Answer:
Explanation:
For image formation in objective lens
object distance u = 14 +1 = 15 mm
focal length f = 14 mm .
image distance v = ?
lens formula
[tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]
Putting the values
[tex]\frac{1}{v} +\frac{1}{15} =\frac{1}{14}[/tex]
v = 210 mm .
B )
magnification = v / u
= 210 / 15
= 14
size of image = 14 x 1.1 mm
= 15.4 mm
= 15 mm approx
C )
For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .
21 mm is the answer .
D )
overall magnification =
[tex]\frac{210}{15} \times \frac{D}{f_e}[/tex]
D = 25 cm , f_e = focal length of eye piece
= 14 x 250 / 21
= 166.67
= 170 ( in two significant figures )
(a) The distance of the image v=220mm
(b) SIze of the image 15 mm
(c) Distance of lens be from the image produced by the objective lens 21 mm
(d) overall magnification of the microscope 170
What is objective lens?The objective lens of a microscope is the one at the bottom near the sample. At its simplest, it is a very high-powered magnifying glass, with very short focal length. This is brought very close to the specimen being examined so that the light from the specimen comes to a focus inside the microscope tube
For image formation in objective lens
object distance u = 14 +1 = 15 mm
focal length f = 14 mm .
image distance v = ?
By using lens formula
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
Putting the values
[tex]\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{14}[/tex]
v = 210 mm .
B ) Magnification is the ratio of the size of the image to the size of the an object.
[tex]\rm magnification = \dfrac{v} { u}[/tex]
[tex]M= \dfrac{210} { 15}[/tex]
M= 14
size of image = 14 x 1.1 mm
= 15.4 mm
= 15 mm approx
C )
For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .
21 mm is the answer .
D )
overall magnification =
[tex]\dfrac{210}{15}\times \dfrac{D}{f_e}[/tex]
D = 25 cm , f_e = focal length of eye piece
[tex]= 14 \times \dfrac{ 250} { 21}[/tex]
= 166.67
= 170 ( in two significant figures )
Hence all the answers are:
(a) The distance of the image v=220mm
(b) SIze of the image 15 mm
(c) Distance of lens be from the image produced by the objective lens 21 mm
(d) overall magnification of the microscope 170
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With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor would the following change?
A. Kinetic energy when passing through the equilibrium position.
B. Speed when passing through the equilibrium position.
Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4
a person Travels along a straight road for half the distance with velocity V1 and the remaining half the distance with velocity V2 the average velocity is given by
Answer: (V1+V2)/2
Explanation: This is because basically with the question they are trying to say u(initial velocity) is V1 and v(final velocity) is V2 as the journey starts off with V1 and ends with V2 so therefore we know an equation where average velocity=(u+v)/2. So here it’s (V1+V2)/2
You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 38.5∘ from the ground with an initial speed of 27.5 m/s. At what height above its initial position does the softball hit the wall? Ignore any effects of air resistance.
The ball's horizontal position in the air is
[tex]x=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t[/tex]
It hits the wall when [tex]x=17.5\,\mathrm m[/tex], which happens at
[tex]17.5\,\mathrm m=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t\implies t\approx0.813\,\mathrm s[/tex]
Meanwhile, the ball's vertical position is
[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ t-\dfrac g2t^2[/tex]
where [tex]g[/tex] is the acceleration due to gravity, 9.80 m/s^2.
At the time the ball hits the wall, its vertical position (relative to its initial position) is
[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ(0.813\,\mathrm s)-\dfrac g2(0.813\,\mathrm s)^2\approx\boxed{10.7\,\mathrm m}[/tex]
The target variable is the speed of light v in the glass, which you can determine from the index of refraction n of the glass. Which equations will you use to find n and v?
Answer:
n= speed of light in vacuum/ speed of light in the other medium.
Explanation:
If light is moving from medium 1 into medium 2 where medium 1 is vacuum (approximated to mean air) and we are required to find the velocity of light; then we can confidently write;
n= speed of light in vacuum/ speed of light in the other medium.
Hence;
n= c/v
Where;
n= refractive index of the material
c= speed of light in vacuum
v = speed of light in another medium.
Note that the refractive index is the amount by which a transparent medium decreases the speed of light.
An electrostatic paint sprayer contains a metal sphere at an electric potential of 25.0 kV with respect to an electrically grounded object. Positively charged paint droplets are repelled away from the paint sprayer's positively charged sphere and towards the grounded object. What charge must a 0.168-mg drop of paint have so that it will arrive at the object with a speed of 18.8 m/s
Answer:
The charge is [tex]Q = 2.177 *10^{-9} \ C[/tex]
Explanation:
From the question we are told that
The electric potential is [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]
The mass of the drop is [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]
The speed is [tex]v = 18.8 \ m/s[/tex]
Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question is mathematically represented as
[tex]Q = \frac{m v^2 }{ 2 * V }[/tex]
Substituting values
[tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]
[tex]Q = 2.177 *10^{-9} \ C[/tex]
Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of static friction are μA = 0.16 and μB = 0.23. Determine the incline angle θ for which both blocks begin to slide. Also find the required stretch or compression in the connecting spring for this to occur. The spring has a stiffness of k = 2.1 lb/ft .
Answer:
[tex]\theta=10.20^{\circ}[/tex]
[tex]\Delta l=0.10 ft[/tex]
Explanation:
First of all, we analyze the system of blocks before starting to move.
[tex]\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0[/tex]
[tex]\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]
[tex]16sin(\theta)-2.91cos(\theta)=0[/tex]
[tex]tan(\theta)=0.18[/tex]
[tex]\theta=arctan(0.18)[/tex]
[tex]\theta=10.20^{\circ}[/tex]
Hence, the incline angle θ for which both blocks begin to slide is 10.20°.
Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.
[tex]P_{A}sin(\theta)-F_{fA}-F_{spring}=0[/tex]
Where:
[tex]F_{spring} = k\Delta l=2.1\Delta l[/tex]
[tex]P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0[/tex]
[tex]\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}[/tex]
[tex]\Delta l=0.10 ft[/tex]
Therefore, the required stretch or compression in the connecting spring is 0.10 ft.
I hope it helps you!
(a) The inclined angle for which both blocks begin to slide is 10.3⁰.
(b) The compression of the spring is 0.22 ft.
The given parameters;
mass of block A, = 11 lbmass of block B, = 5 lbcoefficient of static friction for A, = 0.16coefficient of static friction for B, = 0.23 spring constant, k = 2.1 lb/ftThe normal force on block A and B:
[tex]F_n_A = m_Agcos \ \theta\\\\F_n_B = m_Bgcos \ \theta[/tex]
The frictional force on block A and B:
[tex]F_f_A = \mu_s_AF_n_A \\\\F_f_B = \mu_s_BF_n_A[/tex]
The net force on the blocks when they starts sliding;
[tex](m_Ag sin \theta+ m_Bgsin\theta) - (F_f_A + F_f_B) = 0\\\\m_Ag sin \theta+ m_Bgsin\theta = F_f_A + F_f_B\\\\m_Ag sin \theta+ m_Bgsin\theta = \mu_Am_Agcos\theta \ + \ \mu_Bm_Bgcos\theta\\\\gsin\theta(m_A + m_B) = gcos\theta (\mu_Am_A + \mu_Bm_B)\\\\\frac{sin\theta}{cos \theta} = \frac{\mu_Am_A\ + \ \mu_Bm_B}{m_A\ + \ m_B} \\\\tan\theta = \frac{(0.16\times 11) \ + \ (0.23 \times 5)}{11 + 5} \\\\tan\theta = 0.1819\\\\\theta = tan^{-1}(0.1819)\\\\\theta = 10.3 \ ^0[/tex]
The change in the energy of the blocks is the work done in compressing the spring;
[tex]\Delta E = W\\\\F_A (sin \theta )d- \mu F_n d= \frac{1}{2} kd^2\\\\F_A sin\theta \ - \ \mu F_A cos\theta = \frac{1}{2} kd\\\\d = \frac{2F_A(sin\theta - \mu cos \theta) }{k} \\\\d = \frac{2\times 11(sin \ 10.3\ - \ 0.16\times cos \ 10.3) }{2.1} \\\\d = 0.22 \ ft[/tex]
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You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.11 mm and place your screen 8.63 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.71 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength lambda expressed in nanometers?
Answer:
λ = 605.80 nm
Explanation:
These double-slit experiments the equation for constructive interference is
d sin θ = m λ
where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.
In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm
Let's use trigonometry to find the angle
tan θ = y / L
as the angles are very small
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
we substitute in the first equation
d y / L = m λ
λ = d y / m L
let's calculate
λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)
λ = 6.05805 10⁻⁷ m
let's reduce to nm
λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)
λ = 605.80 nm
Light of wavelength 520 nm is used to illuminate normally two glass plates 21.1 cm in length that touch at one end and are separated at the other by a wire of radius 0.028 mm. How many bright fringes appear along the total length of the plates.
Answer:
The number is [tex]Z = 216 \ fringes[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 520 \ nm = 520 *10^{-9} \ m[/tex]
The length of the glass plates is [tex]y = 21.1cm = 0.211 \ m[/tex]
The distance between the plates (radius of wire ) = [tex]d = 0.028 mm = 2.8 *10^{-5} \ m[/tex]
Generally the condition for constructive interference in a film is mathematically represented as
[tex]2 * t = [m + \frac{1}{2} ]\lambda[/tex]
Where t is the thickness of the separation between the glass i.e
t = 0 at the edge where the glasses are touching each other and
t = 2d at the edge where the glasses are separated by the wire
m is the order of the fringe it starts from 0, 1 , 2 ...
So
[tex]2 * 2 * d = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]
=> [tex]2 * 2 * (2.8 *10^{-5}) = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]
=>
[tex]m = 215[/tex]
given that we start counting m from zero
it means that the number of bright fringes that would appear is
[tex]Z = m + 1[/tex]
=> [tex]Z = 215 +1[/tex]
=> [tex]Z = 216 \ fringes[/tex]
Consider a hydraulic lift that uses an input piston with an area of 0.5m2. An input force of 15N is exerted on this piston. If the output piston has an area of 3.5m? What is the output force?
Answer:
The output force of the piston is 105 N.
Explanation:
Given;
the area of the input piston, A₁ = 0.5 m²
the input force of the piston, F₁ = 15 N
the area of the output piston, A₀ = 3.5 m²
the output force of the piston, F₀ = ?
The pressure of the hydraulic lift is given by;
[tex]P = \frac{F}{A}[/tex]
where;
P is the hydraulic pressure
F is the piston force
A is the area of the piston
[tex]P = \frac{F}{A} \\\\\frac{F_o}{A_o} = \frac{F_i}{A_i} \\\\F_o = \frac{F_iA_o}{A_i} \\\\F_o = \frac{15*3.5}{0.5} \\\\F_o = 105 \ N[/tex]
Therefore, the output force of the piston is 105 N.
7. A sound wave begins traveling through a thin metal rod at one end with a speed that is 15 times the speed of sound in air. If an observer at the other end of the rod hears the sound twice, one from the sound traveling through the rod and one from the sound traveling through the air, with a time delay of 0.12 s, how long is the rod? The speed of sound in air is 343 m/s.
Answer:
L = 44,096 m
Explanation:
The speed of the sound wave is constant therefore we can use the relations of uniform kinematics
v = x / t
the speed of the wave in the bar is
v = 15 v or
v = 15 343
v = 5145 m / s
The sound at the bar goes the distance
L = v t
Sound in the air travels the same distance
L = v_air (t + 0.12)
as the two recognize the same dissonance,
v t = v_air (t +0.12)
t (v- v_air) = 0.12 v_air
t = 0.12 v_air / (v -v_air)
l
et's calculate
t = 0.12 343 / (5145 - 343)
t = 8.57 10-3 s
The length of the bar is
L = 5145 8.57 10-3
L = 44,096 m
Two football teams, the Raiders and the 49ers are engaged in a tug-of-war. The Raiders are pulling with a force of 5000N. Which of the following is an accurate statement?
A. The tension in the rope depends on whether or not the teams are in equilibrium.
B. The 49ers are pulling with a force of more than 5000N because of course they’d be winning.
C. The 49ers are pulling with a force of 5000N.
D. The tension in the rope is 10,000N.
E. None of these statements are true.
Answer:
E. None of these statements are true.
Explanation:
We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.
for A, the tension on the rope does not depend on if both teams pull are in equilibrium.
for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.
for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.
for D, just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.
we go with option E.
A 1000 kg car experiences a net force of 9500 N while slowing down from 30 m/s to 16 m/s. How far does it travel while slowing down?
Answer:
33.89 m
Explanation:
We must first obtain the acceleration of the car from;
F=ma
Where
F= force= 9500 N
m= mass of the car= 1000kg
a= acceleration
a= F/m= 9500/1000
a= 9.5 m/s^2
From;
V^2=u^2 + 2as
Where;
V= final velocity
u= initial velocity
s= distance covered
a= acceleration
s= v^2 -u^2/2a
s= (30)^2 -(16)^2/2×9.5
s= 900 - 256/19
s= 644/19
s= 33.89 m
The distance is 33.89 m
The first step is to calculate the acceleration
F= ma
force= 9500N
mass= 1000 kg
9500= 1000 × a
a= 9500/1000
= 9.5 m/s
v²= u² + 2as
30²= 16² + 2(9.5)(s)
900= 256 + 19s
900-256= 19s
644= 19s
s= 644/19
s= 33.89 m
Hence the distance traveled by the car is 33.89 m
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If you wanted to make your own lenses for a telescope, what features of a lens do you think would affect the images that you can see
Answer:
Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
Explanation:
The length of the telescope is
L = f_ocular + f_objetive
the magnification of the telescope is
m = - f_objective / f_ocular
These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
You want the current amplitude through a 0.450 mH inductor (part of the circuitry for a radio receiver) to be 1.50 mA when a sinusoidal voltage with an amplitude of 13.0 V is applied across the inductor. What frequency is required?
Answer:
3.067MHzExplanation:
The formula for calculating the voltage across an inductor is expressed as
[tex]V_l = IX_l\\\\Since\ X_l = 2\pi fL\\V_l = I(2\pi fL)[/tex]
Given parameters
current amplitude I = 1.50mA = 1.5*10⁻³A
inductance L = 0.450mH = 0.450*10⁻³H
Voltage across the inductor [tex]V_l[/tex] = 13.0V
Required
frequency f
Substituting the given parametres into the formula, we have;
[tex]V_l = I(2\pi fL)\\\\13 = 1.50*10^{-3}(2*3.14*f*0.450*10^{-3})\\\\13 = 4.239*10^{-6}f\\\\f = \frac{13}{4.239*10^{-6}} \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz[/tex]
Hence, the frequency required is 3.067MHz
A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exerts a resistive force with magnitude proportional to the square of the speed with k
Answer:
The velocity is 40 ft/sec.
Explanation:
Given that,
Force = 3200 lb
Angle = 30°
Speed = 64 ft/s
The resistive force with magnitude proportional to the square of the speed,
[tex]F_{r}=kv^2[/tex]
Where, k = 1 lb s²/ft²
We need to calculate the velocity
Using balance equation
[tex]F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}[/tex]
Put the value into the formula
[tex]3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}[/tex]
Put the value of k
[tex]3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}[/tex]
[tex]1600-v^2=m\dfrac{d^2v}{dt^2}[/tex]
At terminal velocity [tex]\dfrac{d^2v}{dt^2}=0[/tex]
So, [tex]1600-v^2=0[/tex]
[tex]v=\sqrt{1600}[/tex]
[tex]v=40\ ft/sec[/tex]
Hence, The velocity is 40 ft/sec.
If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First what are the mass units?
Answer:
The mas unit is the the 'Kilogram' written as 'kg'
Volume is 10 L
Explanation:
The complete question is
If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First, what are the mass units?
Second, what is the volume
mass units is the 'Kilogram', written as 'kg'
density = mass/volume = 100 kg/L
the mass = 1000 kg
volume = mass/density = 1000/100 = 10 L
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled
Answer:
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
Explanation:
The diffraction of a slit is given by the expressions
a sin θ = m λ
where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.
sin θ = m λ / a
If this equation
a ’= 2 a
we substitute
2 a sin θ'= m λ
sin θ'= (m λ / a) 1/2
sin θ ’= sin θ / 2
We can use trigonometry to find the width
tan θ = y / L
as the angle is small
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we substitute
y ’/ L = y/L 1/2
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
3. What color of laser light shines through a diffraction grating with a line density of 500 lines/mm if the third maxima from the central maxima (m=3) is at an angle of 45°?
Answer:
Wavelength is 471 nm
Explanation:
Given that,
Lines per unit length of diffraction grating is 500 lines/mm.
The third maxima from the central maxima (m=3) is at an angle of 45°
We need to find the color of laser light shines through a diffraction grating.
The condition for maxima is :
[tex]d\sin\theta=m\lambda[/tex]
d = 1/N, N = number of lines per mm
[tex]\lambda=\dfrac{1}{Nm}\sin\theta\\\\\lambda=\dfrac{10^{-3}}{500\times 3}\sin(45)\\\\\lambda=4.31\times 10^{-7}\\\\\text{or}\\\\\lambda=471\ nm[/tex]
Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J = cr2 = 9.00 ✕ 106 A/m4 r2. What is the current (in A) through the inner section of the wire from the center to r = 0.5R?
Answer:
The current is [tex]I = 8.9 *10^{-5} \ A[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]
The current density is [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]
The distance we are considering is [tex]r = 0.5 R = 0.001585[/tex]
Generally current density is mathematically represented as
[tex]J = \frac{I}{A }[/tex]
Where A is the cross-sectional area represented as
[tex]A = \pi r^2[/tex]
=> [tex]J = \frac{I}{\pi r^2 }[/tex]
=> [tex]I = J * (\pi r^2 )[/tex]
Now the change in current per unit length is mathematically evaluated as
[tex]dI = 2 J * \pi r dr[/tex]
Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows
[tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]
[tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]
substituting values
[tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]
[tex]I = 8.9 *10^{-5} \ A[/tex]
Electrons are accelerated through a voltage difference of 270 kV inside a high voltage accelerator tube. What is the final kinetic energy of the electrons?
Each electron winds up with kinetic energy of
(270 keV)
plus
(whatever KE it had when it started accelerating).
The advantage of a hydraulic lever is A : it transforms a small force acting over a large distance into a large force acting over a small distance. B : it transforms a small force acting over a small distance into a large force acting over a large distance. C : it allows you to exert a larger force with less work. D : it transforms a large force acting over a large distance into a small force acting over a small distance. E : it transforms a large force acting over a small distance into a small force acting over a large distance.
Answer:
A) it transforms a small force acting over a large distance into a large force acting over a small distance.
Explanation:
The hydraulic lever works based on Pascal's law of transmission of pressure through a fluid. In the hydraulic lever, the pressure transmitted is the same.
Pressure transmitted P = F/A
where F is the force applied
and A is the area over which the force is applied.
This pressure can be manipulated on the input end as a small force applied over a small area, and then be transmitted to the output end as a large force over a large area.
F/A = f/a
where the left side of the equation is for the output, and the right side is for the input.
The volume of the displaced fluid will be the same on both ends of the hydraulic lever. Since we know that
volume V = (area A) x (distance d)
this means that the the piston on the input smaller area of the hydraulic lever will travel a greater distance, while the piston on the larger output area of the lever will travel a small distance.
From all these, we can see that the advantage of a hydraulic lever is that it transforms a small force acting over a large distance into a large force acting over a small distance.
An electron experiences a force of magnitude F when it is 5 cm from a very long, charged wire with linear charge density, lambda. If the charge density is doubled, at what distance from the wire will a proton experience a force of the same magnitude F?
Answer:
The distance of the proton is [tex]r_p =10 \ cm[/tex]
Explanation:
Generally the force experience by the electron is mathematically represented as
[tex]F_e = \frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e}[/tex]
Where [tex]\lambda _e[/tex] is the charge density of the charge wire before it is doubled
Also the force experience by the proton is mathematically represented as
[tex]F_p = \frac{q * \lambda_p }{ 2 \pi * \epsilon_o * r_p}[/tex]
Given that the charge density is doubled i.e [tex]\lambda_p = 2 \lambda_e[/tex] and that the the force are equal then
[tex]\frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e} = \frac{q * 2 \lambda_e }{ 2 \pi * \epsilon_o * r_p}[/tex]
[tex]\frac{ \lambda_e }{ r_e} = \frac{ 2 \lambda_e }{ r_p}[/tex]
[tex]r_p * \lambda_e =2 \lambda_e * r_e[/tex]
[tex]r_p =2 r_e[/tex]
Now given from the question that [tex]r_e[/tex] the distance of the electron from the charged wire is 5 cm
Then
[tex]r_p =2 (5)[/tex]
[tex]r_p =10 \ cm[/tex]
The difference between a DC and an AC generator is that
a. the DC generator has one unbroken slip ring.
b. the AC generator has one unbroken slip ring
c. the DC generator has one slip ring splitin two halves.
d. the AC generator has one slip ring split in two halves.
e The DC generator has twounbroken sip rings
Answer:
The AC generator has one unbroken slip ring
Explanation:
In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.
An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.
We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.
It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.
The accommodation limits for a nearsighted person's eyes are 20.0 cm and 82.0 cm. When he wears his glasses, he can see faraway objects clearly. At what minimum distance is he able to see objects clearly
Answer;
26.45cm
See attached file for explanation
Suppose you observed the equation for a traveling wave to be y(x, t) = A cos(kx − ????t), where its amplitude of oscillations was 0.15 m, its wavelength was two meters, and the period was 2/15 s. If a point on the wave at a specific time has a displacement of 0.12 m, what is the transverse speed of that point?
Answer:
15m/sExplanation:
The equation for a traveling wave as expressed as y(x, t) = A cos(kx − [tex]\omega[/tex]t) where An is the amplitude f oscillation, [tex]\omega[/tex] is the angular velocity and x is the horizontal displacement and y is the vertical displacement.
From the formula; [tex]k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f[/tex] where;
[tex]\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency[/tex]
Before we can get the transverse speed, we need to get the frequency and the wavelength.
frequency = 1/period
Given period = 2/15 s
Frequency = [tex]\frac{1}{(2/15)}[/tex]
frequency = 1 * 15/2
frequency f = 15/2 Hertz
Given wavelength [tex]\lambda[/tex] = 2m
Transverse speed [tex]v = f \lambda[/tex]
[tex]v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s[/tex]
Hence, the transverse speed at that point is 15m/s
In a physics lab, light with a wavelength of 490 nm travels in air from a laser to a photocell in a time of 17.5 ns . When a slab of glass with a thickness of 0.800 m is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light a time of 21.5 ns to travel from the laser to the photocell.What is the wavelength of the light in the glass? Use 3.00×108 m/s for the speed of light in a vacuum. Express your answer using two significant figures.
Answer:
196 nm
Explanation:
Given that
Value of wavelength, = 490 nm
Time spent in air, t(a) = 17.5 ns
Thickness of glass, th = 0.8 m
Time spent in glass, t(g) = 21.5 ns
Speed of light in a vacuum, c = 3*10^8 m/s
To start with, we find the difference between the two time spent
Time spent on glass - Time spent in air
21.5 - 17.5 = 4 ns
0.8/(c/n) - 0.8/c = 4 ns
Note, light travels with c/n speed in media that has index of refraction
(n - 1) * 0.8/c = 4 ns
n - 1 = (4 ns * c) / 0.8
n - 1 = (4*10^-9 * 3*10^8) / 0.8
n - 1 = 1.2/0.8
n - 1 = 1.5
n = 1.5 + 1
n = 2.5
As a result, the wavelength of light in a medium with index of refraction would then be
490 / 2.5 = 196 nm
Therefore, our answer is 196 nm
A semi-circular loop consisting of one turn of wire is place in the x-y plane. A constant magnetic field B=1.7T points along the negative z-axis(into the page), and a current I=0.7A flows counterclockwisefrom the positive z-axis. The net magnetic force on the circular section of the loop points in what direction? What is the net magnetice force on the circular section of the loop?
Answer:
The direction of net magnetic force on the circular section of the loop is in the positive y-axis
The net magnetic force on the circular section of the loop is 3.74 N
Explanation:
The magnetic field strength [tex]B[/tex] = 1.7 T
the current [tex]I[/tex] = 0.7 A
The diameter of the loop = 2 m
the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2
==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m
The force on the semi-circular is given as
F = [tex]BIl[/tex] sin ∅
but the loop is perpendicular to the field, therefore
sin ∅ = sin 90° = 1
F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N
The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".
According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis
An L-R-C series circuit has L = 0.450 H, C=2.50×10^−5F, and resistance R.
Required:
a. What is the angular frequency of the circuit when R = 0?
b. What value must R have to give a decrease in angular frequency of 10.0 % compared to the value calculated in Part a.
Answer:
298rad/s and 116.96 ohms
Explanation:
Given an L-R-C series circuit where
L = 0.450 H,
C=2.50×10^−5F, and resistance R= 0
In this situation we have a simple LC circuit with angular frequency
Wo = 1√LC
= 1/√(0.450)(2.50×10^-5)
= 1/√0.00001125
= 1/0.003354
= 298rad/s
B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.
Wi/W° = (100-10)/100
Wi/W° = 90/100
Wi/W° = 0.90 ............... 1
Angular frequency of oscillation
The complete aspect of the solution is attached, please check.
a. The angular frequency of the circuit when R = 0 Ohms is 294.12 rad/s.
b. The value R must have to give a decrease in angular frequency of 10.0 % compared to the initial value is equal to 116.96 Ohms.
Given the following data:
Inductance, L = 0.450 HenryCapacitance, C = [tex]2.50\times 10^{-5}[/tex] Faradsa. To determine the angular frequency of the circuit when R = 0 Ohms:
Mathematically, the angular frequency of a LC circuit is given by the formula:
[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega =\frac{1}{\sqrt{0.450 \times 2.50\times 10^{-5}}} \\\\\omega =\frac{1}{\sqrt{1.125 \times 10^{-5}}} \\\\\omega = \frac{1}{0.0034} \\\\\omega = 294.12\;rad/s[/tex]
b. To find the value R must have to give a decrease in angular frequency of 10.0 % compared to the value calculated above:
The mathematical expression is given as follows:
[tex]\frac{\omega_f}{\omega_i} = \frac{100-10}{100} \\\\\frac{\omega_f}{\omega_i} =\frac{90}{100} \\\\\frac{\omega_f}{\omega_i} =0.9[/tex]
[tex](\frac{\omega_f}{\omega_i})^2 = 1 - \frac{R^2C}{4L} \\\\0.90^2=1 - \frac{R^2C}{4L}\\\\R=\sqrt{\frac{4L(1-0.81)}{C}} \\\\R=\sqrt{\frac{4\times 0.450 \times (0.19)}{2.50\times 10^{-5}}}\\\\R = \sqrt{\frac{0.342}{2.50\times 10^{-5}} }\\\\R =\sqrt{13680}[/tex]
R = 116.96 Ohms.
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3. Which of the following accurately describes circuits?
O A. In a parallel circuit, the same amount of current flows through each part of the circuit
O B. In a series circuit, the amount of current passing through each part of the circuit may vary
O C. In a series circuit, the current can flow through only one path from start to finish
O D. In a parallel circuit, there's only one path for the current to travel.
Answer:
Option (c)
Explanation:
In a Series circuit, as the components are connected end-to-end ,the current can flow through only one path from start to finish.
(C.) is the only correct statement in the list of choices.
In a series circuit, the current can flow through only one path from start to finish.
A car moving at 36 m/s passes a stationary police car whose siren has a frequency of 500 Hz. What is the change in the frequency (in Hz) heard by an observer in the moving car as he passes the police car? (The speed of sound in air is 343 m/s.)
Answer:
Change in the frequency (in Hz) = 104.96 Hz
Explanation:
Given:
Speed of sound in air (v) = 343 m/s
Speed of car (v1) 36 m/s
Frequency(f) = 500 Hz
Find:
Change in the frequency (in Hz)
Computation:
Frequency hear by the observer(before)(f1) = [f(v+v1)] / v
Frequency hear by the observer(f1) = [500(343+36)] / 343
Frequency hear by the observer(f1) = 552.48 Hz
Frequency hear by the observer(after)(f2) = [f(v-v1)] / v
Frequency hear by the observer(f2) = [500(343-36)] / 343
Frequency hear by the observer(f2) = 447.52 Hz
Change in the frequency (in Hz) = f1 - f2
Change in the frequency (in Hz) = 552.48 Hz - 447.52 Hz
Change in the frequency (in Hz) = 104.96 Hz
