Answer:
Explanation:
λ
given λ = 700 nm
a) for first maxima, d*sin(θ) = λ
sin(theta) = λ/d
= 700*10^-9/(0.025*10^-3)
= 0.028
theta θ = sin^-1(0.028)
= 1.60 degrees
b) given R = 1 m,
delta_y = λ*R/d
= 700*10^-9*1/(0.025*10^-3)
= 0.028 m or 2.8 cm
c) for first maxima, d*sin(θ) = λ
sin(θ) = λ/d
= 700*10^-9/(2.5*10^-3)
= 0.00028
theta = sin^-1(0.00028)
= 0.0160 degrees
d) R = 25 mm = 0.025 m
δ_y = λ*R/d
= 700*10^-9*0.025/(2.5*10^-3)
= 7*10^-6 m or 7 micro m
e) No. But the position of maxima and minima will be shifted.
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.
Learn more about Kinetic Energy in Collisions here:https://brainly.com/question/16933340
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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.
A skateboarder rolls horizontally off the top of a
staircase at a speed of 18 ms-1
and at the bottom of
the stairs which has a horizontal length of 9 m as
shown.
Calculate the skater's vertical displacement during
the jump?
Given Information:
Initial horizontal speed = Vx = 18 m/s
horizontal distance = R = 9 m
Required Information:
Vertical displacement = h = ?
Answer:
Vertical displacement = h = 1.23 m
Explanation:
The horizontal distance covered by the skater is given by
R = Vx*t
Where R is horizontal distance, Vx is the initial horizontal speed and t is the time taken for the jump.
t = R/Vx
t = 9/18
t = 0.5 seconds
The vertical displacement covered by the skater is given by
h = Vy*t + ½gt²
Where Vy is the initial vertical speed of the skater and is zero since the skater jumps horizontally, g is the gravitational acceleration and h is the vertical displacement.
h = 0*t + ½gt²
h = ½gt²
h = ½(9.81)(0.5)²
h = 1.23 m
Therefore, the vertical displacement of the skater is 1.23 m
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.
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:
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
Monochromatic light of wavelength λ illuminates a pair of thin parallel slits of width a separated by a distance d at normal incidence, producing an interference pattern on a distant screen. a) (10 points) Explain two modifications to the experiment which would cause the maxima (bright spots) of the diffraction pattern to move away from the center. b) (10 points) Explain two modifications to the experiment which would cause the minima (dark spots) of the interference pattern to move towards the center.
Answer:
Explanation:
a ) If x be the position of n the bright fringe on the screen , following formula holds .
x = n (λD / d) ; λ is wave length , D is screen distance and d is slit separation .
If we increase the value of λ or wave length, x will increase so central fringe along with all the fringes will shift away from the centre .
If we increase the value of D or screen distance , it will also increase x , so fringes along with central fringe will shift away from the center.
b ) Fringes ,whether bright or dark , both shift together , either towards or away from the center .
So to move the dark spots towards the center , we need to do the opposite to what we did in the first case , ie decrease the wavelength or decrease the screen distance .
You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind and you perceive the frequency as 1310 Hz. You are relieved that he is in pursuit of a different speeder when he continues past you, but now you perceive the frequency as 1240 Hz. What is the frequency of the siren in the police car? The speed of sound in air is 343 m/s.
Answer:
[tex]f_{police}=1268.7 Hz[/tex]
Explanation:
We can use Doppler equation to find the frequency of the siren.
First of all we have the police car moving behind the car. Hence, the frequency detected by the car will be:
[tex]f_{car1}=f_{police}(\frac{v_{s}-v_{car}}{v_{s}-v_{police}})[/tex] (1)
Now, when the police car is moving in front of the car, the frequency detected by the car will be:
[tex]f_{car2}=f_{police}(\frac{v_{s}+v_{car}}{v_{s}+v_{police}})[/tex] (2)
By solving equation (1) and equation (2) for [tex]v_{police}[/tex] we have:
[tex]v_{police} = 44.67 m/s[/tex]
Knowing that:
f(car1) = 1310 Hzf(car2) = 1240 HzVs = 343 m/sV(car) = 35 m/sFinally, we just need to put this value into the first equation to find frequency of the police car.
[tex]f_{police}=f_{car}(\frac{v_{s}-v_{police}}{v_{s}-v_{car}})[/tex]
[tex]f_{police}=1310(\frac{343-44.7}{343-35})[/tex]
[tex]f_{police}=1268.7 Hz[/tex]
I hope it helps you!
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.
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]
A force of 6600 N is exerted on a piston that has an area of 0.010 m2
What area is required for a second piston to exert a force of 9900 N?
Fi F2
Use
A Ą
A. 0.015 m2
B. 150 m2
C. 0.0066 m2
D. 66 m2
Answer:
Choice A: approximately [tex]0.015\; \rm m^2[/tex], assuming that the two pistons are connected via some confined liquid to form a simple machine.
Explanation:
Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:
[tex]\displaystyle P_1 = \frac{F_1}{A_1} = \frac{6.600\times 10^3\; \rm N}{1.0\times 10^{-2}\; \rm m^{2}} = 6.6\times 10^{5}\; \rm N \cdot m^{-2}[/tex].
By Pascal's Principle, because the first piston exerted a pressure of [tex]6.6\times 10^{5}\; \rm N \cdot m^{-2}[/tex] on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.
Assume that the second piston is part of that wall. The pressure on the second piston will also be [tex]6.6\times 10^{5}\; \rm N \cdot m^{-2}[/tex]. In other words:
[tex]P_2 = P_1 = 6.6\times 10^{5}\; \rm N \cdot m^{-2}[/tex].
To achieve a force of [tex]9.900 \times 10^3\; \rm N[/tex], the surface area of the second piston should be:
[tex]\displaystyle A_2 = \frac{F_2}{P_2} = \frac{9.900\times 10^{3}\; \rm N}{6.6\times 10^5\; \rm N \cdot m^{-2}} \approx 0.015\; \rm m^{2}[/tex].
Which letter (A, B, or C) shows where you should apply the most effort to lift the stone?
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 thin, circular disk of radius R = 30 cm is oriented in the yz-plane with its center as the origin. The disk carries a total charge Q = +3 μC distributed uniformly over its surface. Calculate the magnitude of the electric field due to the disk at the point x = 15 cm along the x-axis.
Answer:
electric field due to the disk at the point x = 15 cm along the x-axis is;
E = 3.31 x 10^(9) N/C
Explanation:
We are given;
Radius;r = 30cm = 0.3m
Charge; Q = +3 μC = +3 x 10^(-6) C
Point, x = 15 cm = 0.15m
The formula for electric field due to the disk on the x-axis is given by;
E = [Q/(2ε₀•π•r²)] * [1 - (x/√(x² + r²))]
Where;
Q, x and r are as stated earlier
ε₀ is the permittivity of free space and has a constant value of 8.85 x 10⁻¹² C²/N.m
Thus, plugging in the relevant values, we have;
E = [3 x 10^(-6)/(8.85 x 10⁻¹² x π x 0.3²)] * [1 - (0.15/√(0.15² + 0.3²))]
E = 3.31 x 10^(9) N/C
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 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.
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.
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
Supervillain Prof. Marcia Grail is experimenting with energy and electric charge to better create a new doomsday weapon (and take over the world). She gathers a charge of 1.11 x 10^-10 C and places it in the center of a large empty spherical vacuum chamber of radius 250m (the place she fixes this immovable charge can be thought of as the origin of coordinates). The potential at a point 1m away in the positive y-direction is 1 V. What is the potential 1m away in the positive x-direction
Answer:
Check the explanation
Explanation:
given
charge of [tex]1.11 * 10^{-10}[/tex] C
radius 250m
The potential at a point 1m
and 1 V
mass 2 micrograms
charge of +2 Coulombs
the potential is V = K Q / r
V = 9 X 109 X 1.11 X 10-10 / 1
V = 9 X 109 X 1.11 X 10-10
V = 0.999 volt
nearly it is V = 1 v
the energy stored is U = q X V
U = 2 X 1
U = 2 J
the energy is stored in this configuration is U = 2 J
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.
A magnetic field between the poles of the electromagnet is uniform at any time, but its magnitude is increasing at the rate of 0.020T/s. The area of the conducting loop in the field is 120 cm2 , and the total circuit resistance, including the meter is 5.0 ohms. (a)Find the induced emf and the induced current in the circuit.(b) If the loop is replaced by one made of an insulator, what effect does this have on the induced emf and induced current?
Answer:
a) -2.4*10^-4 V
4.8*10^-5 A
b) emf = 0
induced current = 0
Explanation:
a) The induced emf is given by the following formula:
[tex]emf=-\frac{d\Phi_B}{dt}=-\frac{d(AB)}{dt}[/tex] ( 1 )
A: area of the loop = 120 cm^2 = 120 (10^-2)^2 = 0.012m^2
The area is constant and dB/dt = 0.020T/s
By replacing in the values of the parameters in the equation (1) you obtain:
[tex]emf=-A\frac{dB}{dt}=-(0.012m^2)(0.020T/s)=-2.4*10^{-4}V[/tex]
The induced current is:
[tex]I=\frac{emf}{R}=\frac{2.4*10^{-4}V}{5.0\Omega}=4.8*10^{-5}A[/tex]
b) If the loop is made of an insulator, electrons in the wire does not feel the change in the magnetic flux. Due to that, there is no a work over the electrons, and consequantly, there is neither emf nor induced current.
(a) The magnitude of the induced emf is 2.4 x 10⁻⁴ V.
(b) The induced current is 4.8 x 10⁻⁵ A.
Induced emfThe magnitude of the induced emf is determined by applying Faraday's law of electromagnetic induction.
emf = dΦ/dt
emf = BA/t
where;
A is the area = 120 cm² = 0.012 m²emf = (0.02 x 0.012)
emf = 2.4 x 10⁻⁴ V
Induced currentThe induced current is calculated as follows;
emf = IR
I = emf/R
I = (2.4 x 10⁻⁴) / (5)
I = 4.8 x 10⁻⁵ A
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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.
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.
A satellite travels around the Earth in a circular orbit. What is true about the forces acting in this situation? A. The resultant force is the same direction as the satellite’s acceleration. B. The gravitational force acting on the satellite is negligible. C. There is no resultant force on the satellite relative to the Earth. D. The satellite does not exert any force on the Earth.
Answer:
A. The resultant force in the same direction as the satellite’s acceleration.
Explanation:
Launching a satellite in the space and then placing it in orbit around the Earth is a complicated process but at the very basic level it works on simple principles. Gravitational force pulls the satellite towards Earth whereas it acceleration pushes it in straight line.
The resultant force of gravity and acceleration makes the satellite remain in orbit around the Earth. It is condition of free fall where the gravity is making the satellite fall towards Earth but the acceleration doesn't allow it and keeps it in orbit.
In a circular orbit around the Earth, the resultant force acting on a satellite is in the same direction as its acceleration.
Explanation:In a satellite orbiting the Earth in a circular orbit, there are several forces at play. The gravitational force between the satellite and the Earth provides the centripetal force that keeps the satellite in its orbit. The centripetal force acts towards the center of the circular orbit, while the satellite's acceleration is directed towards the center as well. Therefore, option A is correct: the resultant force is in the same direction as the satellite's acceleration.
The gravitational force acting on the satellite is not negligible; in fact, it is crucial in providing the necessary centripetal force. Therefore, option B is incorrect.
Option C is incorrect as well. There is a resultant force acting on the satellite relative to the Earth, which is responsible for keeping the satellite in its circular orbit.
Lastly, option D is also incorrect. According to Newton's third law of motion, the satellite exerts an equal and opposite force on the Earth, keeping the Earth and the satellite in orbit around their common center of mass.
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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.
A laser beam is incident at an angle of 30.2° to the vertical onto a solution of corn syrup in water. (a) If the beam is refracted to 18.82° to the vertical, what is the index of refraction of the syrup solution? (b) Suppose the light is red, with wavelength 632.8 nm in a vacuum. Find its wavelength in the solution. nm (c) Find its frequency in the solution. Hz (d) Find its speed in the solution.
Answer:
a) n2 = 1.55
b) 408.25 nm
c) 4.74*10^14 Hz
d) 1.93*10^8 m/s
Explanation:
a) To find the index of refraction of the syrup solution you use the Snell's law:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex] (1)
n1: index of refraction of air
n2: index of syrup solution
angle1: incidence angle
angle2: refraction angle
You replace the values of the parameter in (1) and calculate n2:
[tex]n_2=\frac{n_1sin\theta_1}{sin\theta_2}=\frac{(1)(sin30.2\°)}{sin18.82\°}=1.55[/tex]
b) To fond the wavelength in the solution you use:
[tex]\frac{\lambda_2}{\lambda_1}=\frac{n_1}{n_2}\\\\\lambda_2=\lambda_1\frac{n_1}{n_2}=(632.8nm)\frac{1.00}{1.55}=408.25nm[/tex]
c) The frequency of the wave in the solution is:
[tex]v=\lambda_2 f_2\\\\f_2=\frac{v}{\lambda_2}=\frac{c}{n_2\lambda_2}=\frac{3*10^8m/s}{(1.55)(408.25*10^{-9}m)}=4.74*10^{14}\ Hz[/tex]
d) The speed in the solution is given by:
[tex]v=\frac{c}{n_2}=\frac{3*10^8m/s}{1.55}=1.93*10^8m/s[/tex]
For each star, determine how its light would be shifted. Not all choices may be used, and some may be used more than once. A red dwarf moving away from Earth at 39.1 km/s
A yellow dwarf moving transversely at 15.1 km/s
A red giant moving towards Earth at 23.3 km/s
A blue dwarf moving away from Earth at 25.9 km/s
A red dwarf moving transversely at 14.1 km/s
When a star moves towards or away from Earth, its light is shifted to longer or shorter wavelengths, respectively. The red dwarf moving away from Earth and the blue dwarf moving away from Earth would have their light shifted towards longer wavelengths, resulting in a redshift.
Explanation:The Doppler effect is the change in the observed frequency of sound or light waves due to the relative motion between the source of the waves and the observer.
When a star moves towards or away from Earth, its light is shifted to a longer or shorter wavelength, respectively.
In this case, the red dwarf moving away from Earth at 39.1 km/s would have its light shifted towards longer wavelengths, resulting in a redshift.
Similarly, the blue dwarf moving away from Earth at 25.9 km/s would also experience a redshift.
On the other hand, the red giant moving towards Earth at 23.3 km/s would have its light shifted towards shorter wavelengths, resulting in a blueshift.
The yellow dwarf moving transversely at 15.1 km/s would not exhibit any shift in its light because its motion is perpendicular to the line of sight.
The red dwarf moving transversely at 14.1 km/s would also not show any shift in its light.
A closed loop conductor with radius 1.5 m is located in a changing magnetic field. If the max emf induced in the loop is 7.0 V, what is the max rate which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies
Answer:
0.99 T/s
Explanation:
Solution
From the example given, we recall that,
The emf induced in the loop is V = 7.0
Closed loop conductor with r = 1.5 m
∅B = BA
The emf = d∅B/ dt
which is
d/dt (BA)
so,
emf = A dB/ dt
emf =πr² dB/ dt
Now,
dB/dt = emf /πr²
= 7/π * (1.5)²
Therefore dB/dt = 0.99 T/s
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.
The electric field between the square plates of a parallel-plate capacitor has magnitude E. The potential across the plates is maintained at constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. (i)The magnitude of the electric field between the plates after being pulled is equal to A) 4E. B) 2E. C) E. D) E/2. E) E/4. (ii) If the plates were charged but not connected to the battery as they were pulled apart to double the separation, which of the above answers give the correct electric field ?
Answer:
D) E/2
Explanation:
i) To find the magnitude of electric field between the plates you use:
[tex]E=\frac{\Delta V}{d}[/tex]
where V is the voltage and d the separation between plates. If d is doubled you obtain:
[tex]E'=\frac{\Delta V}{2d}=\frac{1}{2}\frac{\Delta V}{d}=\frac{1}{2}E[/tex]
That is, the magnitude of the electric field is halved.
Then, the answer is D.
ii) The magnitude of E' is E/2 because the charge on the plates generates the field E.
Then, the answer is D again
A horizontal spring with a 10000 N/m spring constant is compressed 0.08 m, and a 12-kg block is placed against it. When the block is released, the block shoots forward along a horizontal surface that exerts 8 N friction force on the block. How far from the original position does the block travel before coming to a stop
Answer:
4.04m
Explanation:
First you calculate the velocity of the block when it leaves the spring. You calculate this velocity by taking into account that the potential energy of the spring equals the kinetic energy of the block, that is:
[tex]U=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(10000N/m)(0.08m)^2}{12kg}}=2.3\frac{m}{s}[/tex]
To find the distance in which the block stops you use the following expression (net work done by the friction force is equal to the difference in the kinetic energy of the block):
[tex]W_{T}=\Delta K\\\\F_f d=\frac{1}{2}m[v^2-v_o^2]\\\\d=\frac{mv^2}{2F_f}[/tex]
where Ff is the friction force. By replacing the values of the parameters you obtain:
[tex]d=\frac{(12kg)(2.3m/s)^2}{2(8N)}=3.96m[/tex]
hence, the distance to the original position is 3.96m+0.08m=4.04m
The 12 kg block travels a distance of 4 meters before coming to a stop. This is calculated using the principles of conservation of energy and work done by a force, in this case, the friction force.
Explanation:The question involves a situation related to Physics, specifically kinematics and mechanics. In order to calculate how far the block travels, we first need to understand how the forces in this situation work. When the block is released, the potential energy stored in compressed spring converts into kinetic energy which propels the block forward. However, due to the presence of friction, this kinetic energy gradually diminishes causing the block to eventually come to a stop.
The first step is to find the initial speed of the block just when it is released. We can use the principle of conservation of energy for this: the spring potential energy is equal to the initial kinetic energy. For potential energy in a spring, we use the formula PE = 0.5 * k * x^2, where k is the spring constant and x is the amount of compression (so PE = 0.5 * 10000 * 0.08^2 = 32 Joules). Kinetic energy is given by KE = 0.5 * m * v^2, where m is the mass and v is the velocity. Equating these (since initial PE = initial KE), we get v = sqrt((2 * PE) / m) = sqrt((2 * 32) / 12) = approx 4.08 m/s.
Now, knowing the initial speed, we can calculate how far the block travels before friction brings it to stop. The work done by the friction force will equal the initial kinetic energy of the block: Work = Friction force * distance = KE. Solving this equation for distance gives us distance = KE / Friction force = 32 Joules / 8 N = 4 meters.
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