tex]I_t[/tex] = 1,200 kg·m²
Mass of the person, m = 73.0 kg
The point at which the feet starts to slip, r₁ = 3.00 m
Required:
The friction between the bottom of the feet and the surface of the turntable.
Solution:
The moment of inertia of the person and the turntable combined, J, can be
found by considering the person as a point mass.
Therefore;
At the rim, J₀ = [tex]I_t[/tex] + m·r² = 1,200 + 73×6² = 3828
J₀ = 3,828 kg·m²
At 3.00 m from the center, J₁ = [tex]I_t[/tex] + m·r² = 1,200 + 73×3² = 1,857
J₁ = 1,857 kg·m²
Angular momentum, L = J·ω
Whereby the angular momentum is conserved, we have;
L₀ = L₁
J₀·ω₀ = J₁·ω₁
[tex]\displaystyle \omega_1 = \mathbf{ \frac{J_0 \cdot \omega_0}{J_1}}[/tex]
Which gives;
To remain at equilibrium, the friction force, [tex]F_f[/tex] = The centrifugal force, [tex]F_c[/tex]
[tex]F_f =\mu_s \cdot W = \mu_s \cdot m \cdot g[/tex][tex]\displaystyle F_c = \frac{m \cdot v^2}{r} = \frac{m \cdot (\omega \cdot r)^2}{r} = m \cdot r \cdot \omega^2[/tex]
Where;
[tex]\mu_s[/tex] = The coefficient of static friction
g = Acceleration due to gravity which is approximately 9.81 m/s²
r = The radius of rotation
ω = The angular speed
[tex]F_f[/tex] = [tex]F_c[/tex]
Therefore;
[tex]\displaystyle F_f = \mu_s \cdot m \cdot g = \mathbf{m \cdot r \cdot \omega^2}[/tex]Therefore, at 3.00 m from the center of the turntable, we have;
[tex]\displaystyle F_f = \mu_s \cdot m \cdot g = \mathbf{ m \cdot r_1 \cdot \omega_1^2}[/tex]
[tex]\displaystyle\mu_s \cdot g = r_1 \cdot \omega_1^2[/tex]
[tex]\displaystyle\mu_s = \frac{r_1 \cdot \omega_1^2}{g} = \mathbf{\frac{r_1 \cdot \left(\displaystyle \frac{J_0 \cdot \omega_0}{J_1}\right)^2}{g}}[/tex]
Which gives;
[tex]\displaystyle\mu_s =\frac{3 \times \left(\displaystyle \frac{3,828 \times \frac{2 \cdot \pi}{8} }{1,857}\right)^2}{9.81} \approx \mathbf{0.802}[/tex]
The coefficient of static friction between the bottom of the feet and the surface of the turntable, at 3.00 m from the center, [tex]\mu_s[/tex] ≈ 0.802.
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A lawn mower engine running for 20 m i n does 4, 5 6 0, 0 0 0 J of work. What is the power output of the engine?
Answer:[tex]3800\ W[/tex]
Explanation:
Given
Lawn mover running for [tex]t=20\ min [/tex]
and does [tex]W=4560\times 10^3\ J[/tex]
We know Power is rate of work i.e.
[tex]P=\frac{\text{Work}}{\text{time}}[/tex]
[tex]P=\frac{4560\times 10^3}{20\times 60}[/tex]
[tex]P=3800\ W[/tex]
Thus Power output is [tex]3800\ W[/tex]
1. Laser beam with wavelength 632.8 nm is aimed perpendicularly at opaque screen with two identical slits on it, positioned horizontally, and close enough so that both of them fall in the beam cross section. By shifting vertically the screen, each slits can be illuminated independently, allowing the other to be disregarded. On an observation screen positioned 1m further from the opaque screen, the diffraction patterns from the independent slit illumination were found identical, with minima 6.5 mm apart. The interference pattern maxima from simultaneous illumination of both slits were 0.53 mm apart. What are the double slit characteristics (slit width, slit separation)
To find the double-slit characteristics of width and separation, one must use the data from a single-slit diffraction pattern to calculate the slit width and then the interference pattern from both slits to find the slit separation.
The student's question is about the characteristics of a double-slit diffraction pattern produced when a laser beam with a specific wavelength is projected through two slits. To deduce the slit width and separation, we use the information provided about the diffraction and interference patterns observed.
Firstly, we use the minima separation from the single slit diffraction to find the slit width using the diffraction formula d sin(θ) = mλ, where d is the slit width, m is an integer denoting the order of the minima, λ is the wavelength, and θ is the angle of the minima. Since the screen is 1 meter away and the minima are 6.5 mm apart, we can approximate sin(θ) as the ratio of minima separation to the distance to the screen to solve for the slit width d.
Next, we apply the formula for double-slit interference, d sin(θ) = nλ (where d is now the slit separation, n is an integer denoting the order of the maxima) to the observed interference maxima separation to calculate the slit separation.
Slit separation: [tex]\(97.23 \times 10^{-6}\)[/tex] m. Slit width: [tex]\(1194.34 \times 10^{-6}\)[/tex] m. Calculated using interference formula and provided data.
To solve this problem, we can use the double-slit interference formula:
[tex]\[ \Delta y = \frac{\lambda L}{d} \][/tex]
where:
- [tex]\( \Delta y \)[/tex] is the distance between adjacent maxima or minima on the observation screen,
- [tex]\( \lambda \)[/tex] is the wavelength of the laser beam,
- [tex]\( L \)[/tex] is the distance between the screen and the slits, and
- [tex]\( d \)[/tex] is the slit separation.
Given:
- Wavelength [tex]\( \lambda = 632.8 \)[/tex] nm = [tex]\( 632.8 \times 10^{-9} \)[/tex] m,
- Distance between the screen and the slits [tex]\( L = 1 \)[/tex] m = 1 m,
- Distance between adjacent minima [tex]\( \Delta y_{\text{min}} = 6.5 \)[/tex] mm = [tex]\( 6.5 \times 10^{-3} \)[/tex] m, and
- Distance between adjacent maxima [tex]\( \Delta y_{\text{max}} = 0.53 \)[/tex] mm = [tex]\( 0.53 \times 10^{-3} \)[/tex] m.
First, let's find the slit separation [tex]\( d \)[/tex] using the minima data:
[tex]\[ d = \frac{\lambda L}{\Delta y_{\text{min}}} \][/tex]
Then, let's find the slit width [tex]\( w \)[/tex] using the maxima data:
[tex]\[ w = \frac{\lambda L}{\Delta y_{\text{max}}} \][/tex]
Let's calculate these values.
First, let's calculate the slit separation [tex]\(d\)[/tex] using the minima data:
[tex]\[ d = \frac{\lambda L}{\Delta y_{\text{min}}} = \frac{(632.8 \times 10^{-9} \, \text{m})(1 \, \text{m})}{6.5 \times 10^{-3} \, \text{m}} \][/tex]
[tex]\[ d \approx \frac{(632.8)(1)}{6.5} \times 10^{-6} \, \text{m} \][/tex]
[tex]\[ d \approx 97.23 \times 10^{-6} \, \text{m} \][/tex]
Now, let's calculate the slit width [tex]\(w\)[/tex] using the maxima data:
[tex]\[ w = \frac{\lambda L}{\Delta y_{\text{max}}} = \frac{(632.8 \times 10^{-9} \, \text{m})(1 \, \text{m})}{0.53 \times 10^{-3} \, \text{m}} \][/tex]
[tex]\[ w \approx \frac{(632.8)(1)}{0.53} \times 10^{-6} \, \text{m} \][/tex]
[tex]\[ w \approx 1194.34 \times 10^{-6} \, \text{m} \][/tex]
So, the slit separation [tex]\(d\)[/tex] is approximately [tex]\(97.23 \times 10^{-6}\)[/tex] m, and the slit width [tex]\(w\)[/tex] is approximately [tex]\(1194.34 \times 10^{-6}\)[/tex] m.
Situation: two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.Question: What is the minimum separation r_min that the electrons reach?Express your answer in term of q, m, v, and k (where k=).r_min =
Answer:
[tex]r_{min}=\frac{kq^2}{5m_ev^2}[/tex]
Explanation:
The total kinetic energy of both electrons will be electrostatic potential energy, when the electrons reach the minima distance due to electrostatic repulsion. Then, you have:
[tex]E_{k}=U_E\\\\\frac{1}{2}m_ev_1^2+\frac{1}{2}m_ev_2^2=k\frac{q^2}{r_{min}}[/tex]
me: mass of the electron
q: charge of the electron
k: Coulomb's constant
you take into account that v2=3v1=3v and do rmin the subject of the formula:
[tex]\frac{1}{2}m_e[v^2+9v^2]=5m_ev^2=k\frac{q^2}{r_{min}}\\\\r_{min}=\frac{kq^2}{5m_ev^2}[/tex]
The minimum separation rmin of two electrons moving towards each other due to their electric repulsion can be calculated using the equation rmin = 2q2 / (mkv2), where q is the charge of the electron, m is its mass, v is the initial speed of one electron, and k is the electrostatic constant.
Explanation:When two electrons with mass m and charge q move towards each other along the x-axis, their repulsion causes them to slow down. Eventually, they reach a point where their repulsion is strong enough to push them away from each other. This point of minimum separation, denoted by rmin, can be found by equating the electrical force of repulsion to the centripetal force of motion. Since the electrical force between two point charges q1 and q2 separated by a distance r is given by F = k(q1q2)/r2, where k is the electrostatic constant, the equation for rmin is:
rmin = 2q2/ (mkv2)
where v is the initial speed of one electron towards the other.
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1. An object (m = 500 g) with an initial speed of 0.2 m/s collides with another object (m = 1.5 kg) which was at rest before the collision. Calculate the resulting speed for an inelastic collision (when they stick together). 2. A small object (m = 200 g) collides elastically with a larger object (m = 1000 g), which was at rest before the collision. The incoming speed of the smaller object was 1.0 m/s. The speed of the larger object after the collision is 0.33 m/s. Calculate the resulting speed and determine the direction for the smaller object after the collision when it rebounds. (Watch out for the directions of the motions and use respective signs for the velocities and momentums.)
Answer:
Explanation:
1 )
We shall apply conservation of momentum law to solve the problem.
mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .
.5 x .2 = (1.5 + .5)V
V = .05 m /s
2 ) We shall use formula for velocity of object after elastic collision as follows
v₁ = [tex]\frac{(m_1-m_2)}{(m_1+m_2)} u_1+\frac{2m_2u_2}{(m_1+m_2)}[/tex]
m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.
Putting the values
= [tex]\frac{(200-1000)}{(1000+200)} \times 1 +\frac{2 \times1000\times0}{(1000+200)}[/tex]
= - .66 m /s
Since the sign is negative so it will be in opposite direction .
1. The resultant speed for an inelastic collision is =.05 m /s
2. When The direction for the smaller object the collision when it rebounds is = -.66 m /s
What is Momentum of Law?
Conservation of momentum is refer to a major law of physics that states that the momentum of a system is constant if no external forces are acting on the system. when It is embodied in Newton’s First Law or The Law of Inertia.
1. We are involving the conservation of momentum law to solve the problem.
mv is = ( M +m) V, m, and M are masses of the small and large objects, v is the velocity of a small object before the collision and also V is the velocity of both the objects together after the collision.
Then .5 x .2 = (1.5 + .5)V
Therefore, V = .05 m /s
2. We shall use the formula for the velocity of the object behind the elastic collision as follows
After that, v₁ = (m₁-m₂)/(m₁+m₂)u1 + 2m₂u₂/(m₁+m₂)
m₁ and m₂ are masses of the first and second objects u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.
Putting the values
Now, = (200-1000)/(1000+200) × 1 + 2×1000×0/(1000+200)
Thus, = - .66 m /s
When the sign is negative so it will be in opposite direction.
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A flexible balloon contains 0.400 molmol of an unknown polyatomic gas. Initially the balloon containing the gas has a volume of 7.50 m3 and a temperature of 26.0∘C. The gas first expands isobarically until the volume doubles. Then it expands adiabatically until the temperature returns to its initial value. Assume that the gas may be treated as an ideal gas with Cp=33.26J/mol⋅K and γ=4/3.
What is the total heat Q supplied to the gas in the process?
The total heat supplied to the gas in the process is calculated by considering the two expansion processes separately.
Explanation:The total heat supplied to the gas in the process can be calculated by considering the two expansion processes separately.
In the first expansion, which is isobaric, the gas volume doubles. Since the pressure remains constant, the work done on the gas is given by W = PΔV, where P is the pressure and ΔV is the change in volume. The work done is equal to the heat supplied to the gas.
In the second expansion, which is adiabatic, the temperature returns to its initial value. In an adiabatic process, the work done on the gas is given by W = (γ - 1)ΔU, where γ is the ratio of specific heats and ΔU is the change in internal energy. Since the temperature returns to its initial value, the change in internal energy is zero, and therefore, no heat is supplied to the gas in this process.
Rank the following meteorites in terms of their age, from youngest to oldest:1) a meteorite containing 500 238U isotopes and 100 206Pb isotopes.2) a meteorite containing 400 238U isotopes and 100 206Pb isotopes.3) a meteorite containing 400,000 238U isotopes and 400,000 206Pb isotopes.4) a meteorite containing 4,000 238U isotopes and 200 206Pb isotopes.5) a meteorite containing 3,000 238U isotopes and 1,000 206Pb isotopes.
Answer:
Explanation:
When 238U which is radioactive turns into 206Pb , it becomes stable and no further disintegration is done . Hence in the initial period ratio of 238U undecayed and 206Pb formed will be very high because no of atoms of 238U in the beginning will be very high. Gradually number of 238U undecayed will go down and number of 206Pb formed will go up . In this way the ratio of 238U and 206Pb in the mixture will gradually reduce to be equal to one or even less than one .
In the given option we shall calculate their raio
1 ) ratio of 238U and 206Pb = 5
2 ) ratio of 238U and 206Pb = 4
3 )ratio of 238U and 206Pb = 1
4 ) ratio of 238U and 206Pb = 20
5 )ratio of 238U and 206Pb = 3
lowest ratio is 1 , hence this sample will be oldest.
Ranking from youngest to oldest
4 , 1 , 2 , 5 , 3 .
A submarine is 2.84 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the submarine so that the beam strikes the surface of the water 2.34 102 m from the shore. A building stands on the shore, and the laser beam hits a target at the top of the building. The goal is to find the height of the target above sea level.
Answer:
468 m
Explanation:
So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.
The length CE is length AE - length AC = 284 - 234 = 50 m
We can calculate the angle ECD:
[tex]tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2[/tex]
[tex]\hat{ECD} = tan^{-1} 2 = 63.43^o[/tex]
This is also the angle ACB, so we can find the length AB:
[tex]tan(\hat{ACB}) = \frac{AB}{AC} = \frac{AB}{234}[/tex]
[tex]2 = \frac{AB}{234}[/tex]
[tex]AB = 2*234 = 468 m[/tex]
So the height of the building is 468m
A heavy lab cart moves with kinetic energy K init on a track and collides with a lighter lab cart that is initially at rest. The carts bounce off each other but the collision is not perfectly elastic, causing the two-cart system to lose kinetic energy K lost. A student wonders if the fraction of kinetic energy lost from the two-cart system during the collision (Klost/Kini) depends on the speed of the first cart before the collision and plans to perform an experiment.
The student hypothesizes that a greater fraction of kinetic energy is lost from the system during the collision when the speed of the first cart is greater. Briefly state one reason the hypothesis might be correct
A greater initial speed of the heavy cart could lead to a higher initial kinetic energy, which could result in a greater fraction of the kinetic energy being lost during a non-perfectly elastic collision. This could support the student's hypothesis.
Explanation:The student's hypothesis could be correct because the fraction of kinetic energy lost from the two-cart system during the collision could indeed depend on the initial speed of the heavy cart. If the first, heavier cart is moving at a higher speed before the collision, it would have a higher initial kinetic energy (K init). When it collides with the second cart, more kinetic energy could be transformed into other forms of energy such as heat or sound, especially since the collision is not perfectly elastic. This means a greater fraction of kinetic energy could be lost (K lost) in the process, supporting the student's hypothesis.
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The higher initial speed of the first cart leads to greater deformation and energy conversion to non-mechanical forms, resulting in a larger fraction of kinetic energy loss may be the reason the hypothesis might be correct.
The hypothesis that a greater fraction of kinetic energy is lost from the system during the collision when the speed of the first cart is greater might be correct due to the nature of inelastic collisions and the role of deformation and heat generation during such collisions.
One reason for this is that the energy loss in a collision often includes contributions from factors like the deformation of the carts and the generation of heat, both of which can increase with the relative speed at the moment of collision. As the speed of the first cart increases, the impact forces during the collision are higher, leading to greater deformation and more energy being converted into internal energy (such as heat and sound) rather than being retained as kinetic energy in the system.
This means that when the speed of the first cart is higher, a larger portion of the initial kinetic energy is likely to be transformed into non-mechanical forms of energy, resulting in a greater fraction of kinetic energy being lost from the two-cart system.
An enemy spaceship is moving toward your starfighter with a speed of 0.400 c , as measured in your reference frame. The enemy ship fires a missile toward you at a speed of 0.700 c relative to the enemy ship. If you measure the enemy ship to be 7.00×106 km away from you when the missile is fired, how much time t, measured in your frame, will it take for the missile to reach you
Two students, standing on skateboards, are initially at rest, when they give each other a shove! After the shove, one student (77kg) moves to the left at 5.7m/s while the other student (59kg) moves right.
Answer:
The other student (59kg) moves right at 7.44 m/s
Explanation:
Given;
mass of the first student, m₁ = 77kg
mass of the second student, m₂ = 59kg
initial velocity of the first student, u₁ = 0
initial velocity of the second student, u₂ = 0
final velocity of the first student, v₁ = 5.7 m/s left
final velocity of the second student, v₂ = ? right
Apply the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(77 x 0) + (59 x 0) = (-77 x 5.7) + (59 x v₂)
0 = - 438.9 + 59v₂
59v₂ = 438.9
v₂ = 438.9 / 59
v₂ = 7.44 m/s to the right
Therefore, the other student (59kg) moves right at 7.44 m/s
Final answer:
The physics problem involves two students on skateboards applying a force to each other and moving in opposite directions. To find the velocity of the second student, the conservation of momentum is used, considering no external forces act on them.
Explanation:
The subject of this question is physics, specifically relating to the principle of conservation of momentum in a system of two students standing on skateboards. When the two students push off each other, they start moving in opposite directions due to this principle. Conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this scenario, the system comprises the two students on skateboards.
To find how fast the second student moves, one would use the formula:
Total momentum before the shove = Total momentum after the shove
0 = (mass of student 1 \(\times\) velocity of student 1) + (mass of student 2 \(\times\) velocity of student 2)
We have the mass and velocity of student 1 (77 kg, moving left at 5.7 m/s), and the mass of student 2 (59 kg), but we need to calculate the velocity of student 2 after the push to ensure momentum is conserved.
A capacitor of capacitance 4 µF is discharging through a 2.0-MΩ resistor. At what time will the energy stored in the capacitor be one-third of its initial value?
Answer:
Explanation:
Given that,
The capacitance of the capacitor is
C = 4 µF
Discharging through a resistor of resistance
R = 2 MΩ
Time the energy stored in the capacitor be one-third of its initial energy
i.e. u(E)_ final = ⅓u(E)_initial
U / Uo = ⅓
Energy stored in a capacitor (discharging) can be determined using
U = Uo•RC•exp(-t/RC)
U / Uo = -RC exp(-t/RC)
U / Uo = ⅓
RC = 2 × 10^6 × 4 × 10^-6 = 8s
⅓ = 8 exp( -t / 8)
Divide both side by -8
1/24 = exp(-t/8)
Take In of both sides
In(1/24) = In•exp(-t/8)
-3.1781 = -t / 8
t = -3.1781 × -8
t = 25.42 seconds
It will take 25.42 seconds for he capacitor to be ⅓ of it's initial energy
A conducting wire formed in the shape of a right triangle with base b = 34 cm and height h = 77 cm and having resistance R = 2.2 Ω, rotates uniformly around the y-axis in the direction indicated by the arrow (clockwise as viewed from above (loooking down in the negative y-direction)). The triangle makes one complete rotaion in time t = T = 1.6 seconds. A constant magnetic field B = 1.6 T pointing in the positive z-direction (out of the screen) exists in the region where the wire is rotating.1)What is ?, the angular frequency of rotation?2)What is Imax, the magnitude of the maximum induced current in the loop?3)At time t = 0, the wire is positioned as shown. What is the magnitude of the magnetic flux ?1 at time t = t1 = 0.45 s?4)What is I1, the induced current in the loop at time t = 0.45 s? I1 is defined to be positive if it flows in the negative y-direction in the segment of length h.
Answer:
a) 3.92 rad/s
b) 0.373 A
d) 0.018 A
Explanation:
a) The angular frequency of rotation is given by:
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{(1.6s)}=3.92\frac{rad}{s}[/tex]
b) The maximum induced current in the loop is given by:
[tex]I_{max}=\frac{emf_{max}}{R}=\frac{AB\omega}{R}[/tex]
R: resistance
A: area of the triangle loop = bh/2 = (0.34m)(0.77m)/(2) = 0.1309m^2
B: magnitude of the magnetic field
[tex]I_{max}=\frac{(0.13m^2)(1.6T)(3.92rad/s)}{2.2\Omega}=0.373A[/tex]
d) For t = 0.45s you have:
[tex]I(t)=\frac{ABcos(\omega t)}{R}\\\\I(0.45)=\frac{(0.13m^2)(1.6T)cos(3.92rad/s \ (0.45s))}{2.2\Omega}=-0.018A[/tex]
But I1 is defined to be positive if it flows in the negative y-direction.
hence, I for t=0.45 s is 0.018A
(1) The angular frequency is 3.92 rad/s
(2) the value of I(max) = 0.373A
(3) Image is required
(4) The current at 0.45s is -0.018A
Induced current:Given a conducting wire formed in the shape of a right triangle with:
Base b = 34 cm
Height h = 77 cm
Resistance R = 2.2 Ω
(1) Time period of rotation is T = 1.6s
The angular frequency is given by:
ω = 2π/T
ω = 2π/1.6
ω = 3.92 rad/s
(2) The magnetic field applied is B = 1.6T, perpendicular to the plane of the triangle.
the induced current is given by:
[tex]I_{ind}=\frac{EMF_{ind}}{R}[/tex]
where R is the resistance
[tex]I_{max}=\frac{EMF_{max}}{R}\\\\I_{max}=\frac{BA\omega}{R}\\\\I_{max}=\frac{1.6\times(1/2)\times34\times77\times3.92}{2.2}\\\\[/tex]
I(max) = 0.373A
(3) The image for the question is not attached.
(4) The instantaneous value of induced current at time t = 0.45s is given by:
[tex]I_{ind}=\frac{EMF_{ind}}{R}\\\\I(t)=\frac{BA(cos\omega t)}{R}\\\\I(t)=\frac{(1/2)\times34\times77\times1.6\timescos(3.92\imes0.45)}{2.2}[/tex]
I(t) = -0.018 A
the negative sign indicated that the current flows in the negative y-direction.
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If 3.2*10^20 electrons pass through a wire in 4s, what would be the electrical current in the wire?
Answer: 12.8 A
Explanation:
Current is defined as the rate of flow of electric charge based on the formula:
I(current) = deltaQ(change in charge)/deltat(change in time).
First, however, we must convert the number of electrons into the number of coulombs. Based on the fact that the charge of 1 electron or 1 elementary charge is equal to 1.6*10^-19 C, we can calculate:
3.2*10^20 e = 1.6*10^-19*3*10^20 C = 51.2 C.
Now we use: I = Q/t = 51.2/4 = 12.8 A.
Hope this helped.
can someone help asap with this
Answer:
B
Explanation:
i had a test on this and got it correct
You desire to observe details of the Statue of Freedom, the sculpture by Thomas Crawford that is the crowning feature of the dome of the United States Capitol in Washington, D.C. For this purpose, you construct a refracting telescope, using as its objective a lens with focal length 89.7 cm. In order to acheive an angular magnification of magnitude 5.53 , what focal length f e should the eyepiece have?
Answer:[tex]f_e=16.22\ cm[/tex]
Explanation:
Given
Focal length of objective lens is [tex]f_o=89.7\ cm[/tex]
angular magnification is [tex]m_a=-5.53[/tex]
Angular magnification depends on the focal length of objective and eye piece
i.e.
[tex]m_a=-\frac{f_o}{f_e}[/tex]
[tex]f_e=-\frac{f_o}{m_a}[/tex]
[tex]f_e=-\frac{89.7}{-5.53}[/tex]
[tex]f_e=16.22\ cm[/tex]
The eyepiece should have a focal length of approximately 16.22 cm to achieve an angular magnification of magnitude 5.53.
To determine the required focal length of the eyepiece for a refracting telescope, one can use the formula for the angular magnification M of the telescope, which is given by:
[tex]\[ M = \frac{f_o}{f_e} \][/tex]
where [tex]f_o[/tex] is the focal length of the objective lens and[tex]\( f_e \)[/tex] is the focal length of the eyepiece.
Given that the focal length of the objective lens [tex]\( f_o \)[/tex] is 89.7 cm, and the desired angular magnification M is 5.53, we can rearrange the formula to solve for the focal length of the eyepiece[tex]\( f_e \)[/tex]:
[tex]\[ f_e = \frac{f_o}{M} \][/tex]
Substituting the given values:
[tex]\[ f_e = \frac{89.7 \text{ cm}}{5.53} \][/tex]
[tex]\[ f_e \ = 16.22 \text{ cm} \][/tex]
Therefore, the eyepiece should have a focal length of approximately 16.22 cm to achieve the desired angular magnification of 5.53 when used with an objective lens of focal length 89.7 cm.
When does gravitational lensing occur? High concentrations of dark matter cause length contraction of nearby objects. The gravity from a distant object (like a quasar) distorts the appearance of nearby galaxies. Telescope lenses are distorted by gravitational forces from dark matter. A massive object (like a galaxy cluster) bends the light from an object (like a quasar) that lies behind it.
Answer:
A massive object (like a galaxy cluster) bends the light from an object (like a quasar) that lies behind it.
Explanation:
A massive object, like a galaxy cluster, is able to deform the space-time shape as a consequence of its own gravity, so the light that it is coming from a source that is behind it in the line of sight will be bend or distorts in a way that will be magnified, making small arcs around the cluster with the image of the background object.
This technique is useful for astronomers since they make research of faraway objects (at hight redshift) that otherwise will difficult to detect with a telescope.
what component of a wet-dry vacuum cleaner is not present in a typical canister style residential cleaner
Answer:
A float switch
Explanation:
A float switch is a component of a wet-dry vacuum cleaner that is not present in a typical canister style residential cleaner.
A float switch is a device used to detect the level of liquid within a wet-dry vacuum cleaner. The switch may be used as an indicator to control other devices.
Some float switches contain a two-stage switch. The associated pump is activated as the liquid rises to the trigger point of the first stage. The second stage will be triggered if the liquid continues to rises.
A wet-dry vacuum cleaner contains a water collection bin or moisture-resistant collection system, and a float mechanism to handle liquid spills, which are not present in typical canister-style residential vacuum cleaners.
The component of a wet-dry vacuum cleaner that is not present in a typical canister-style residential cleaner is specifically designed to handle liquid spills in addition to dry debris. This design typically includes a separate water collection bin or moisture-resistant collection system, which is absent in regular canister vacuums that are only suitable for dry pickup. Wet-dry vacuums also often contain a float mechanism that stops suction once the water reaches a certain level to prevent the motor from ingesting water.