Answer:
a) 3×10^18 Hz
b) The electromagnetic wave is an x-ray
Explanation:
We know that the speed of an electromagnetic wave is given by;
c= λf
Where;
c= speed of electromagnetic waves = 3×10^8 ms-1
f= frequency of electromagnetic waves= the unknown
λ = wavelength of the electromagnetic wave= 1 ×10^-10 m
Hence;
f = c/λ= 3×10^8/ 1×10^-10
f= 3×10^18 Hz
b) The electromagnetic wave is an x-ray.
(a) The frequency of the electromagnetic radiation is [tex]3\times 10^{18} \ Hz[/tex]
(b) The electromagnetic radiation is ultra violet ray.
The given parameters;
distance of the atom, λ = 1 x 10⁻¹⁰ mThe frequency of the electromagnetic radiation is calculated as follows;
[tex]c =f \lambda \\\\f = \frac{c}{\lambda} \\\\f = \frac{3\times 10^8}{1\times 10^{-10}} \\\\f = 3\times 10^{18} \ Hz[/tex]
The frequency range of ultra violet ray is between 10¹⁴ to 10¹⁸ m.
Thus, the electromagnetic radiation is ultra violet ray.
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Two long wires hang vertically. Wire 1 carries an upward current of 1.20 A. Wire 2, 20.0 cm to the right of wire 1, carries a downward current of 4.20 A. A third wire, wire 3, is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force.
(a) Is this situation possible?
A. Yes
B. No
Is it possible in more than one way?
A. Yes
B. No
(b) Describe the position of wire 3.
distance ________ cm
direction: left of wire 1
(c) Describe the magnitude and direction of the current in wire 3.
magnitude
direction
down
Answer:
a) NO, YES that cable 3 should be placed to the left of cable 1
b) a = 8cm the current is UP
Explanation:
The force between two cables that carries a current is given through the magnetic field created between them, it has the expression
F₁ = μ₀ I₁ I₂ / 2π a l
where I₁ and I₂ are the current in each wire, at the separation between them and l the length of wire 1 on which the force is applied.
The direction of the force this given is given by the vector product between the current in wire 1 and the field created by 2, but we can summarize it:
* if the two currents are in the same direction the force is attractive
* if the two currents are contrary, the force is repulsive
With these relationships we can examine the different situations presented
a) Between cable 1 and 3 the force is zero. If cable 1 the current is upward in cable 3 it must be downward, so that the force is repulsive
Between wire 2 and 3. If wire 2 the current is downward, in wire 3 the current must be upward, for a repulsive force.
We see that in the two conditions the current in cable 3 has different directions, which is impossible, therefore this situation cannot occur
the answer is NO
The only way this is possible is that cable 3 should be placed to the left of cable 1
Answer YES
b) and c) Your question is unclear, I interpret that you want the position of cable 3 so that the force is zero on cable 3
we will assume that the current in cables 1 and 2 is ascending and that in cable 3 is descending
We write Newton's second law at the point of cable 3, the acceleration is zero
F13 - F23 = 0
F13 = F23
Let's calculate the forces
F₁₃ = μ₀ I₁ I₃ /2π a l
F₂₃ = μ₀ I₂ I₃ /2π (d + a) l
we substitute and simplify
I₁ / a = I₂ / (d + a)
where we have assumed that the length of all cables is the same
We solve to find
I₁ (d + a) = I₂ a
a (I₂ - I₁) = I₁ d₁
a = I₁ / (I₂ - I₁) d
let's calculate
a = 1.20 / (4.2 -1.2) 20
a = 8cm
to the left of wire 1 and the direction is the current is UP
so that it is an attractive force between 1 and 3 and a repulsive force between 2 and 3
Which of these forces help protons and neutrons to stay at the center of the Atom
The measured value of the latent heat of vaporization of helium (at 1 atm) is
84 J/mol. Use this to estimate the binding energy of helium atoms in the liquid.
a. 2.6 x 10-6 eV
b. 3.6 x 10-4 eV
c. 5.1 x 10-4 eV
d. 8.6 x 10-3 eV
e. 3.6 x 10-2 eV
Answer:
d. 8.6 x 10-3 eV
Explanation:
Binding energy is energy that binds two atoms together . Energy is required to separate them . In the process of evaporation , energy is required to separate each molecules/ atoms of liquid so that they are separated and then get evaporated . That is why latent heat of evaporation is required .
1 mole of helium will contain 6.02 x 10²² atoms of it .
binding energy per atom = 84 / 6.02 x 10²² J
= 13.953 x 10⁻²² J
= 13.953 x 10⁻²² / 1.6 x 10⁻¹⁹ eV .
= 8.6 x 10⁻³ eV
a pistol fires a bullet towards a target located 175m away. The bullet is traveling at 320 m/s. How long does it take for the bullet to hit the target?
Time = distance/speed
Time = (175 m) / (320 m/s)
Time = 0.547 second
A volleyball is released three times from three different heights: 5 m, 10 m, and 20 m. Which statement is correct about the volleyball?
The volleyball accelerates more quickly toward the ground when released from a higher height.
The volleyball accelerates the same amount toward the ground each time.
The volleyball accelerates more slowly toward the ground when released from a higher height.
The volleyball may accelerate more quickly or slowly depending on its mass.
A student is setting up an experiment with 10 different balls. The student wants to measure which ball hits the ground with the largest force when released from a box 5 feet above the ground. Which quantity should be measured to correctly calculate the force with which each ball hits the ground?
the initial velocity of each ball
the mass of each ball
the rate at which the velocity of each ball increases
the final velocity of each ball right before hitting the ground
Explanation:
1)
The answer is
The volleyball accelerates the same amount toward the ground each time.
Because gravitational acceleration is constant irrespective of altitude(height) or mass of the object.
2)
The answer is
the mass of each ball
because gravitational force is given by
F = mg
since g is a constant, he only needs to measure the mass if each ball
The concepts of free fall and Newton's second law allow to find the correct answers
1) Acceleration is constant
2) The mass of the body
The game of volleyball consists of throwing the balls from the serve to pass it to the other court, this is a two-dimensional movement type projectile launch, in this type of movement the acceleration is the constant on the y axis, it is called acceleration of the gravity and vale (g = 9.8 m / s²)
1) Let's review the different statements of the first part regarding acceleration.
a) The ball accelerates faster as it goes down
False. Acceleration is constant
b) The ball accelerates the same amount
True. Acceleration is constant and directed on the vertical axis
c) The ball accelerates slower or faster
False. Acceleration is constant
The vertical launch of kinematics, establishes that the ball as it descends it goes faster with an acceleration equal to the acceleration of gravity, in this part it is asked to look for the force of the ball when it reaches the ground, for this the second is used Newton law
F = m a
where in this case the acceleration is the acceleration of gravity
a = g
F = m g
Let's review the different claims
a) Initial velocity
False. To calculate the force you only need the mass since acceleration is constant
b) The mass
True. The product of the mass and the acceleration of gravity gives the force with which the ball hits the ground
c) Increasing speed
False. Increasing speed allows calculating acceleration not force
d) The final speed
False. Force does not depend on the speed of the body
In conclusion using the concepts of free fall and Newton's second law we can find the correct answers:
1) Acceleration is constant
2) Body mass
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The coefficient of static friction between a 3.00 kg crate and the 35.0o incline is 0.300. What minimum force F must be applied perpendicularly to the incline to prevent the crate from sliding down
Answer:
So the minimum force is
32.2Newton
Explanation:
To solve for the minimum force, let us assume it to be F (N)
So
F=mgsinA
But
=>>>> coefficient of static friction x (F + mgcosA
=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)
So making F subject of formula
F + 24.0 = 56.2
F = 32.2N
Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of the force that one particle exerts on the other?
Answer:
-384.22N
Explanation:
From Coulomb's law;
F= Kq1q2/r^2
Where;
K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2
q1 and q2 = magnitudes of the both charges
r= distance of separation
F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2
F= -495.65 × 10^-3/ 1.29 × 10^-3
F= -384.22N
When stretched beyond it's elastic limit, a metal rod such as steel
Answer:
When an elastic material is stretched, depending on the stress, it might reach a point beyond which it will no longer return to its original size and shape. This point is called the elastic limit.
When the material (such as a metal rod (e.g steel, copper)) is stretched beyond its elastic limit, Hooke's law is no longer obeyed. i.e strain is no longer directly proportional to stress.
A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 1.5 m away. He then wants to take a picture of an object that is 30 cm away.
How far must the lens move to focus on this second object?
Answer:
The distance is [tex]z = 0.008 \ m[/tex]
Explanation:
From the question we are told that
The focal length is [tex]f = 50 \ mm = 50*10^{-3} \ m[/tex]
Generally the lens equation is mathematically represented as
[tex]\frac{1}{u} + \frac{1}{v} = \frac{1}{f}[/tex]
At image distance u = 1.5 m
[tex]\frac{1}{1.5} + \frac{1}{v} = \frac{1}{50 *10^{-3}}[/tex]
=> [tex]\frac{1}{50 *10^{-3}} - \frac{1}{1.5} = \frac{1}{v}[/tex]
=>[tex]v = 0.052 \ m[/tex]
At image distance [tex]u = 30\ cm = 0.30 \ m[/tex]
[tex]\frac{1}{0.3} + \frac{1}{v_1} = \frac{1}{50 *10^{-3}}[/tex]
=> [tex]\frac{1}{50 *10^{-3}} - \frac{1}{0.30 } = \frac{1}{v_1}[/tex]
=> [tex]v_1 = 0.06 \ m[/tex]
The distance the lens need to move is evaluate as
[tex]z = |v - v_1|[/tex]
[tex]z = |0.052 - 0.06|[/tex]
[tex]z = 0.009 \ m[/tex]
It is found that an engine rejects 100.0 J while absorbing 125.0 J each cycle of operation.
(a) What is the efficiency of the engine?
(b) How much work does it perform per cycle?
A source vibrating at constant frequency generates a sinusoidal wave on a string under constant tension. If the power delivered to the string is tripled, by what factor does the amplitude change
Answer:
n = 1,732 the amplitude must be increased by a factor of 1,732
Explanation:
The power delivered by a wave is given by
P = E / t
P = ½ μ w² v A²
let's apply this expression to our case the power tripled
3P₀ = ½ μ w² v A’²
let's write the amplitude function of a initial amplitude
A ’= n A₀
where n is a number
3 P₀ = (½ μy w² v A₀²) n²
3P₀ = P₀ n²
n = √ 3
n = 1,732
therefore the amplitude must be increased by a factor of 1,732
A cheetah can accelerate from rest to 25.0 m/s in 6.22 s. What is the cheetah's average speed for the first 3.11 s of its sprint
Answer:
18.75m/sExplanation:
Given data
initial speed ,u= o m/s
final speed v= 25 m/s
time t= 3.11 seconds
Applying the first equation of motion we have
[tex]v= u+at\\\\v-u=at\\\\a= \frac{v-u}{t} \\\\a= \frac{25-0}{6.22} \\\\a= 4.02 m/s^2[/tex]
Also applying [tex]v=u+at[/tex] with initial speed set at 0, we can also find speed at 3.11 seconds
[tex]v=0+4.02*3.11\\v= 12.50 m/s[/tex]
Hence the average speed in the first 3.11 seconds is
[tex]\bar v= \frac{v}{2} \\\\\bar v=\frac{ 12.50}{2} \\\\ \bar v= 6.25 m/s[/tex]
Applying the expression [tex]\bar v=\bar u+at[/tex] we can now fing the average speed in the second 3.11 second
[tex]\bar v=\bar u+at\\\\ \bar v=6.25+4.02*3.11\\\\ \bar v=18.75m/s[/tex]
Water flows through a 0.5 cm diameter pipe connected to a 1 cm diameter pipe. Compared to the speed of the water in the 0.5 cm pipe, the speed in the 1 cm pipe is
A
Answer:
Speed of water in the 0.5cm diameter will be faster because it has a smaller area
Since area x radius ² so if radius is reduced by 0.5 speed is increased by 4times in the 0.5 diameter pipe
An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into the water a pair of concentric metallic cylinders at the end of a cable and taking data to determine the resistance between these electrodes as a function of depth. The water between the two cylinders forms a cylindrical shell of inner radius r_a, outer radius r_b, and length L much larger than r_b. The scientist applies a potential difference Delta V between the inner and outer surfaces, producing an outward radial current I. Let rho represent the resistivity of the water.
(a) Find the resistance of the water between the cylinders in terms of L, rho, r_a, and r_b.
(b) Express the resistivity of the water in terms of the measured quantities L, r_a, r_b, Delta V and I.
Answer:
a) R = ρ₀ L /π(r_b² - R_a²) , b) ρ₀ = V / I π (r_b² - R_a²) / L
Explanation:
a) The resistance of a material is given by
R = ρ l / A
where ρ is the resistivity, l is the length and A is the area
the length is l = L and the resistivity is ρ = ρ₀
the area is the area of the cylindrical shell
A = π r_b² - π r_a²
A = π (r_b² - r_a²)
we substitute
R = ρ₀ L /π(r_b² - R_a²)
b) The potential difference is related to current and resistance by ohm's law
V = i R
we subsist the expression of resistance
V = I ρ₀ L /π (r_b² - R_a²)
ρ₀ = V / I π (r_b² - R_a²) / L
An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the sign of the value in your answers.)
(a) What is the magnification?
(b) What is the image height?
m
Explanation:
Given that,
Size of object, h = 0.066 m
Object distance from the lens, u = 0.210 m (negative)
Focal length of the converging lens, f = 0.14 m
If v is the image distance from the lens, we can find it using lens formula as follows :
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m[/tex]
(a) Magnification,
[tex]m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2[/tex]
(b) Magnification, [tex]m=\dfrac{h'}{h}[/tex]
h' is image height
[tex]-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m[/tex]
Hence, this is the required solution.
Consider a conducting rod of length 34 cm moving along a pair of rails, and a magnetic field pointing perpendicular to the plane of the rails.
Required:
At what speed (in m/s) must the sliding rod move to produce an emf of 0.75V in a 1.75T field?
Answer:
v = 1.26 m/s
Explanation:
Given that,
Length of the conducting rods = 34 cm = 0.34 m
Magnetic field, B = 1.75 T
Induced emf is 0.75 V
we need to find the speed with which it slides. Due to the motion of the rod an emf is induced in it which is given by :
[tex]\epsilon=Blv[/tex]
v is the speed of rod
[tex]v=\dfrac{\epsilon}{Bl}\\\\v=\dfrac{0.75}{1.75\times 0.34}\\\\v=1.26\ m/s[/tex]
So, the speed of the sliding rod is 1.26 m/s.
Scientists pose a ___ to examine cause and effect relationships.
Answer:
hypothesis
Explanation:
I think this is right I dont remember
An electron from a Ti ^ + 2 hydrogen ion leaps from one orbit with radius 13.25 angstrom to another orbit with radius 2.12 angstrom. determine the energy (Joule) e produced in said transition and the wavelength (in cm)
Answer:
E = 0.2276 10⁻¹⁹ J , λ = 8.73374 10⁻⁴ cm
Explanation:
Bohr's atomic model can be used for hydrogen-type atoms, that is, they have a single electron in their last orbit, this is the case of doubly ionized Titanium.
It is much easier to work in EV units.
r = a₀ / Z n²
E = - 13.606 (Z² / n²)
where ao is the Bohr radius, which is the ground state orbital of hydrogen a₀ = 0.0529 nm, Z is the atomic number of titanium, and n is an integer that represents the different states of the system.
a) Let's look for the energy, for this we look for the integer numbers of these orbits
r = 13.25 A = 1.325 nm
r = a₀ / Z n²
n = √(Z r / a₀)
the atomic number of titanium is Z = 22
n = √ (22 1.325 / 0.0529)
n = 23
r = 2.12 A = 0.212 nm
n = √ (22 0.212 / 0.0529)
n = 9
now we can calculate the energy of the transition
E = 13.606 (1 / [tex]n_{f}^2[/tex] - 1 / [tex]x_{o}^2[/tex])
E = 13.606 (1/9² - 1/23²)
E = 13.6060 (0.010455)
E = 0.1423 eV
Let's reduce to J
E = 0.1423 eV (1.6 10⁻¹⁹ J / 1eV) = 0.2276 10⁻¹⁹ J
To find the wavelength of the transition, use the Planck relation
E = h f
the relationship of frequency and speed of light
c = λ f
we substitute
E = h c /λ
λ = hc / E
λ = 6.626 10⁻³⁴ 3 10⁸ / 0.2276 10⁻¹⁹
λ = 8.73374 10⁻⁶ m
λ = 8.73374 10⁻⁴ cm
Neutrons travelling at 0.400 m/s are directed through a pair of slits having 1.00 m separation. an array of detectors is placed 10.0 m from the slits.
a) What is the de Broglie wavelength of the neutrons?
b) How far off axis if the first zero-intensity point on the detector array?
c) When a neutron reaches a detector, can we say which slit the neutrons passed through? Explain.
Answer:
9.91*10^-7 m
4.955*10^-6 m
Explanation:
Given that
v = 0.4 m/s
d = 1 m
L = 10 m
h = 6.62*10^-34 Js
m(neutron) = 1.67*10^-27 kg
To find the debroglie wavelength of the neutron, we use the formula
λ = h/mv
Now, we plug in the values we have listed.
λ = 6.62*10^-34 / (1.67*10^-27 * 0.4)
λ = 6.62*10^-34 / 6.68*10^-28
λ = 9.91*10^-7 m
b)
y1 = L (m + ½) λ/d, where m = 0
y1 = L (0 + ½) λ/d
y1 = L (½) λ/d
y1 = L/2 * λ/d or
y1 = Lλ/2d
now, we substitute the values for each of them, we have
y1 = (10 * 9.91*10^-7) / (2 * 1)
y1 = 9.91*10^-6 / 2
y1 = 4.955*10^-6 m
c) no, we can not say the neuron passed through one slit
Remove the wood block from the water. Place the brick block into the water and let it sink. How much water does the brick block displace? Hint: This is the liquid level with the block minus the liquid level before the block is added.
Answer:
the volume desalinated that is placed is equal to the volume of the block.
Explanation:
For this exercise we use the Archimedean principle which states that the thrust is equal to the weight of the desalted liquid.
B = ρ g V
When we have the block of wood the volume of water desalinated in the volume of the block under water, this is because the wood floats in the water
When placing the block of clay (brick), it sinks, so the volume desalinated that is placed is equal to the volume of the block.
V_body = V_waer = l to h
Maggie completed a 10000-m race at an average speed of 160
m/min. If Tom took 12.5 fewer minutes to complete the race,
what was Tom's average speed?
Answer: 200m/min
Explanation:
Divide 10000m by 160m/min, you will get the answer 62.5. You then subtract 12.5 from 62.5 to understand what you will need your answer for the other person’s speed will be. 10000m divided by 50min is 200m/min.
A westward-moving bicycle increased its speed from 2.0m/s to 12.0m/s in 4.0 seconds. What is the magnitude and direction of the acceleration?
Answer:
2.5 m/s^2
Explanation:
Acceleration = (Final Velocity - Initial Velocity) / TIme
Acceleration = 12 - 2 / 4
=> Acceleration = 10 /4
=> Accelration = 5/2
=> Acceleration = 2.5 m/s^2
So, Acceleration is 2.5 m/s^2
The acceleration of the bicycle which is the change in the velocity of the bike with change in time will be 2.5m/s in the Westward direction
Initial speed = 2.0 m/s Final speed = 12.0 m/s Change in time, △t = 4.0 secondsWe define acceleration thus :
Acceleration = (△velocity / △time) Change in velocity = (12 - 2) = 10 m/s Acceleration = 10 / 4 = 2.5 m/s²Therefore, the acceleration of the bike will be 2.5m/s² in the West direction
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Goslo and Speedo are twins. They have just celebrated their 20th birthday (Yay!). Speedo travels to the planet Zorth which is 40 light years away, he immediately returns. Neglect any time Speedo spent accelerating. If Speedo travels at 0.997 relative to the Earth, who will measure the proper time?
a. Goslo
b. Speedo
c. Nobody
d. Everybody
Answer:
a. Goslo
Explanation:
given data
Speedo travels = 40 light year
Speedo travels relative to the Earth = 0.997
solution
Goslo will measure here proper time and this time is expres as
time T = Speedo travels ÷ Speedo travels relative to the Earth ..........................1
put here value
time T = [tex]\frac{40}{0.997}[/tex]
time T = 40.1 year
so Goslo will measure the proper time
One train is at the 20 Km position, another is at the 200Km position, both are approaching, the first with a speed of 30Kmh-1 and the second with a speed of 48 Kmh-1. If they leave at 7:12, what time do they cross? Answer: t = 2.30 H Time: 9:30 am.
Explanation:
The distance between them is 200 km − 20 km = 180 km.
The relative velocity is 30 km/h − (-48 km/h) = 78 km/h.
The time it takes is 180 km / (78 km/h) = 2.31 hours, or 2 hrs 18 min.
Therefore, the trains meet at 9:30 AM.
light of wavelength 610 nm is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central meximum and the first minimum on the other side is 1.23°. What is the width of the slit?
Answer:
The width of the slit is [tex]d = 5.68 *10^{-5} \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 610 \ nm = 610 *10^{-9} \ m[/tex]
The angle is [tex]\theta = 1.23 ^o[/tex]
Generally the angle between the first minimum on one side and that the central maximum is evaluated as
[tex]\theta _1 = \frac{\theta}{2}[/tex]
=> [tex]\theta _1 = \frac{1.23}{2}[/tex]
=> [tex]\theta _1 = 0.615 ^o[/tex]
Generally the condition for minimum diffraction is mathematically represented as
[tex]d sin \theta_1 = n\lambda[/tex]
For first minimum n = 1
=> [tex]d = \frac{n \lambda }{ sin (\theta_1)}[/tex]
=> [tex]d = \frac{1 * 610 *10^{-9}}{ sin (0.615)}[/tex]
=> [tex]d = 5.68 *10^{-5} \ m[/tex]
What is process in which long thin strips of flexible ribbon are produced by pouring onto a flat surface?
Answer:
Tape casting
Explanation:
Tape casting is also known as knife coating. This process is used specially in the production of thin surfaces of a ceramic work.
Tape casting is also the process which which involves long thin strips of flexible ribbon being produced by pouring onto a flat surface. It is then left to dry by applying a high temperature and then ready for use.
What is the wavelength of the electromagnetic radiation needed to eject electrons from a metal?
Answer:
λ = hc/(eV + h[tex]f_{0}[/tex])
Explanation:
Let the work function of the metal = ∅
the kinetic energy with which the electrons are ejected = E
the energy of the incident electromagnetic wave = hf
Then, we know that the kinetic energy of the emitted electron will be
E = hf - ∅
because the energy of the incident electromagnetic radiation must exceed the work function for electrons to be ejected.
This means that the energy of the incident e-m wave can be written as
hf = E + ∅
also, we know that the kinetic energy of the emitted electron E = eV
and the work function ∅ = h[tex]f_{0}[/tex]
we can they combine all equations to give
hf = eV + h[tex]f_{0}[/tex]
we know that f = c/λ
substituting, we have
hc/λ = eV + h[tex]f_{0}[/tex]
λ = hc/(eV + h[tex]f_{0}[/tex]) This is the wavelength of the e-m radiation needed to eject electrons from a metal.
where
λ is the wavelength of the e-m radiation
h is the Planck's constant = 6.63 x 10^-34 m^2 kg/s
c is the speed of e-m radiations in a vacuum = 3 x 10^8 m/s
e is the charge on an electron
V is the voltage potential on the electron
[tex]f_{0}[/tex] is the threshold frequency of the metal
What is the difference between
Dobereiner's triads and Newland's octaves
Answer:
Newlands law of octave was the first logically basis on the atomic weight. Demerits of dobereiner's triads system: - This system fails because it does not hold good for all the elements discovered till date time. It is limited to only few elements.
Explanation:
Planet Tatoone is about 1.7 AU from its Sun. Approximately how long will it take for light to travel from the Sun to Tatoone in minutes? Use 3 × 108 m/s for the speed of light.
Answer:
The value is [tex]t = 14.129 \ minutes[/tex]
Explanation:
From the question we are told that
The distance of planet Tatoone is [tex]d = 1.7 \ AU = 1.7 *1.496* 10^{11}=2.543*10^{11} \ m[/tex]
The speed of light is [tex]c = 3.0*10^{8} \ m/ s[/tex]
Generally the time taken is mathematically represented as
[tex]t = \frac{d}{c}[/tex]
=> [tex]t = \frac{2.543*10^{11}}{3.0*10^{8} }[/tex]
=> [tex]t = 847.7 \ s[/tex]
Now converting to minutes
[tex]t = \frac{847.7}{60}[/tex]
=> [tex]t = 14.129 \ minutes[/tex]
A rectangular trough, 1.6 m long, 0.60 m wide, and 0.35 m deep, is completely full of water. One end of the trough has a small drain plug right at the bottom edge.
When you pull the plug, at what speed does water emerge from the hole?
Answer:
v = 2.61 m/s
Explanation:
Given that,
Length of a rectangular trough, l = 1.6 m
Breadth of a rectangular trough, b = 0.6 m
Depth of a rectangular trough, h = 0.35 m
We need to find the speed with which water emerge from the hole. It is a concept of efflux. The velocity of efflux is given by :
[tex]v=\sqrt{2gh}\\\\v=\sqrt{2\times 9.8\times 0.35}\\\\v=2.61\ m/s[/tex]
So, water will emerge from the hole with a speed of 2.61 m/s.