Answer:
y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]
Explanation:
Let's solve this exercise in parts. Let's start by finding the wavelength of the electrons accelerated to v = 10 103 V, let's use the DeBroglie relation
λ= [tex]\frac{h}{p} = \frac{h}{mv}[/tex]
Let's use conservation of energy for speed
starting point
Em₀ = U = e V
final point
Em_f = K = ½ m v²
Em₀ = Em_f
eV = ½ m v²
v =[tex]\sqrt{\frac{2eV}{m} }[/tex]
we substitute
λ= [tex]\sqrt{ \frac{h^2 m}{2eV}}[/tex]
the diffraction phenomenon determines the minimum resolution, for this we find the first zero of the spectrum
a sin θ = m λ
first zero occurs at m = 1, also these experiments are performed at very small angles
sin θ = θ
θ = λ / a
This expression is valid for linear slits, in the microscope the slits are circular, when solving the polar coordinates we obtain
θ = 1.22 λ / D
where D is the diameter of the opening
we substitute
θ = [tex]\frac{1.22}{D}[/tex] \sqrt{ \frac{h^2 m}{2eV}}
this is the minimum angle that can be seen, if the distance is desired suppose that the distance of the microscope is L, as the angles are measured in radians
θ = y / L
when substituting
where y is the minimum distance that can be resolved for this acceleration voltage
y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]
Two masses are suspended by cord that passes over a pulley with negligible mass. The cord also has negligible mass. One of the masses, m1, has a mass of 7.0 kg and the other mass, m2, has a mass of 3.0 kg. The pulley turns on a shaft through the center of the pulley and supports the pulley and all the masses. The vertical force of the shaft on the pulley that supports the whole system is
Answer: F = 98N
Explanation: The shaft have to sustain the pulley, the cord and the two masses. The pulley and the cord have negligible masses, so, they have negligible weight.
The two masses have two vertical forces acting on them: force of traction because of the cord and force due to gravitational force, also known as weight.
So, the vertical force the shaft has to support is the sum of the weight of each mass:
[tex]F_{net}=F_{g}_{1}+F_{g}_{2}[/tex]
[tex]F_{net}=m_{1}.g+m_{2}.g[/tex]
[tex]F_{net}=g(m_{1}+m_{2})[/tex]
[tex]F_{net}=9.8(7+3)[/tex]
[tex]F_{net}=[/tex] 98
The vertical force that supports the whole system is 98 N.
Using Figure 2, what is the momentum of Train Car A before the collision?
A
180,000 kg*m/s
B
0 kg*m/s
C
11,250 kg*m/s
D
4 kg*m/s
Answer:
Option A. 180000 Kgm/s.
Explanation:
From the question given above, the following data were obtained:
For Train Car A:
Mass of train car A = 45000 Kg
Velocity of train car A = 4 m/s
Momentum of train car A =?
For Train Car B:
Mass of train car B = 45000 Kg
Velocity of train car B = 0 m/s
Momentum is simply defined as the product of mass and velocity. Mathematically, it can be expressed as:
Momentum = mass × velocity
With the above formula, the momentum of train car A before collision can be obtained as follow:
Mass of train car A = 45000 Kg
Velocity of train car A = 4 m/s
Momentum of train car A =?
Momentum = mass × velocity
Momentum = 45000 × 4
Momentum of train car A = 180000 Kgm/s
Solar flares are disturbances on the Sun’s surface that send large amounts of energy away from the Sun’s surface. These disturbances can often interfere with technology here on Earth. What statement BEST describes the cause of solar flares?
Answer:
with that boom and that boom glare boom that boom glare with that boom the tha ba glare boom that boom glare with that.
Explanation:
The current supplied by a battery in a portable device is typically about 0.122 A. Find the number of electrons passing through the device in two hours.
5. Friction has two components __________ and ___________.
Answer:
static friction and kinetic friction.
Explanation:
How can you describe the orbital period (planetary year) of each planet?
Answer:
How can you describe the orbital period (planetary year) of each planet?
A year is defined as the time it takes a planet to complete one revolution of the Sun, for Earth this is just over 365 days. This is also known as the orbital period. Unsurprisingly the the length of each planet's year correlates with its distance from the Sun.
Explanation:
Answer: The orbital period is how long it takes a plane to fulfill its "revolution"(365 days)
Explanation:
The all-digital touch-tone phones use the summation of two sine waves for signaling. Frequencies of these sine waves are defined as 697, 770, 852, 941, 1209, 1336, 1477, and 1633 Hz. Since the sampling rate used by the telecommunications is 8000 Hz, convert those eight analog frequencies into digital frequencies of radians and cycles.
Two objects are located on an airtrack (you may assume there is no friction). The magnitude of the charge on object A is 5 times higher than on object B, and object A also has a 6 times higher mass than object B (the picture is not scale or necessarily in the correct direction). Each object is accelerating in the direction of the arrows.
a. Draw a system schema.
b. Draw the force body diagrams for each charge and identify all newton 3d law pairs.
c. Write (in symbolic form) expressions for the net force on each object.
d. Find how many times faster/slower the acceleration of object B is compared to object A?
Answer:
Explanation:
From the information given:
The schematic diagram for the system and the force body diagrams for each charge can be seen in the image attached below
(c)
The symbolic expressions for the net force on each object is as follows:
[tex]N_A = m_Ag \\ \\ F_{net}, A= F_A = \dfrac{kQ_AQ_B}{r^2}[/tex]
[tex]N_B = m_Bg \\ \\ F_{net}, B= F_B = \dfrac{kQ_AQ_B}{r^2}[/tex]
(d) From above
[tex]F_A =F_B[/tex]
[tex]m_Aa_A = m_Ba_B \\ \\ (6m_B)a_A = m_B a_B[/tex]
[tex]a_B = 6a_A[/tex]
The resolution of a lens can be estimated by treating the lens as a circular aperture. The resolution is the smallest distance between two point sources that produce distinct images. This is similar to the resolution of a single slit, related to the distance from the middle of the central bright band to the firstorder dark band; however, the aperture is circular instead of a rectangular slit which introduces a scale factor. Suppose the Hubble Space Telescope, 2.4 m in diameter, is in orbit 90.4 km above Earth and is turned to look at Earth. If you ignore the effect of the atmosphere, what is the resolution of this telescope for light of wavelength 557 nm?
Answer:
y = 2.56 10⁻² m
Explanation:
The resolution of this telescope is given by the Rayleigh criterion, for the phenomenal diffraction the first minimum for a linear slit is in
a sin θ = λ
in general the angles are very small, so we approximate
sin θ = θ
we substitute
θ = λ / a
in the case of circular slits we must use polar coordinates, which introduces a numerical factor, leaving the equation
θ = 1.22 [tex]\frac{\lambda }{D}[/tex]
where D is the diameter of the circular opening
In this case they indicate the lens diameter D = 2.4 m, the observation distance r = 90.4 km = 90.4 10³ m
how angles are measured in radians
θ = y / r
we substitute
y / r = 1.22\frac{\lambda }{D}
y = 1.22 \frac{\lambda r }{D}
let's calculate
y = [tex]1.22 \frac{ 557 \ 10^{-9} \ \ 90.4 \ 10^{3} }{2.4}[/tex]
y = 2.56 10⁻² m
this is the minimum distance that can differentiate two objects on Earth
4.The equation for Kinetic Energy is 1/2 x m x v2. Calculate the KE when the mass is 5 kg and the velocity is 4 m/s.
a.2 joules
b.10 joules
c.20 joules
d.40 joules
Explanation:
KE = ½mv²
= ½ × 5 × 4²
= ½ × 5 × 16
= ½ × 80
= 40 J
A girl standing on a bridge throws a stone vertically downward with an initial velocity of 15.0 m/s into the river below. If the stone hits the water 2.00 seconds later, what is the height of the bridge above the water
Answer:
the height of the bridge above the water is 49.6 m.
Explanation:
Given;
initial velocity of the stone, u = 15 m/s
time of motion of the stone, t = 2 s
The height of the bridge above the water is calculated from the following kinematic equation as follows;
h = ut + ¹/₂gt²
h = (15 x 2) + ¹/₂(9.8)(2²)
h = 30 + 19.6
h = 49.6 m
Therefore, the height of the bridge above the water is 49.6 m.
PLEASEEEE HELP MEEEEE!!!!
Answer:
D. Yes
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A lake has a surface area of 410 m2 and a volume of 1140 m3 . Suppose that during the day, sunlight with a power averaging 820 W/m2 shines on the lake, and that about 10% of this power is absorbed in the lake, producing heat. Assume that the temperature of the lake water stays constant because the absorbed solar power is exactly balanced by heat lost due to evaporation of water from the lake surface. What is the evaporation rate, in g/s
Answer:
Explanation:
Total solar energy falling on total surface per second
= 410 x 820 W
= 336200 W
10 % of 336200 = 33620 J is converted into heat which is absorbed by lake water . But its temperature does not rise because heat is used up in evaporating water in the form of vapor .
Total heat released during evaporation = 33620 J
Let evaporation rate be m gram /s
heat absorbed by m gram water = m x latent heat of evaporation
= m x 2260 J .
Given ,
m x 2260 = 33620
m = 14.87 g /s .
A scientist reports a measurement of the temperature of the surface of a newly discovered planet as negative 20 Kelvin. What conclusion can you draw from this report
Answer:
The temperature of this newly discovered planet violates the third law of thermodynamic, there is a mistake in this value.
Explanation:
The third law of the Thermodynamic says:
At zero kelvin all molecular movement stops, which means that the entropy will be zero at this temperature.
So we can say there is no thermodynamic system that has temperature values less than 0 K.
The conclusion of the report will be.
The temperature of this new planet violates the third law of thermodynamic, there is a mistake in this value.
I hope it helps you!
A 12 V battery is connected across a device with variable resistance. As the resistance of the device increases, determine whether the following quantities increase, decrease, or remain unchanged. HINT (a) The current through the device. increases decreases remains unchanged (b) The voltage across the device. increases decreases remains unchanged (c) The power consumed by the device. increases decreases remains unchanged
Answer:
a) DECREASE, b) INCREASE, c) power remains constant
Explanation:
The resistance and the battery are connected in series.
a) How the current changes when the resistance changes
V = I R
I = V / R
I = 12 / R
if the resinification increases, the current must DECREASE
b) when you check the expression
V= I R
, if R increases the voltage of the INCREASE
c) the power in an electric circuit is
P = I V
Let's analyze this expression, the voltage increases linearly with the increase in resistance and the current decreases linearly with the increase in r, for which the two effects are compensated and the dissipated power remains constant
If the magnetic force is 3.5 × 10–2 N, how fast is the charge moving?
Answer:
D
Explanation:
Took it on edg
The speed of the charge at the given magnetic force and field is determined as 1.1 x 10⁴ m/s.
Speed of the chargeThe speed of the charge is calculated as follows;
F = qvBsinθ
v = F/qBsinθ
where;
F is the magnetic forceB is magnetic fieldv is speed of the chargev = (3.5 x 10⁻²)/(8.4 x 10⁻⁴ x 6.7x 10⁻³ x sin35)
v = 10,842.33 m/s
v ≅ 1.1 x 10⁴ m/s
Learn more about speed of charge here: https://brainly.com/question/24116470
#SPJ2
Question: How did NASA use Newton's Laws to land the Perseverance lander,
safely
on Mars?
Please help asap.
Answer:
If the thrust is increased, the aircraft accelerates and the velocity increases. This is the second part sited in Newton's first law; a net external force changes the velocity of the object. The drag of the aircraft depends on the square of the velocity. So the drag increases with increased velocity.
A stone of mass 1kg is thrown at 10m/s upwards making an angle of 37°with the horizontal from a building that is 20m high. Using the law of conservation of energy calculate the speed wjen the stone hits the ground.
Answer:
31.68 m/s
Explanation:
The law of conservation of energy states that energy is not lost or gained it is just converted, in this example, since it is not given any resistance from the wind, you'd have two variables Speed on the Y-axis and on the X-axis, since both of them would result in the same decrease and increase with against gravity, it doesn't matter the value of both.
As the stone continues to go upwards it will continue to lose speed due to de-acceleration from the gravity acting on it, similarly, it will continue to gain Potential energy, instead of kinetic energy, when it reaches its highest point the speed on Y will be "0" and the free fall will start, since the up and down movement will be equal in time the and acceleration would be equal -9.81 m/s and 9.81 m/s because the only acceleration you have is gravity, you only need to calculate how much speed will gain a rock accelerating at 9.81 m/s falling 20 m:
[tex]H=\frac{1}{2}g*t^{2} \\\frac{20}{1/2*9.81} =t^{2} \\\\t^{2} =4.08\\t=\sqrt{4.08} \\t=2.21[/tex]
Now we just add the time accelerating:
[tex]Vf=Vo+at\\Vf=10 m/s+ 2.21*9.81\\Vf=10 m/s+21.68\\Vf=31.68 m/s[/tex]
The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons).
(a) What is the force between the two alpha particles when they are 6.60 ✕ 10−15 m apart? N
(b) What is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u. m/s2
Answer:
a) F = 21.16 N, b) a = 3.17 10²⁸ m / s
Explanation:
a) The outside between the alpha particles is the electric force, given by Coulomb's law
F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]
in that case the two charges are of equal magnitude
q₁ = q₂ = 2q
let's calculate
F = [tex]9 \ 10^9 \ \frac{ (2 \ 1.6 \ 10^{-19} )^2 }{ (6.60 \ 10^{-15} )^2 }[/tex]
F = 21.16 N
this force is repulsive because the charges are of the same sign
b) what is the initial acceleration
F = ma
a = F / m
a = [tex]\frac{21.16}{4.0026 \ 1.67 \ 10^{-27} }[/tex]21.16 / 4.0025 1.67 10-27
a = 3.17 10²⁸ m / s
this acceleration is in the direction of moving away the alpha particles
Giving 50 points exactly 50 point and making brainles and if answered wrong I will report and get you kicked out
Draw a picture showing how Heat is added, released, and transferred from one object to another. Also, draw a picture explaining how sublimation or deposition works. please you can draw in your book and take a picture and post it
Answer:
i hope this helps i did'nt quiet understand the secound one. I hope youcan see my picture well .
Explanation:
sincerily, MEMC3891
When a wave hits an object,energy from the wave is both absorbed and reflected off the object
A. True
B. False
P.S pls help
A 1 800-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.60 m before coming into contact with the top of the beam, and it drives the beam 13.6 cm farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.
Answer:
F = 614913.88 N
Explanation:
We are given;
Mass of pile driver; m = 1800 kg
Height of fall of pole driver; h = 4.6 m
Depth driven into beam; d = 13.6 cm = 0.136 m
Now, from energy equations and applying to this question, we can write that;
Workdone = Change in potential energy
Formula for workdone is; W = F × d
While the average potential energy here is; W = mg(h + d)
Thus;
Fd = mg(h + d)
Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.
Making F the subject, we have;
F = mg(h + d)/d
F = 1800 × 9.81 × (4.6 + 0.136)/0.136
F = 614913.88 N
Earth has seasons because _____.
it rotates on its axis as it moves around the sun
the temperature of the sun changes
its axis is tilted
the distance between Earth and the sun changes
Answer:
c, its axis is tilted
maybe
As it works its way around the sun, its tilted axis exposes different parts of earth.
C would be it because the roation of Earth on its axis doesn't have anything to do with the exposer of the revolution on its axis
if you make a sound by tapping on a glass of water what is the order of vibration
A. glass - water - air
C. air - water - glass
B. water - glass - air
Answer:
A.) glass water air
Explanation:
hope this helps :) have a great day!!
Answer:
A.) glass - water - air
good luck, i hope this helps :)
A constant electric field of 5.00 N/C points along the positive x-direction. An electron, initially at rest, moves a distance of 2.00 m in this space. How fast is the electron moving after its 2.00 m journey
Answer:
1.875 x 10⁶ m /s .
Explanation:
Force on electron = E e where E is electric field and e is charge on electron
acceleration generated = Ee / m where m is mass of the electron .
Putting the values
acceleration generated = 5 x 1.6 x 10⁻¹⁹ / 9.1 x 10⁻³¹
= .879 x 10¹² m /s²
v² = u² + 2 as , initial velocity u = 0 , displacement s = 2 m
v² = 0 + 2 x .879 x 10¹² x 2
v = 1.875 x 10⁶ m /s .
Solids have a definite shape and volume this is because
There are two metallic spheres and a positive-charge generator. Throughout the entire problem, one sphere is always closest to the generator (call it C), and one is always farther away (call it F). Both spheres start electrically neutral and are in contact with each other. These spheres are now brought near the positivecharge generator (call it G), but do not touch it. The two metallic spheres C and F are then separated from each other, and then individually moved away from the positive-charge generator.
Immediately after the two spheres are moved away from the positive-charge generator, the sphere farther away from the charge generator is charged_______
a. positive
b. negative
c. neutral
d. cannot be determined
Answer:
The correct answer is C
Explanation:
In this problem we must remember that charges of the same sign repel and of the same sign attract.
Generator G induces a charge in the closest spheres C, otherwise the charge is negative, therefore in the furthest sphere F, with the ac that is in contact, a positive charge is formed, they do not indicate that the spheres It is in contact with Earth, so the charges remain on them.
As the spheres move away from the generator, the load is redistributed, they return to their initial load, that is, they remain Neutral since no load left or entered the spheres, there was only a redistribution due to the proximity of the generator.
The correct answer is C
A locomotive is accelerating at 1.60 m/s2. It passes through a 20.0-m-wide crossing in a time of 2.40 s. After the locomotive leaves the crossing, how much time is required until its speed reaches 31.6 m/s
Answer:
The time to reach 31.6 m/s is 14.54 s.
Explanation:
Given;
acceleration of the locomotive, a = 1.60 m/s²
distance traveled by the locomotive, s = 20 m
time of motion, t = 2.4 s
final velocity of the locomotive, v = 31.6 m/s
The speed of the of the locomotive as it leaves the 20 m wide crossing;
[tex]v = \frac{d}{t} \\\\v = \frac{20}{2.4} \\\\v = 8.333 \ m/s[/tex]
The time to reach 31.6 m/s is calculated from the equation below;
v = u + at
31.6 = 8.333 + 1.6t
1.6t = 31.6 - 8.333
1.6t = 23.267
t = 23.267 / 1.6
t = 14.54 s
Therefore, the time to reach 31.6 m/s is 14.54 s.
pls help everything is in the pic
Answer:
c
Explanation:
You are working in a shoe test laboratory measuring the coefficients of friction for running shoes on a variety of surfaces. The shoes are pushed against the surface with a downward force of 400 N, and a sample of the surface material is then pulled out from under the shoe by a machine. The machine pulls with a force of 300 N before the material begins to slide. When the material is sliding, the machine only has to pull with a force of 200 N to keep the material moving. What is the coefficient of static friction between the shoe and the material
Answer:
0.75
Explanation:
Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.
Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).
So, μ = F/N
= 300 N/400 N
= 3/4
= 0.75
So, the coefficient of static friction μ = 0.75