Answer:
a) t = 9.16*10^{-18} s
b) y = 0.402 mm
Explanation:
(a) To find the time that the particle takes to pass trough the region between parallel plates, you take into account that the horizontal component of the velocity is constant in all trajectory of the particle. Then, you use the following formula:
[tex]t=\frac{x}{v_x}\\\\[/tex]
x: length of the sides of the plates = 0.22m
v_x: horizontal component of the velocity = 2.4*10^6 m/s
[tex]t=\frac{0.22m}{2.4*10^6m/s}\\\\t=9.16*10^{-8}s=91.6ns[/tex]
(b) To find the vertical displacement of the particle you first calculate the acceleration of the particle generated by the electric force:
[tex]F_e=ma\\\\qE=ma\\\\a=\frac{qE}{m}\\\\a=\frac{(1.6*10^{-19}C)(33*10^3N/C)}{5.5*10^{-26}C}=9.6*10^{10}\frac{m}{s^2}[/tex]
where you have used that the charge is 1.6*10^-19 C (charge of an electron).
With the values of the acceleration and time you use the following kinematic equation to calculate the vertical displacement:
[tex]y=v_{oy}t+\frac{1}{2}at^2\\\\v_{oy}=0m/s\\\\y=\frac{1}{2}(9.6*10^{10}\frac{m}{s^2})(9.16*10^{-8}s)^2=4.02*10^{-4}m=0.402mm[/tex]
Following are the solution to the given points:
When calculating the time required the particle will pass through the zone between plates, remember that such a horizontal component of the particle's velocity is fixed throughout the particle's route. The following formula is then used:
[tex]\to t=\frac{x}{v_x}[/tex]
[tex]\to x:[/tex] plate length = 0.22m
[tex]\to v_x :[/tex]velocity's horizontal component[tex]= 2.4\times 10^{6} \ \frac{m}{s}\\\\[/tex]
[tex]\to t=\frac{0.22\ m}{ 2.4 \times 10^6 \frac{m}{s}}= 9.16 \times 10^{-8} \ s = 91.6 \ ns[/tex]
To determine the particle's vertical displacement, firstly compute the particle's acceleration due to electric force:[tex]\to F_e=ma\\\\\to qE=ma\\\\\to a=\frac{qE}{m}\\\\[/tex]
[tex]=\frac{1.6 \times 10^{-19}\ C \times 33 \times 10^3 \frac{N}{C}}{5.5 \times 10^{-26} C} \\\\ =\frac{1.6 \times 10^{-16} \times 33 \times 10^{26}}{5.5} \\\\ =\frac{1.6 \times 33 \times 10^{10}}{5.5} \\\\ =\frac{16 \times 33 \times 10^{10}}{55} \\\\ = 9.6 \times 10^{10} \ \frac{m}{s^2}\\\\[/tex]
when you have used that charge [tex]1.6\times 10^{-19}\ C[/tex]electronic charge).The following kinematic formula is used to determine vertical displacement using the values of acceleration and time:[tex]\to y=v_{oy} t+\frac{1}{2} at^2\\\\\to v_{oy}= 0 \ \frac{m}{s}\\\\\to y=\frac{1}{2} (9.6 \times 10^{10} \frac{m}{s^2}) (9.6 \times 10^{-8}\ s)^2[/tex]
[tex]= 4.02 \times 10^{-2} \ m \\\\ = 0.402 \ mm\\\\[/tex]
Find out more information about the displacement:
brainly.com/question/11705032
Yellow light with wavelength 600 nm is travelling to the left (in the negative x direction) in vacuum. The light is polarized along the z direction. (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave. (c) If the wave were to represent blue light instead of yellow light, how would your pictures in parts a and b change? If there is no change, say so explicitly.
Answer: (a) and (b) => check attached file.
(c). Picture (a) and (b) will both remain the same.
Explanation:
IMPORTANT: The solution to the question (a) and (b) that is (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave is there in the ATTACHED FILE/PICTURE.
It is also worthy of note to know that in anything Electromagnetic wave, the magnetic field, the Electric Field and their direction of propagation are perpendicular to each other.
Therefore, knowing the fact above we can say that in yellow light, the magnetic field is in the y-direction and the Electric Field is in the z-direction.
Hence, the solution to option C is given below;
(C).If the wave were to represent blue light instead of yellow light, picture (a) will remain the same because both light are Electromagnetic wave, although the wavelength will have to change. Picture (b) will also remain the same because they are both Electromagnetic waves and possess similar properties.
A"boat"is"moving"to"the"right"at"5"m/s"with"respect"to"the"water."A"wave"moving"to"the"left,"opposite"the"motion"of"the"boat."The"waves"have"2.0"m"between"the"top"of"the"crests"and"the"bottom"of"the"troughs."The"period"of"the"wave"is"8.3"s"and"their"wavelength"is"110"m."At"one"instant"the"boat"sits"on"a"crest"of"the"wave,"20"seconds"later,"what"is"the"vertical"displacement"of"the"boat
Answer:
0.99m
Explanation:
Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:
[tex]v=\lambda f=\lambda\frac{1}{T}=(110m)\frac{1}{8.3s}=13.25\frac{m}{s}[/tex]
the relative velocity is:
[tex]v'=13.25m/s-5m/s=8.25\frac{m}{s}[/tex]
This velocity is used to know which is the distance traveled by the boat after 20 seconds:
[tex]x'=v't=(8.25m/s)(20s)=165m[/tex]
Next, you use the general for of a wave:
[tex]f(x,t)=Acos(kx-\omega t)=Acos(\frac{2\pi}{\lambda}x-\omega t)[/tex]
you take the amplitude as 2.0/2 = 1.0m.
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{8.3s}=0.75\frac{rad}{s}[/tex]
by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:
[tex]f(165,20)=1.0m\ cos(\frac{2\pi}{110m}(165)-(0.75\frac{rad}{s})(20s))\\\\f(165,20)=0.99m[/tex]
How are the elements in the same row similar
Answer:
All elements in a row have the same number of electron shells. Each next element in a period has one more proton and is less metallic than its predecessor. Arranged this way, groups of elements in the same column have similar chemical and physical properties, reflecting the periodic law.
A turntable has a moment of inertia of 3.00 x 10-2 kgm2 and spins freely on a frictionless bearing at 25.0 rev/min. A 0.300 kg ball of putty is dropped vertically on the turntable and sticks at a point 0.10m from the center. The total moment of inertia of the system increases, and the turntable slows down. But by what factor does the angular momentum of the system change after the putty is dropped onto the turntable
Answer:
There will be no change in the angular momentum of the system.
Explanation:
Total angular momentum of the system will remain unchanged . We can apply law of conservation of momentum because no external torque is acting on the system . There is increase in the momentum of inertia due to dropping of ball of putty . In order to conserve angular momentum , the system decreases its angular velocity . Hence the final angular momentum remains unchanged .
Help ill give you brainliest !!!
Answer:
1. B
2. A
3. C
4. B
5. A
6. Muscular strength is different than muscular endurance because of the fact that muscular strength is the amount of force that can be exerted in one instance. Muscular endurance is how long that you can exert that force without being completely exhausted.
7. Some benefits to strength training is the increase in muscular endurance. There is also the benefit of better muscular strength.
Explanation:
a 1200 kg trailer is hitched to a 1400 kg car. the car and trailer are traveling at 72 km.h when the driver applies the brakes on both the car and the trailer. knowing that the braking forces exerted on the car and the trailer are 5000 N and 4000 N respectively, determine (a) the distance traveled by the car and trailer before they come to a stop and (b) the horizontal component of the force exerted by the trailer hitch
Answer:
a) 8.67m
b) 1000N
Explanation:
(a) To find the distance you use the second Newton Law for both car and trailer, in order to calculate the dis-acceleration of the system:
[tex]F=ma\\\\a_=\frac{F}{m}=\frac{5000N+4000N}{1400kg+1200kg}=3.46\frac{m}{s^2}[/tex]
once you have this value, you use the the following kinematic equation to calculate the distance traveled by both car and trailer:
[tex]v^2=v_o^2-2ax\\\\x=\frac{-v^2+v_o^2}{2a}[/tex]
v: final velocity=0
vo: initial velocity = 72km/h = 60 m/s
by replacing the values of these parameters you obtain for x:
[tex]x=\frac{-0m/s+60m/s}{2(3.46m/s^2)}\\\\x=8.67m[/tex]
(b) The horizontal component of the force exerted by the trailer hitch is given by:
[tex]F_T=5000N-4000N=1000N[/tex]
When we apply the energy conversation principle to a cylinder rolling down an incline without sliding, we exclude the work done by friction because: A. there is no friction present B. the angular velocity of the center of mass about the point of contact is zero C. the coefficient of kinetic friction is zero D. the linear velocity of the point of contact (relative to the inclined surface) is zero E. the coefficient of static and kinetic friction are equal
Answer:
D. the linear velocity of the point of contact (relative to the inclined surface) is zero
Explanation:
The force of friction emerges only when there is relative velocity between two objects . In case of perfect rolling , there is no sliding so relative velocity between the surface and the point of contact is zero . In other words the velocity of point of contact becomes zero , even though , the whole body is in linear motion . It happens due point of contact having two velocities which are equal and opposite . One of the velocity is in forward direction and the other velocity which is due to rotation is in backward direction . So net velocity of point of contact becomes zero . Due to absence of sliding , displacement due to friction becomes zero . Hence work done by friction becomes zero.
A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500.0 N/m (assume that the spring is massless). The block is held in position such that the spring is compressed 4.00 cm shorter than its undisturbed length. The block is suddenly released and allowed to slide away on the frictionless surface. Find the speed the block will be traveling when it leaves the spring.
Answer:
6 m/s
Explanation:
Given that :
mass of the block m = 200.0 g = 200 × 10⁻³ kg
the horizontal spring constant k = 4500.0 N/m
position of the block (distance x) = 4.00 cm = 0.04 m
To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.
i.e [tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2[/tex]
[tex]kx^2 = mv^2[/tex]
[tex]4500* 0.04^2 = 200*10^{-3} *v^2[/tex]
[tex]7.2 =200*10^{-3}*v^{2}[/tex]
[tex]v^{2} =\frac{7.2}{200*10^{-3}}[/tex]
[tex]v =\sqrt{\frac{7.2}{200*10^{-3}}}[/tex]
v = 6 m/s
Hence,the speed the block will be traveling when it leaves the spring is 6 m/s
A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 3.7 mm will experience only elastic deformation when a tensile load of 1890 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.
Answer:
L= 276.4 mm
Explanation:
Given that
E= 180 GPa
d= 3.7 mm
F= 1890 N
ΔL= 0.45 mm
We know that ,elongation due to load F in a cylindrical bar is given as follows
[tex]\Delta L =\dfrac{FL}{AE}[/tex]
[tex]L=\dfrac{\Delta L\times AE}{F}[/tex]
Now by putting the values in the above equation we get
[tex]L=\dfrac{0.45\times 10^{-3}\times \dfrac{\pi}{4}\times (3.7\times 10^{-3})^2\times 108\times 10^9}{1890}\ m[/tex]
L=0.2764 m
L= 276.4 mm
Therefore the length of the specimen will be 276.4 mm
5.00 kg of liquid water is heated to 100.0 °C in a closed system. At this temperature, the density of liquid water is 958 kg/m3 . The pressure is maintained at atmospheric pressure of 1.01 x 105 Pa. A moveable piston of negligible weight rests on the surface of the water. The water is then converted to steam by adding an additional amount of heat to the system. When all of the water is converted, the final volume of the steam is 8.50 m3 . The latent heat of vaporization of water is 2.26 x 106 J/kg. Calculate how much work is done and the change in the internal energy during this isothermal process.
Answer:
1.04 x 107 J.
Explanation:
We can use the following method to do the calculation
Total energy given to water to convert intosteam
dQ = m* l
dQ = 5.00* 2.26 * 106
= 1.13* 107 J
Work done at constantpressure dW = P* dV
Initialvolume V1 = 5.00kg / 958
= 5.22* 10-3 m3
Finalvolume = 8.50 m3
=> dW = 1.01* 105 * ( 8.50 - 5.22 * 10-3)
= 8.58* 105 J
First law of thermodynamicsis dQ = ΔU + dW
Change in internalenergy ΔU = 1.13* 107 - 8.58 *105
= 1.04 x 107 J as our answer
A piston with stops containing water goes through an expansion process through the addition of heat. State 1 the pressure is 200 kPa and the volume is 2 m3. After half of the heat has been delivered the piston hits the stops corresponding to a volume of 5 m3. After all the heat has been delivered, state 2, the pressure is 1000 kPa with the piston resting on the stops. What is the work?
Answer:
The work will be "600 kJ/kg".
Explanation:
(1-a) ⇒ Constant Pressure
(a-2) ⇒ Constant Volume
The given values are:
In state 1,
Pressure, P₁ = 200 kPa
Volume, V₁ = 2m³
In state 2,
Pressure, P₂ = 1000 kPa
Volume, V₁ = 5m³
Now,
In process (1-a), work will be:
⇒ W₁₋ₐ = P₁(Vₐ - V₁)
On putting the values, we get
⇒ W₁₋ₐ = 200(5-2)
⇒ = 200(3)
⇒ = 600 kJ/kg
In process (a-2), work will be:
⇒ Wₐ₋₂ = 0
∴ (The change in the volume will be zero.)
So,
⇒ Total work = (W₁₋ₐ) + (Wₐ₋₂)
⇒ = 600 + 0
⇒ = 600 kJ/kg
Three packing crates of masses, M1 = 6 kg, M2 = 2 kg
and M3 = 8 kg are connected by a light string of
negligible mass that passes over the pulley as shown.
Masses M1 and M3 lies on a 30o
incline plane which
slides down the plane. The coefficient of kinetic friction
on the incline plane is 0.28.
Determine the acceleration of the system.
Answer:
a = 2.5 m / s²
Explanation:
This is an exercise of Newton's second law, in this case we fix a coordinate system with the x axis parallel to the plane with positive direction
Let's write the second law for bodies in the inclined plane
W₁ₓ + W₃ₓ - fr = (m₁ + m₃) a
N₁ - [tex]W_{1y}[/tex] + N₃- W_{3y} = 0
N₁ + N₃ = W_{1y} + W_{3y}
let's use trigonometry to find the weight components
sin 30 = Wₓ/ W
Wₓ = W sin 30
cos 30 = W_{y} / W
W_{y} = W cos 30
we substitute
N₁+ N₃ = W₁ cos 30 + W₃ cos 30
W₁ₓ + W₃ₓ - μ (m₁ + m₃) g cos30 = (m₁ + m₃) a
a = (m₁g sin 30 + m₃g sin 30 - μ (m₁ + m₃) g cos 30) / (m₁ + m₃)
a = g sin 30 - μ g cos30
let's calculate
a = 9.8 sin 30 - 0.28 9.8 cos 30
a = 4.9 - 2,376
a = 2.5 m / s²
A convex mirror of focal length 33 cm forms an image of a soda bottle at a distance of 19 cm behind the mirror.If the height of the image is 7.0 cm,where is the object located,and how tall is it? What is the magnification of the image? Is the image virtual or real? Is the image inverted or upright? Draw a ray diagram to confirm your results.
Answer:
Image distance = 44.8cm, Image height = 16.5cm, Magnification = 0.42
The image is a virtual and upright image.
Explanation:
The nature of image formed by an object placed in front of a convex mirror is always diminished, virtual and erect.
The focal length f and the image distance are always NEGATIVE beacause the image is formed behind the mirror.
Given f = -33.0cm, v = -19.0cm
using thr mirror formula to get the object distance u, we have;
[tex]\frac{1}{f}=\frac{1}{u} + \frac{1}{v}\\ \frac{1}{u}=\frac{1}{f} - \frac{1}{v}\\\frac{1}{u}=\frac{1}{-33} - \frac{1}{-19}\\\frac{1}{u}=\frac{-19+33}{627} \\\frac{1}{u}=\frac{14}{627} \\u=\frac{627}{14} \\u = 44.8cm[/tex]
To calculate the image height, we will use the magnification formula
M = [tex]\frac{image\ height}{object\ height}=\frac{image\ distance}{object\ distance} \\[/tex]
M = [tex]\frac{Hi}{HI}=\frac{v}{u}[/tex]
Given Hi = 7.0cm
v = 19.0cm
u = 44.8cm
HI = 7*44.8/19
HI = 16.5cm
The object height is 16.5cm
Magnification = v/u = 19.0/44.8 = 0.42
SInce the image is formed behind the mirror, the image is a VIRTUAL and UPRIGHT image
A student performs an experiment that involves the motion of a pendulum. The student attaches one end of a string to an object of mass M and secures the other end of the string so that the object is at rest as it hangs from the string. When the student raises the object to a height above its lowest point and releases it from rest, the object undergoes simple harmonic motion. As the student collects data about the time it takes for the pendulum to undergo one oscillation, the student observes that the time for one swing significantly changes after each oscillation. The student wants to conduct the experiment a second time. Which two of the following procedures should the student consider when conducting the second experiment?
a) Make sure that the length of the string is not too long.
b) Make sure that the mass of the pendulum is not too large.
c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.
d) Make sure that the experiment is conducted in an environment that has minimal wind resistance.
Answer:
the answers the correct one is cη
Explanation:
In this simple pendulum experiment the student observes a significant change in time between each period. This occurs since an approximation used is that the sine of the angle is small, so
sin θ = θ
with this approach the equation will be surveyed
d² θ / dt² = - g / L sin θ
It is reduced to
d² θ / dt² = - g / L θ
in which the time for each oscillation is constant, for this approximation the angle must be less than 10º so that the difference between the sine and the angles is less than 1%
The angle is related to the height of the pendulum
sin θ = h / L
h = L sin θ.
Therefore the student must be careful that the height is small.
When reviewing the answers the correct one is cη
Considering the approximation of simple harmonic motion, the correct option is:
(c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.
Simple Harmonic MotionAccording to Newton's second law in case of rotational motion, we have;
[tex]\tau = I \alpha[/tex]
Applying this, in the case of a simple pendulum, we get;
[tex]-mg\,sin\,\theta =mL^2 \,\frac{d^2 \theta}{dt^2}[/tex]
On, rearranging the above equation, we get;
[tex]mL^2 \,\frac{d^2 \theta}{dt^2} + mg\,sin\,\theta=0\\\\\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} sin \,\theta=0[/tex]
Now, if angular displacement is very small, i.e.; the bob of the pendulum is only raised slightly.
Then, [tex]sin\, \theta \approx \theta[/tex]
[tex]\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} \,\theta=0[/tex]
This is now in the form of the equation of a simple harmonic motion.
[tex]\frac{d^2 \theta}{dt^2} +\omega^2 \,\theta=0[/tex]
Comparing both these equations, we can say that;
[tex]\omega = \sqrt{\frac{g}{L}}[/tex]
[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]
This relation for the time period can only be obtained if the angular displacement is very less.
So, the correct option is;
Option (c): Make sure that the difference in height between the pendulum's release position and rest position is not too large.
Learn more about simple harmonic motion here:
https://brainly.com/question/26114128
A ball with a mass of 4 kg is initially traveling at 2 m/s and has a 5 N force applied for 3 s. What is the initial momentum of the ball?
Answer:
The initial momentum of the ball is 8 kg-m/s.
Explanation:
Given that,
Mass of the ball is 4 kg
Initial speed of the ball is 2 m/s
Force applied to the ball is 5 N for 3 seconds
It is required to find the initial momentum of the ball. Initial momentum means that the product of mass and initial velocity of the ball. It is given as :
[tex]p_i=mu\\\\p_i=4\ kg\times 2\ m/s\\\\p_i=8\ kg-m/s[/tex]
So, the initial momentum of the ball is 8 kg-m/s.
What do you think will be different about cars in the future? Describe a change that is already being developed or that you think should be invented.
Answer:
Flying cars.
Explanation:
2. Air at a temperature of 20 ºC passes through a pipe with a constant velocity of 40 m/s. The pipe goes through a heat exchanger in which hot gases outside the pipe heat up the air to 820 ºC. It then enters a turbine with a velocity of 40 m/s and expands till the temperature falls to 620 ºC. The air stream loses 4.3 kW heat in the turbine. If the air flow rate is 2.5 kg/s, calculate (a) How much heat is transferred to the air in the heat exchanger. (b) The power output of the turbine.
Answer:
a) Q = 1436 kW
b) P ≈ 776 kW
Explanation:
Let's begin by listing out the given parameters:
T1 = 20 °C, u = 40 m/s, T2 = 820 °C, P = 4.3 kW, m = 2.5 kg/s, T3 = 510 °C, V1 = 40 m/s,
V2 = 40 m/s, V3 = 55 m/s, ṁ = 2.5 kg/s
To solve the question, we make this assumption that the size of the pipe is constant
a) No change in velocity implies that heat added is isochoric
Q = m * C * ΔT
Cv of air at 300 K(≈20 °C) = 0.718
Q = 2.5 * 0.718 * (820 − 20)
Q = 1436 kW
b) P = ṁ * Cp * ΔT + ṁ * (V2² - V3²) ÷ 2000] - Ql
V2² - V3² = 55² - 40² = 1425
ΔT = T2 - T3 = 820 - 510 = 310 °C
Cp of air at 300 K(≠20 °C) = 1.005 kJ/kgK
Ql = 4.3 kW
P = 2.5 * (1.005 * 310) + 2.5 * (1425 ÷ 2000) - 4.3
P = 778.875 + 1.78125 - 4.3 = 776.35625
P ≈ 776 kW
What types of mediums are involved in the energy transfer
Answer:
In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.
A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2300 kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.75, and the cars slide to a halt at a point 5.54 m west and 6.19 m south of the impact point. How fast was sedan traveling just before the collision? How fast was SUV traveling just before the collision?
Answer:
Explanation:
momentum of sedan of 1600 kg = 1600x v , where v is its velocity
momentum of suv of 2300 kg = 2300 x u where u is its velocity .
force of friction = ( 1600 + 2300 ) x 9.8 x .75 ( fiction = μ mg )
= 28665 N
distance by which friction acted = √ (5.54² + 6.19²)
= 8.3 m
work done by friction
= 28665 x 8.3
= 237919.5 J
Total kinetic energy of cars = work done by friction
1/2 x 1600 x v² + 1/2 x 2300 u² = 237919.5
16 v² + 23 u² = 4758.4
1600 x v / 2300 u = 6.19 / 5.54
v / u = 1.6
v = 1.6 u
putting this equation in fist equation
40.96 u² + 23 u² = 4758.4
= 63.96 u² = 4758.4
u² = 74.4
u = 8.62 m /s
v = 13.8 m /s
Which is the correct representation of the right-hand rule for a current flowing to the right?
Answer:
The third image
Explanation:
The one with the thumb pointing to the right
Answer:
3, correct on Edge 2020
You could use an analytical or triple beam balance to determine a ___ called ____
A)
physical property; mass.
B)
chemical property, mass.
C)
physical property; weight.
D)
physical property; density.
Answer:
a and b are the correct answers
Explanation:
Answer:
A) physical property; mass.
Explanation:
took the test
Electric fields are MOST associated with ________.
Einstein developed much of his understanding of relativity through the use of gedanken, or thought, experiments. In a gedanken experiment, Einstein would imagine an experiment that could not be performed because of technological limitations, and so he would perform the experiment in his head. By analyzing the results of these experiments, he was led to a deeper understanding of his theory. In each the following gedanken experiments, Albert is in the exact center of a glass-sided freight car speeding to the right at a very high speed vvv relative to you. Albert has a flashlight in each hand and directs them at the front and rear ends of the freight car. Albert switches the flashlights on at the same time.
In Albert's frame of reference, which beam of light travels at a greater speed, the one directed toward the front or the one toward the rear of the train, or do they travel at the same speed? Which beam travels faster in your frame of reference? Enter the answers for Albert's frame of reference and your frame of reference separated by a comma using the terms front, rear, and same. For example, if in Albert's frame of reference the beam of light directed toward the front of the train travels at a greater speed and in your frame of reference the two beams travel at the same speed, then enter front,same.
Answer:
For eintein's frame of reference, both beam travel at the same speed.
For my own frame of reference, both beams travel at the same speed.
Explanation:
According to special relativity, the speed of light is the same in all direction on all reference frame. If not for this law we will assume the from beam will have a relative speed that will be the speed of light plus the speed of the fright car. This is not so and it violates the speed limit of light which according to the first law is the highest speed possible and nothing can go beyond that.
Each propeller of the twin-screw ship develops a full-speed thrust of F = 285 kN. In maneuvering the ship, one propeller is turning full speed ahead and the other full speed in reverse. What thrust P must each tug exert on the ship to counteract the effect of the ship's propellers?
Answer:
tug_tug = 570 10³ l
Explanation:
In this problem, each propeller creates a force that makes the boat rotate, so the tugs have to create a die of equal magnitude rep from the opposite direction
∑ τ = 0
F1 la+ (-F1) (-l) = τ-tug
τ-tug = 2 f1 l
τ-tug = 2 28510³ l
tug_tug = 570 10³ l
where the is the distance from the propane axis to the point where the ship turns
This force may be less depending on where the tug is.
To understand thermal linear expansion in solid materials. Most materials expand when their temperatures increase. Such thermal expansion, which is explained by the increase in the average distance between the constituent molecules, plays an important role in engineering. In fact, as the temperature increases or decreases, the changes in the dimensions of various parts of bridges, machines, etc., may be significant enough to cause trouble if not taken into account. That is why power lines are always sagging and parts of metal bridges fit loosely together, allowing for some movement. It turns out that for relatively small changes in temperature, the linear dimensions change in direct proportion to the temperature.
For instance, if a rod has length L0 at a certain temperature T0 and length L at a higher temperature T, then the change in length of the rod is proportional to the change in temperature and to the initial length of the rod: L - L0 = αL0(T - T0),
or
ΔL = αL0ΔT.
Here, α is a constant called the coefficient of linear expansion; its value depends on the material. A large value of α means that the material expands substantially as the temperature increases; smaller values of α indicate that the material tends to retain its dimensions. For instance, quartz does not expand much; aluminum expands a lot. The value of α for aluminum is about 60 times that of quartz!
Questions:
A) Compared to its length in the spring, by what amount ΔLwinter does the length of the bridge decrease during the Teharian winter when the temperature hovers around -150°C?
B) Compared to its length in the spring, by what amount ΔLsummer does the length of the bridge increase during the Teharian summer when the temperature hovers around 700°C?
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
By which process does the heat from the Sun reach the Earth? (AKS 4b DOK 1) *
A mutation causes a dog to be born with a tail that is shorter than normal.
Which best describes this mutation?
Answer:
A mutation causes a dog to be born with a tail that is shorter than normal. Which best describes this mutation? It is harmful because it obviously affects the dog’s survival. It is harmful because it affects the dog’s physical appearance. It is neutral because it does not obviously affect the dog’s survival. It is beneficial because it affects the dog’s physical appearance.
Explanation:
Answer:
C
Explanation:
:)))
A radiator rests snugly on the floor of a room when the temperature is 10 oC. The radiator is connected to the furnace in the basement by a pipe that is 15 m long. How far off the floor will the radiator be lifted when it is filled with steam at 102 oC? The iron expands 1.0 * 10-5 / oC.
Now that you've done your research on the law of supply, you understand that it basically asserts that how much coffee you'd be willing to supply depends on how much money you can make for each cup.
Which of the following actions would decrease the energy stored in a parallel plate capacitor when a constant potential difference is applied across the plates? (Choose from: Increasing the area of the plates; Decreasing the area of the plates; Increasing the separation between the plates; Decreasing the separation between the plates; Inserting a material with a higher dielectric constant between the plates
Answer:
increasing the separation between the plates
Explanation:
The increase in the vacuum/separation between the plates in a parallel plate capacitor connected to a constant potential difference decreases the energy stored in the capacitor. the increase in the separation of the plates of a parallel plate capacitor reduces the capacitance of the capacitor because
Q(charge) = CV V = VOLTAGE , c = capacitance
E = 1/2 eAV^2/ D ( ENERGY STORED )
where D = distance between plates, e = dielectric, A = area of capacitor , V = potential difference
