Answer:
B (z) = 2,467 10⁻¹¹ cos (1,486 10⁸ x - 5.571 10¹⁵ t )
Explanation:
An electromagnetic wave is a wave that is sustained in the perpendicular fluctuations of the electric and magnetic fields, the equation of the wave is
E (y) = Eo cos (kx –wt)
B (z) = Bo cos (kx-wt)
Let's look for the terms to build these equations. The speed of the wave is given by
c = λ f
The frequency and period are related
f = 1 / T
Let's start by applying this equation our case
f = c /λ
f = 3 10⁸/225 10⁻⁹
f = 1.33 10¹⁵ Hz
The angular velocity and the wave number are
w = 2π f
k = 2π /λ
w = 2π 1.33 10¹⁵ = 8.38 10¹⁵ rad / s
k = 2π / 225 10⁻⁹ = 2.79 10⁷ m⁻¹
It indicates that the period increases by a factor of 1.5, let's look for the new frequency
T = 1.5 T₀
f = 1 / T
f = 1 / 1.5T₀
f = 1 / 1.5 f₀
f = 1 / 1.5 1.33 10¹⁵ = 8.87 10¹⁴ Hz
c = λ f
λ = c / f
λ = 3 10⁸ / 8.87 10¹⁴ = 4,229 10⁻⁸ m
Let's find the new w and k
w = 2π f
w = 2π 8.87 10¹⁴ = 5.571 10¹⁵ rad/s
k = 2π / λ
k = 2π / 4,229 10⁻⁸ = 1,486 10⁸ cm⁻¹
We use the relationship that the fields are in phase
c = E₀ / B₀
B₀ = E₀ / c
B₀ = 7.4 10⁻³ / 3 10 ⁸ = 2.467 10⁻¹¹ T
With these values we can build the equation of the magnetic field
B (z) = 2,467 10⁻¹¹ cos (1,486 10⁸ x - 5.571 10¹⁵ t )
While riding in a moving car, you toss an egg directly upward. Does the egg tend to land behind you,in front of you,or back in your hands if the car is
(a) traveling at a constant speed,
(b) increasing in speed,and
(c) decreasing in speed
Answer:
a) In the hand
b) behind the hand
c) in front of the hand
Explanation:
a)
When traveling at a constant speed when an egg is tossed directly upwards it lands back in our hand. It is so because the instant when the egg is released from the hand the egg has the same velocity as the velocity of the person sitting in the moving car, and hence by the virtue of inertia of motion it land lands back in our hand.
b)
When we toss an egg while the car is accelerating then the egg lands behind the hand because the hand along with the car is accelerating ahead. The instant when the egg is released from the hand the car is accelerating then the car has a lower velocity which the egg acquires in the air, while the egg is in the air the car further gains the velocity and hence the egg lands behind the hand.
C)
When the speed of the car is decreasing then the egg lands in front of the hand because the moment when the egg was tossed into the air it acquired the instantaneous velocity of the car from that moment and then while it was in the air it had the same velocity as initial while the car further decelerated.
Calculate the energy per photon and the energy per mole of photons for radiation of wavelength (a) 200 nm (ultraviolet), (b) 150 pm (X-ray), (c) 1.00 cm (microwave).
Answer: a. E =9.9*EXP(-19)J
1 mole E= 596178J
b. E= 1.32*EXP(-15)J, 1 mole E=795MegaJ
c. E= 1.98*EXP(-23)J
1 mole E = 11.9J
Explanation: The Energy of a photon E, the wavelength are related by
E= h*c/wavelength
h is the Planck's constant 6.6*EXP(-34)J.s
c is speed of light 3*EXP(8)m/s
h*c=1.98*EXP(-25)
Now let's solve
a. E = h*c/wavelength
= h*c/(200*EXP(-9)m
=9.9*EXP(-19)J
1 mole of a photon contian 6.022*EXP(23)photons by advogadro
Now to get the energy of 1 mole of the photon we have
9.9*EXP(-19)*6.023*EXP(23)
=596178J
b. E=h*c/150*EXP(-12)m
=1.32*EXP(-15)J
1 mole will have
1.32*EXP(-15)*6.022*EXP(23)J
=795*EXP(6)J
c. E= h*c/1*EXP(-2)m
=1.98*EXP(-23)J
1 mole of the photon will have
1.98*EXP(-23)J *6.022*EXP(23)
= 11.9J.
You will notice that the longer the wavelength of the photon the lesser the Energy it as.
NOTE: EXP represent 10^
Final answer:
The energy per photon and per mole for ultraviolet radiation at 200 nm, X-ray radiation at 150 pm, and microwave radiation at 1.00 cm can be calculated using Planck's equation, with the resulting energies being 9.939 x 10^-19 J, 1.327 x 10^-15 J, and 1.987 x 10^-23 J respectively, and the energies per mole being 5.98 x 10^5 J/mole, 7.99 x 10^8 J/mole, and 1.20 J/mole respectively.
Explanation:
To calculate the energy per photon and the energy per mole of photons for radiation of various wavelengths, we can use the relationship given by Planck's equation, which relates the energy (E) of a photon to its wavelength (λ):
E = hc / λ
where h is Planck's constant (6.626 x 10-34 J•s), c is the speed of light in a vacuum (3.00 x 108 m/s), and λ is the wavelength of the radiation. To find the energy per mole of photons, we can multiply the energy per photon by Avogadro's number (6.022 x 1023 photons/mole). Let's calculate the energy for each type of radiation:
(a) Ultraviolet (200 nm): E = (6.626 x 10-34 J•s)(3.00 x 108 m/s) / (200 x 10-9 m) = 9.939 x 10-19 J(b) X-ray (150 pm): E = (6.626 x 10-34 J•s)(3.00 x 108 m/s) / (150 x 10-12 m) = 1.327 x 10-15 J(c) Microwave (1.00 cm): E = (6.626 x 10-34 J•s)(3.00 x 108 m/s) / (1.00 x 10-2 m) = 1.987 x 10-23 JTo get the energy per mole, multiply each of these values by Avogadro's number:
Energy per mole for 200 nm UV light: 9.939 x 10-19 J/photon x 6.022 x 1023 photons/mole = 5.98 x 105 J/moleEnergy per mole for 150 pm X-ray light: 1.327 x 10-15 J/photon x 6.022 x 1023 photons/mole = 7.99 x 108 J/moleEnergy per mole for 1.00 cm microwave radiation: 1.987 x 10-23 J/photon x 6.022 x 1023 photons/mole = 1.20 J/moleA 19 cm long, 10 mm diameter string was found to stretch 3.5 mm under a force of 16 N. The string breaks at a force of 42 N. What is the stress in the string What is the strain in the string under a 13 N load?What is the modulus of elasticity of the string? (include units with answer) What is the ultimate strength D. (stress) of the string?
Answer:
Stress = 165,521 Pa
Strain = 0.015
Modulus of Elasticity = 11,034,742.72 Pa
Ultimate Tensile Stress = 534,761 Pa
Explanation:
Stress @ Force = 13 N
F = kΔx .... Eq 1
Evaluate k @ F = 16 N and Δx = 0.0035m
k = 16 N / 0.0035 m = 4571.43 N/m
Stress = Force / Area ..... Eq 2
Stress = 13 N *4 / π (0.01²) = 165,521.1408 Pa
Strain @ Force = 13 N
Strain = Δx / original Length = (F / k) / original Length
Hence,
Strain = (13 N / 4571.43 N / m) / 0.19 m = 0.015
Modulus of Elasticity (E)
E = Stress / Strain
E = 165,521.1408 Pa / 0.015 = 11,034,742.72 Pa
Ultimate Tensile Stress
Max Stress that can be achieved by a material
Stress = Fmax/Area
Stress = 42 N *4 / π (0.01²) = 534,761 Pa
Find the voltage change when: a. An electric field does 12 J of work on a 0.0001-C charge. b. The same electric field does 24 J of work on a 0.0002-C charge.
Explanation:
Given that,
(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :
[tex]V=\dfrac{W}{q}[/tex]
[tex]V=\dfrac{12}{0.0001}[/tex]
[tex]V=12\times 10^4\ Volt[/tex]
(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :
[tex]V=\dfrac{W}{q}[/tex]
[tex]V=\dfrac{24}{0.0002}[/tex]
[tex]V=12\times 10^4\ Volt[/tex]
Therefore, this is the required solution.
Later that day, Martin does the same thing with Josh, who is 2 times heavier than Martin. If the collision is totally inelastic, what height does Josh reach?
a. h/25
b. h/16
c. h/8
d. h/4
e. h/2
f. h
g. None of the above.
Josh, being 2 times heavier than Martin, reaches a height [tex]\( \frac{h}{16} \)[/tex] in a totally inelastic collision, making the correct answer (b).
In a totally inelastic collision, two objects stick together and move with a common final velocity. The conservation of linear momentum can be applied to find the final velocity and subsequently determine the height reached by the combined mass.
Let's denote the mass of Martin as [tex]\(m_M\)[/tex] and the mass of Josh as [tex]\(m_J\)[/tex], and let [tex]\(h_M\)[/tex] and [tex]\(h_J\)[/tex] be the heights they reach, respectively.
The conservation of linear momentum equation is given by:
[tex]\[ m_M \cdot v_M + m_J \cdot v_J = (m_M + m_J) \cdot v_f \][/tex]
where [tex]\(v_M\)[/tex] and [tex]\(v_J\)[/tex] are the initial velocities of Martin and Josh, and [tex]\(v_f\)[/tex] is their final velocity.
Since Martin and Josh stick together after the collision, their final velocity is the same [tex](\(v_f\))[/tex].
Now, given that Josh is 2 times heavier than Martin [tex](\(m_J = 2 \cdot m_M\))[/tex], we can express the initial momentum as:
[tex]\[ m_M \cdot v_M + (2 \cdot m_M) \cdot v_J = (3 \cdot m_M) \cdot v_f \][/tex]
Since the collision is totally inelastic, the final velocity is the same for both, and we can write:
[tex]\[ v_f = \frac{m_M \cdot v_M + 2 \cdot m_M \cdot v_J}{3 \cdot m_M} \][/tex]
Now, let's consider the conservation of energy. The potential energy lost during the collision is converted into kinetic energy, and finally, into potential energy again at the maximum height reached by the combined mass.
[tex]\[ m_M \cdot g \cdot h_M + m_J \cdot g \cdot h_J = \frac{1}{2} \cdot (m_M + m_J) \cdot v_f^2 \][/tex]
Substitute the expression for [tex]\(v_f\)[/tex] from the momentum equation into the energy equation:
[tex]\[ m_M \cdot g \cdot h_M + (2 \cdot m_M) \cdot g \cdot h_J = \frac{1}{2} \cdot (3 \cdot m_M) \cdot \left(\frac{m_M \cdot v_M + 2 \cdot m_M \cdot v_J}{3 \cdot m_M}\right)^2 \][/tex]
Simplify the equation and solve for [tex]\(h_J\)[/tex]. The final result should be [tex]\(h_J = \frac{h}{16}\)[/tex], so the correct answer is (b).
A 1-kg iron block is to be accelerated through a pro- cess that supplies it with 1 kJ of energy. Assuming all this energy appears as kinetic energy, what is the final velocity of the block? b. If the heat capacity of iron is 25.10 J/(mol K) and the molecular weight of iron is 55.85, how large a temperature rise would result from 1 kJ of energy supplied as heat?
Answer:
a) v = 44.72 m/s
b) temperature rise = 2.22 K
Explanation:
A)
Data provided in the question:
Mass of the iron block = 1 kg
Kinetic Energy supplied = 1 KJ = 1000 J
Now,
Kinetic energy = [tex]\frac{1}{2}mv^2[/tex]
here,
v is the velocity
thus,
1000 = [tex]\frac{1}{2}\times1\times v^2[/tex]
or
v² = 2000
or
v = 44.72 m/s
b)
Heat capacity of iron = 25.10 J/(mol K)
Molecular weight of iron = 55.85
Energy supplied = 1 kJ = 1000 J
Now,
Energy supplied = Mass × C × Change in temperature
Here, C = Heat capacity of iron ÷ Molecular weight of iron
= 25.10 ÷ 55.85
= 0.45 J/(g.K)
thus,
1000 = 1 kg × 0.45 × Change in temperature
or
1000 = 1000 g × 0.45 × Change in temperature
or
Change in temperature i.e temperature rise = 2.22 K
A source emits sound uniformly in all directions. There are no reflections of the sound. At a distance of 12 m from the source, the intensity of the sound is 3.5 × 10-3 W/m2. What is the total sound power P emitted by the source?
Answer:
1.58 W
Explanation:
Since the sound spreads uniformly in all directions, it must be in a form of a circle with radius of 12 m. So the area of the circle is
[tex]A = \pi r^2 = \pi 12^2 = 452.389 m^2[/tex]
From the intensity of the sound we can calculate the power at 12 m
[tex]P = AI = 452.389 * 3.5\times10^{-3} = 1.58 W[/tex]
Two force vectors are perpendicular, that is, the angle between their directions is ninety degrees and they have the same magnitude. If their magnitudes are 655 newtons, then what is the magnitude of their sum?
Answer:
[tex]F=926.31N[/tex]
Explanation:
To calculate the magnitude of the sum of the vectors we must find their components on the x and y axes, in this case, since they are perpendicular, one vector is on the x axis and the other is on the y axis:
[tex]F=\sqrt{F_x^2+F_y^2}\\F_x=A_x+B_x=A\\F_y=A_y+B_y=B\\F=\sqrt{A^2+B^2}\\F=\sqrt{(655N)^2+(655N)^2}\\F=926.31N[/tex]
A block attached to a spring with unknown spring constant oscillates with a period of 7.0 s. Parts a to d are independent questions, each referring to the initial situation. What is the period if
a. The mass is halved?
b. The amplitude is doubled?
c. The spring constant is doubled?
d. The mass is doubled?
a) 4.95 s when the mass is halved
b) 7.00 s when the amplitude is doubled
c) 4.95 s when spring constant is doubled
d) 9.9 s when mass is doubled
Given:
T₀=7.0 s
The equation for the time period of the object is
[tex]T=2*\pi *\sqrt{\frac{m}{k} }[/tex] ......................(i)
where m= mass of an object and k= spring constant
For solving a.m= 1/2
∵T=7.0 s
On substituting value in equation (i)
Thus, T=4.95 s
For solving b.Time period does not depend on amplitude thus,
T=7.0 s
For solving c.Spring constant, k=2k
On substituting value in equation (i)
Thus, T=4.95 s
For solving d.Mass=2m
On substituting value in equation (i)
Thus, T=9.9 s
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The period of a mass-spring oscillating system changes based on variations in mass and spring constant but not amplitude. If the mass is halved or the spring constant is doubled, the period decreases to approximately 4.95 s. If the mass is doubled, the period increases to approximately 9.90 s.
Explanation:The period of oscillation for a mass attached to a spring is given by the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. Keeping this formula in mind, let's evaluate how changes to the system will affect the period of oscillation:
a. Halving the mass: Since the period is directly proportional to the square root of the mass, if the mass is halved, the period will be reduced by a factor of the square root of 1/2, which is approximately 0.707. Therefore, the new period would be 7.0 s * 0.707 ≈ 4.95 s.b. Doubling the amplitude: The amplitude does not affect the period of a simple harmonic oscillator, so the period will remain the same at 7.0 s.c. Doubling the spring constant: The period is inversely proportional to the square root of the spring constant. When k is doubled, the period becomes 1/√2 times the original period. Therefore, it will be 7.0 s / √2 ≈ 4.95 s.d. Doubling the mass: Similar to halving the mass, doubling the mass would increase the period by √2. The new period would be 7.0 s * √2 ≈ 9.90 s.
The largest and smallest slits in these experiments are 0.16mm and 0.04mm wide, respectively. The wavelength of the laser light is 650nm. How many wavelengths wide are these slits?
To solve this problem we will convert the values given to international units (meters) and then proceed to calculate the number of wavelengths through the division of the width over the wavelength. This is,
[tex]\lambda = 650 nm =650 * 10^{-9} m[/tex]
[tex]w_1 = 0.16 mm = 0.16 * 10^{-3} m[/tex]
[tex]w_2 = 0.04 mm =0.04 * 10^{-3} m[/tex]
Now the number of wavelengths is the division between the total width over the wavelength therefore
First case,
[tex]\frac{w_1}{\lambda} = \frac{ 0.16 * 10^{-3}}{650 * 10^{-9}} = 4062.5 * 10^6[/tex]
Second case,
[tex]\frac{w_2}{\lambda} = \frac{0.04 * 10^{-3} }{650 * 10^{-9} } = 16250 * 10^6[/tex]
A child is riding a merry-go-round that is turning at 7.18 rpm. If the child is standing 4.65 m from the center of the merry-go-round, how fast is the child moving?A) 5.64 m/s B) 3.50 m/s C) 0.556 m/s D) 1.75 m/s E) 1.80 m/s
Answer:
B) 3.50 m/s
Explanation:
The linear velocity in a circular motion is defined as:
[tex]v=\omega r(1)[/tex]
The angular frequency ([tex]\omega[/tex]) is defined as 2π times the frequency and r is the radius, that is, the distance from the center of the circular motion.
[tex]\omega=2\pi f(2)[/tex]
Replacing (2) in (1):
[tex]v=2\pi fr[/tex]
We have to convert the frequency to Hz:
[tex]7.18rpm*\frac{1Hz}{60rpm}=0.12Hz[/tex]
Finally, we calculate how fast is the child moving:
[tex]v=2\pi(0.12Hz)(4.65m)\\v=3.5\frac{m}{s}[/tex]
Lightning As a crude model for lightning, consider the ground to be one plate of a parallel-plate capacitor and a cloud at an altitude of 650 m to be the other plate. Assume the surface area of the cloud to be the same as the area of a square that is 0.70 km on a side. a. What is the capacitance of this capacitor?b. How much charge can the cloud hold before the dielectric strength of the air is exceeded and a spark (lightning) results?
Answer:
[tex]6.67154\times 10^{-9}\ F[/tex]
13.009503 C
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
k = Dielectric constant of air [tex]3\times 10^6\ V/m[/tex]
Side of plate = 0.7 km
A = Area
d = Distance = 650 m
Capacitance is given by
[tex]C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 700^2}{650}\\\Rightarrow C=6.67154\times 10^{-9}\ F[/tex]
The capacitance is [tex]6.67154\times 10^{-9}\ F[/tex]
Electric field is given by
[tex]Q=CV\\\Rightarrow Q=Ckd\\\Rightarrow Q=6.67154\times 10^{-9}\times 3\times 10^6\times 650\\\Rightarrow Q=13.009503\ C[/tex]
The charge on the cloud is 13.009503 C
The capacitance of the cloud-ground system is 6.67 nF, and the cloud can hold up to 13 C of charge before lightning occurs.
To model lightning using a parallel-plate capacitor, assume the ground is one plate and a cloud at 650 m altitude is the other. Let’s calculate the capacitance and the charge the cloud can hold before a lightning strike occurs.
We have the height h = 650 m and the side length L of the cloud's surface area is 0.70 km (700 m), making the area A = 700 m × 700 m = 490,000 m².The formula for capacitance C of a parallel-plate capacitor is:Therefore, the capacitance of the capacitor formed by the cloud and the ground is 6.67 nF, and the cloud can hold up to 13 C of charge before a lightning strike occurs.
Suppose a cup of coï¬ee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees at t = 20 minutes. Compute the ambient temperature.
Final answer:
To compute the ambient temperature, we can use the principle of heat transfer. By applying the formula Q = mcΔT, we can set up two equations to equate the heat lost by the coffee to the heat gained by the surroundings. Solving these equations will give us the ambient temperature.
Explanation:
To compute the ambient temperature, we can use the principle of heat transfer. The heat lost by the coffee is equal to the heat gained by the surroundings. We can use the formula Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
By applying this formula to the coffee, we can calculate the heat lost from t = 0 to t = 10 minutes and from t = 10 minutes to t = 20 minutes, and then equate it to the heat gained by the surroundings.
Let's assume the ambient temperature is T. Using the formula, we can set up the following equations:
Heat lost from t = 0 to t = 10 minutes = Heat gained by the surroundings: (350 g)(4.18 J/g°C)(100°C - T) = (350 g)(4.18 J/g°C)(T - 25°C) Heat lost from t = 10 minutes to t = 20 minutes = Heat gained by the surroundings: (350 g)(4.18 J/g°C)(70°C - T) = (350 g)(4.18 J/g°C)(T - 25°C)By solving these equations, we can find the value of T, which is the ambient temperature.
When you are painting a car with a sprayer, you get more even coverage (the paint droplets are evenly spread out) and less overspray (there are very few droplets of paint that don’t end up on the vehicle) if you give the droplets a strong electric charge.
a. Why does charging the droplets give more even coverage?
b. Why does charging the droplets help ensure that most of the paint ends up on the car?
PART A) To give an approximation to the microscopic behavior of the drops, we must refer to Coulomb's law for which it is warned that similar charges repel. This spraying technique is used because when it starts to spray and expel the drops, they leave with the same loads and will be repelled, causing the layer of drops to be thinner and not thick drops. The lack of conglomeration of large drops will cause the paint to be finer.
PART B) When the object is loaded differently, even if the paint is shot in a different direction, it will end up adhered to the body by the action of this attraction
Charging paint droplets with an electric charge helps in achieving more even coverage and reducing overspray while painting a car.
Explanation:a. When paint droplets are given a strong electric charge, they repel each other due to their like charges. This causes the droplets to spread out more evenly, resulting in a more uniform coverage on the car's surface.
b. By charging the droplets, they are attracted to the grounded surface of the car. This attraction helps ensure that most of the paint droplets move towards the car and stick to the surface, reducing the amount of overspray.
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Two charges Q =2 C and q=1 nC are 1 m apart. If the electric force beween them is 18 N what is the magnitude of the electric field due to Q at q. The format of the answer must be g
Answer:
[tex]E=1.8\times 10^{10}\ N.C^{-1}[/tex]
Explanation:
Given:
one charge, [tex]Q=2\ C[/tex]another charge, [tex]q=10^{-9}\ C[/tex]distance between the two charges, [tex]r=1\ m[/tex]force between the charges, [tex]F=18\ N[/tex]We know from the Coulomb's law:
[tex]F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}[/tex] ...........(1)
where:
[tex]\epsilon_0=[/tex] permittivity of free space
Also we have electric field due to Q at q (which is at a distance of 1 m):
[tex]E=\frac{1}{4\pi.\epsilon_0} \times \frac{Q}{r^2}\ [N.C^{-1}][/tex] ...........(2)
From (1) & (2)
[tex]E=\frac{F}{q}[/tex]
[tex]E=\frac{18}{10^{-9}}[/tex]
[tex]E=1.8\times 10^{10}\ N.C^{-1}[/tex]
Answer:
magnitude of the electric field = 1.8 × [tex]10^{10}[/tex] N/C
Explanation:
given data
charge Q =2 C
charge q = 1 nC = 1 × [tex]10^{-9}[/tex] C
distance r = 1 m
electric force = 18 N
solution
we know that here electric field due to Q is
E = k × [tex]\frac{Q}{r^2}[/tex] ..............1
here q distance is 1 m
now we apply here Coulomb’s law and get here electric force
F = k ×[tex]\frac{Q*q}{r^2}[/tex] ..........2
we know constant k = 8.988 × [tex]10^{9}[/tex] Nm²/C²
so from above both equation we get
electric filed = [tex]\frac{force}{charge}[/tex] .................3
put here value
electric filed = [tex]\frac{18}{1*10^{-9}}[/tex] = 1.8 × [tex]10^{10}[/tex] N/C
In an experiment you are performing, your lab partner has measured the distance a cart has traveled: 28.4 inch You need the distance in units of centimeter and you know the unit equality 1 inch 2.54 centimeter. By which conversion factor will you multiply 28.4 inch in order to perform the unit conversion?
To convert inches to centimeters, you multiply the number of inches by 2.54 (the conversion factor). Therefore 28.4 inches is approximately 72.1 centimeters.
Explanation:In this physics problem, you are trying to change a distance measurement from inches to centimeters. Given the unit equality, 1 inch is equal to 2.54 centimeters. Therefore, to convert the measurement of 28.4 inches to centimeters, you need to multiply it by the conversion factor of 2.54 (i.e. 1 inch = 2.54 cm). So, you will simply multiply 28.4 inches by 2.54 to get the distance in centimeters: 28.4 inches * 2.54 cm/inch = approximately 72.1 centimeters.
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how many times faster is the speed of sound in the fluid compared to the peak amplitutde of the particle velocity
Answer: 171500 times faster.
Explanation: First let's take our Fluid medium to be air. The speed of sound in air is 343m/s or 343000mm/s.
Now, particle velocity is that which shows the alternating mean velocity of a particle in a medium(say in this case air).
A particle of air exposed to a standard sound pressure of 1 Pascal as a peak velocity amplitude of about 2mm/s.
Recall that the value of speed of sound in Air when converted to mm/s is 343000mm/s.
To get the number of times it's faster than the peak particle Velocity of air. We have to divide it by peak particle Velocity of air which is 2mm/s.
343000/2
=171500 times faster.
I believe this is clear.
Large wind turbines with a power capacity of 8 MW and blade span diameters of over 160 m are available for electric power generation. Consider a wind turbine with a blade span diameter of 100 m installed at a site subjected to steady winds at 8 m/s. Taking the overall efficiency of the wind turbine to be 34 percent and the air density to be 1.25 kg/m3, determine the electric power generated by this wind turbine. Also, assuming steady winds of 8 m/s during a 24 hour period, determine the amount of electric energy and the revenue generated per day for a unit price of $0.09/kWh for electricity.
To determine the electric power generated by the wind turbine, use the formula P = 0.5 x ρ x A x V^3 x η. Calculate the swept area of the blades using A = π x (d/2)^2. Multiply the power by the number of seconds in a day and convert it to kilowatt-hours to find the energy generated per day. Multiply the energy by the unit price of electricity to determine the revenue generated per day.
Explanation:To determine the electric power generated by the wind turbine, we need to use the formula P = 0.5 x ρ x A x V^3 x η, where P is the power, ρ is the air density, A is the swept area of the blades, V is the wind speed, and η is the efficiency of the turbine.
First, calculate the swept area of the blades using the formula A = π x (d/2)^2, where d is the diameter of the blade span.Next, substitute the given values into the formula and calculate the power.To determine the amount of electric energy generated in a day, multiply the power by the number of seconds in a day (24 hours x 60 minutes x 60 seconds), then convert it to kilowatt-hours.To calculate the revenue generated per day, multiply the amount of electric energy by the unit price of electricity.Learn more about Wind Turbine Power Generation here:https://brainly.com/question/34425549
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Two thin conducting plates, each 24.0 cm on a side, are situated parallel to one another and 3.4 mm apart. If 1012 electrons are moved from one plate to the other, what is the electric field between the plates?
Answer:
E=3.307×10⁻⁴N/C
Explanation:
Given data
Length of the plate side L=24.0 cm =0.24 m
Distance between the plates d= 3.4 mm
Number of electron moves from one plate to others n=1012 electrons
Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²
Electron charge e=-1.6×10⁻¹⁹C
To find
Electric field between the plates
Solution
E=σ/ε₀
[tex]E=\frac{(Q/A)}{E}\\ E=\frac{Q}{EA}\\ E=\frac{ne}{EA}\\ E=\frac{1012*()1.6*10^{-19} }{8.85*10^{-12}(0.24)(0.24)}\\ E=3.307*10^{-4}N/C[/tex]
A raft is made of 14 logs lashed together. Each log is 42 cm in diameter and a length of 6.4 m. 42% of the log volume is above the water when no one is on the raft. Determine the following: the specific gravity of the logs.
Answer:
Explanation:
Given
No of logs [tex]n=14[/tex]
diameter of log [tex]d=42\ cm[/tex]
Length of log [tex]L=6.4\ m[/tex]
42 % of log volume(V) is above water when no one is on raft
so 58 % of log volume(V) is submerged in the water
Weight of 14 log
[tex]W=14\times \rho _{log}\times V\times g[/tex]
Buoyancy force on 14 logs [tex]F_b=14\times \rho _{water}\times 0.58V\times g[/tex]
as system is in equilibrium so
[tex]W=F_b[/tex]
[tex]14\times \rho _{log}\times V\times g=14\times \rho _{water}\times 0.58V\times g[/tex]
[tex]\rho _{log}=0.58\rho _{water}[/tex]
[tex]\frac{\rho _{log}}{\rho _{water}}=0.58[/tex]
Specific gravity of log [tex]=0.58[/tex]
You are standing at rest at a bus stop. A bus moving at a constant speed of 5.00 m>s passes you. When the rear of the bus is 12.0 m past you, you realize that it is your bus, so you start to run 2 toward it with a constant acceleration of 0.960 m>s .
A) How far would you have to run before you catch up with the rear of the bus?
B) How fast must you be running then? Find the final speed just as you reach the bus.
C) Would an average college student be physically able to accomplish this? Yes or No
Answer:
A) You have to run 73.8 m before you reach the rear of the bus.
B) When you catch the bus, your velocity will be 11.9 m/s.
C) No, an average college student would not reach the bus.
Explanation:
Hi there!
The equations of velocity and position that we will use are the following:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
v = velocity at time t.
A) First, let´s find the time at which you catch the bus. At this time, the position of the bus and yours are the same:
When you catch the bus:
Bus position = your position
The equation of position of the bus with constant velocity is the following:
x = x0 + v · t
You have a constant acceleration, so your position will be given by the equation of position described above:
x = x0 + v0 · t + 1/2 · a · t²
If we place the origin of the frame of reference at the bus stop, your initial position is zero (and the initial position of the bus is 12.0 m). Since you start from rest, your initial velocity is also zero. Then:
Bus position = your position
x0 + v · t = 1/2 · a · t²
12.0 m + 5.00 m/s · t = 1/2 · 0.960 m/s² · t²
0 = 0.48 m/s² · t² - 5.00 m/s · t - 12.0 m
Solving the quadratic equation for t:
t = 12.4 s
You catch the bus in 12.4 s, now, we have to find the distance you run in that time:
x = x0 + v0 · t + 1/2 · a · t²
Since x0 and v0 = 0
x = 1/2 · a · t²
x = 1/2 · 0.960 m/s² · (12.4 s)²
x = 73.8 m
You have to run 73.8 m before you reach the rear of the bus.
B) Now, let´s use the equation of velocity to find your velocity at t = 12.4 s.
v = v0 + a · t
Since v0 = 0:
v = a · t
v = 0.960 m/s² · 12.4 s
v = 11.9 m/s
When you catch the bus, your velocity will be 11.9 m/s.
C) The record of Usain Bolt is 12.4 m/s. You would have to run almost as fast as Usain Bolt to catch the bus. So the answer is no, an average college student would not reach the bus.
The value of the gravitational acceleration g decreases with elevation from 9.807 m/s2 at sea level to 9.757 m/s2 at an altitude of 13,000 m, where large passenger planes cruise. Determine the percent reduction in the weight of an airplane cruising at 13,000 m relative to its weight at sea level.
Answer:
0.51 %
Explanation:
Since mass is the same at sea level and at 13000 m. And weight is the product of mass and gravitational acceleration g, the percent reduction in the weight of an airplane cruising at 13,000 m relative to its weight at sea level is essentially the percent reduction in the gravitational acceleration of an airplane cruising at 13,000 m relative to its gravitational acceleration at sea level, which is:
1 - 9.757 / 9.807 = 1 - 0.995 = 0.0051 = 0.51 %
How high will a 0.345 kg rock go if thrown straight up by someone who does 111 J of work on it? Neglect air resistance.
To solve this problem we will apply the concepts related to energy conservation. Here we will understand that the potential energy accumulated on the object is equal to the work it has. Therefore the relationship that will allow us to calculate the height will be
[tex]W = PE[/tex]
[tex]W = mgh[/tex]
Here,
m = mass
g = Acceleration due to gravity
h = Height
our values are,
[tex]m = 0.345 kg[/tex]
[tex]g = 9.8m/s^2[/tex]
[tex]W = 111J[/tex]
Replacing,
[tex]W =mgh[/tex]
[tex]111 = (0.345)(9.8)h[/tex]
[tex]h = 32.83m[/tex]
Then the height is 32.83m.
Two blocks, 1 and 2, are connected by a rope R1 of negligible mass. A second rope R2, also of negligible mass, is tied to block 2. A force is applied to R2 and the blocks accelerate forward. Find the ratio of the forces exerted on blocks 1 and 2 by the ropes R1 and R2 respectively. Here m1 = 2.11m2.
Answer:
Explanation:
Given
Two block are connected by rope [tex]R_1[/tex]
[tex]R_2[/tex] rope is attached to block 2
suppose [tex]F_2[/tex] is a force applied to Rope [tex]R_2[/tex]
Applied force [tex]F_2[/tex]=Tension in Rope 2
[tex]F_2=(m_1+m_2)a---1[/tex]
where a=acceleration of system
Tension in rope [tex]R_1[/tex] is denoted by [tex]F_1[/tex]
[tex]F_1=m_1a---2[/tex]
divide 1 and 2 we get
[tex]\frac{F_2}{F_1}=\frac{(m_1+m_2)a}{m_1a}[/tex]
also [tex]m_1=2.11\cdot m_2[/tex]
[tex]\frac{F_2}{F_1}=\frac{2.11m_2+m_2}{2.11m_2}[/tex]
[tex]\frac{F_2}{F_1}=\frac{3.11}{2.11}[/tex]
[tex]\frac{F_1}{F_2}=\frac{2.11}{3.11}[/tex]
The ratio of the forces exerted on blocks 1 and 2 by the ropes R1 and R2 are 2.11 : 3.11.
Tension force:
When an object is connected with a rope such that the force exerted on the rope due to the weight of any object towards the upward direction, then such force is known as tension force.
Let R be the rope by which two blocks are connected. And R' is the rope attached to block 2.
Also, F' is the force applied to the rope R'. Then,
Applied force F' = Tension in the rope R'
[tex]F'=T\\\\ \dfrac{F'}{T}= \dfrac{(m_{1}+m_{2})a} {m_{1}a}[/tex]
here, [tex]m_{1}[/tex] and [tex]m_{2}[/tex] are the masses of ropes such that [tex]m_{1} = 2.11 \times m_{2}[/tex].
Solving as,
[tex]\dfrac{F'}{T}= \dfrac{(m_{1}+m_{2})a} {m_{1}a}\\\\ \dfrac{F'}{T}= \dfrac{(2.11m_{2}+m_{2})a} {2.11 \times m_{2}a}\\\\\\ \dfrac{F'}{T}= \dfrac{3.11}{2.11} [/tex]
or
T/F' = 2.11 / 3.11
Thus, we can conclude that the ratio of the forces exerted on blocks 1 and 2 by the ropes R1 and R2 are 2.11 : 3.11.
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A spaceship ferrying workers to moon Base i takes a straight-line pat from the earth to the moon, a distance of 384,000km, Suppose the spaceship starts from rest and accelerates at 20.0 meter per second squared (m/s2).for the first 15.0min of the trip, and then travels at constant speed until the last 15.0 min, when it slows down at a rate of 20.0 meter per second squared (m/s2),. Just coming to rest as it reaches the moon, (a) what is the maximum speed attained? (b) what fraction of the total distance is traveled at constant speed?? (c) what total time is required for the trip???
Answer:
95.78125%
18000 m/s
22233.33 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
[tex]v=u+at\\\Rightarrow v=0+20\times 15\times 60\\\Rightarrow v=18000\ m/s[/tex]
The velocity of the rocket at the end of the first 15 minutes is 18000 m/s which is the maximum speed of the rocket in the complete journey.
Distance traveled while speeding up
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 20\times 900^2\\\Rightarrow s=8100000\ m[/tex]
Distance traveled while slowing down
[tex]s=18000\times 900+\dfrac{1}{2}\times -20\times 900^2\\\Rightarrow s=8100000\ m[/tex]
Distance traveled during constant speed
[tex]384000000-(2\times 8100000)=367800000\ m[/tex]
Fraction
[tex]\dfrac{367800000}{384000000}\times 100=95.78125\ \%[/tex]
Fraction of the total distance is traveled at constant speed is 95.78125%
Time taken at constant speed
[tex]t=\dfrac{367800000}{18000}=20433.33\ s[/tex]
Total time taken is [tex]900+20433.33+900=22233.33\ s[/tex]
The maximum velocity of the spaceship is 18000 m/s. The spaceship travels 95.78% of the total distance at this constant speed. The total time required for the trip is 370 minutes.
To answer the first part of the question, we can use the physics formula for velocity, v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. As the spaceship starts from rest, u = 0. The acceleration a is given as 20 m/s² and the time t for which the spaceship is accelerating is 15min which is equal to 900 seconds. So, the maximum velocity, v = 0 + 20*900 = 18000 m/s.
For the second part, we understand that the spaceship spends an equal amount of time accelerating and decelerating, and the rest of the time it travels at a constant velocity. Therefore, during acceleration and deceleration, it follows a distance d = 1/2 * a * t², with a = 20 m/s² and t = 900 s. We find the distance covered during acceleration and deceleration, d = 1/2 * 20 * 900² = 8100000 m. Given the total distance from Earth to Moon is 384000000 m, the fraction of the distance at a constant speed is 1 - (2*8100000)/384000000 = 0.9578 or 95.78%.
Finally, to find the total time of the trip, we know that the time spent accelerating and decelerating is 15min + 15min = 30min. The distance covered at constant speed is 0.9578 * 384000000m = 367680000m, and the speed is 18000 m/s. So, the time at constant speed is 367680000 / 18000 = 20400s = 340 min. Therefore, the total time of the trip is 30min + 340min = 370 min.
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An electron moves to the right with a speed of 0.90c rela- tive to the laboratory frame. A proton moves to the right with a speed of 0.70c relative to the electron. Find the speed of the proton relative to the laboratory frame.
Answer:
The speed of the proton relative to the laboratory frame is 0.981c
Explanation:
Given that,
Speed of electron v= 0.90c
Speed of proton u= 0.70c
We need to calculate the speed of the proton relative to the laboratory frame
Using formula of speed
[tex]u'=\dfrac{u+v}{1+\dfrac{uv}{c^2}}[/tex]
Where, u = speed of the proton relative to the electron
v = speed of electron relative to the laboratory frame
Put the value into the formula
[tex]u'=\dfrac{0.70+0.90}{1+\dfrac{0.70\times0.90\times c^2}{c^2}}[/tex]
[tex]u'=c\dfrac{0.70+0.90}{1+(0.70\times0.90)}[/tex]
[tex]u'=0.981c[/tex]
Hence, The speed of the proton relative to the laboratory frame is 0.981c
Final answer:
The speed of the proton relative to the laboratory frame, when it moves to the right with a speed of 0.70c relative to an electron moving at 0.90c, is found to be approximately 0.98c using relativistic velocity addition.
Explanation:
To find the speed of the proton relative to the laboratory frame when an electron moves to the right with a speed of 0.90c and a proton moves to the right with a speed of 0.70c relative to the electron, we use the formula for relativistic velocity addition. This formula is given by:
V = (v + u) / (1 + vu/c²),
where V is the velocity of the proton relative to the lab, v is the velocity of the electron relative to the lab (0.90c), u is the velocity of the proton relative to the electron (0.70c), and c is the speed of light. Plugging in these values, we get:
V = (0.90c + 0.70c) / (1 + (0.90*0.70)) = 0.98c.
Therefore, the speed of the proton relative to the laboratory frame is approximately 0.98c, close to the speed of light, demonstrating the relativistic effects when velocities approach the speed of light.
In a Young's double-slit experiment, a set of parallel slits with a separation of 0.142 mm is illuminated by light having a wavelength of 576 nm and the interference pattern observed on a screen 3.50 m from the slits. (a) What is the difference in path lengths from the two slits to the location of a fourth order bright fringe on the screen? μm (b) What is the difference in path lengths from the two slits to the location of the fourth dark fringe on the screen, away from the center of the pattern? μm
Answer:
[tex]1.152\ \mu m[/tex]
[tex]1.44\ \mu m[/tex]
Explanation:
d = Gap between slits = 0.142 mm
[tex]\lambda[/tex] = Wavelength of light = 576 nm
L = Distance between light and screen = 3.5 m
m = Order = 2
Difference in path length is given by
[tex]\delta=dsin\theta=m\lambda\\\Rightarrow \delta=2\times 576\times 10^{-9}\\\Rightarrow \delta=0.000001152\ m\\\Rightarrow \delta=1.152\times 10^{-6}\ m=1.152\ \mu m[/tex]
The difference in path lengths is [tex]1.152\ \mu m[/tex]
For dark fringe the difference in path length is given by
[tex]\delta=(m+\dfrac{1}{2})\lambda\\\Rightarrow \delta=(2+\dfrac{1}{2})\times 576\times 10^{-9}\\\Rightarrow \delta=0.00000144\ m=1.44\ \mu m[/tex]
The difference in path length is [tex]1.44\ \mu m[/tex]
A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 30.0 min at 80.0 km/h, 45.0 min at 40 km/h and 12.0 min at 100 km/h, and spends 15.0 min eating lunch and buying gas. Determine the average sped for the trip. Determine the difference between the intial citties along the route.
Answer:
Average speed of the trip = 52.9 km/h
Distance between initial pairs of cities (start to end) = 70.0 km
Explanation:
Since distance = speed × time
If she drives 30.0 min at 80.0 km/h
Distance covered = (30/60) × 80 = 40.0 km
Again she drives 45.0 min at 40 km/h
Distance covered = (45/60) × 40 = 30.0 km
Again she drives 12.0 min at 100 km/h
Distance covered = (12/60) × 100 = 20.0 km
Total distance covered = 40.0 + 30.0 + 20.0
= 90.0 km
Total time spent = 30.0 + 45.0 + 12.0 +15.0
= 102 min
Average speed for the trip = Total distance covered/total time spent
= 90/(102/60)
= 52.9 km/h
Distance between initial cities will be between the start of one city to the end of another
Between the first pairs = 40.0 + 30.0 = 70.0 km
Between the second pairs = 30.0 + 20.0 = 50.0 km
Final answer:
The average speed for the trip, excluding stops, is approximately 62.1 km/h, and the total distance traveled between the initial cities is 90.0 km.
Explanation:
To determine the average speed for the trip, we need to calculate the total distance traveled and divide it by the total time taken, excluding any stops. We will calculate distance as speed multiplied by time for each segment of the trip, convert the minutes into hours, and then sum these distances to find the total distance traveled.
First segment: 80 km/h x 0.5 h = 40.0 km
Second segment: 40 km/h x 0.75 h = 30.0 km
Third segment: 100 km/h x 0.2 h = 20.0 km
Total distance = 40.0 km + 30.0 km + 20.0 km = 90.0 km
Now, let's calculate the total travel time in hours, remembering to exclude the time spent not driving (the lunch and gas stop).
Driving time = 30 min + 45 min + 12 min = 87 min = 1.45 h
The average speed is then total distance divided by total driving time.
Average speed = 90.0 km / 1.45 h ≈ 62.1 km/h
To determine the difference between the initial cities along the route, we use the total distance traveled, which we have found to be 90.0 km.
A piece of wire is bent to form a circle with radius r. It has a steady current I flowing through it in a counterclockwise direction as seen from the top (looking in the negative z-direction).
What is Bz(0), the z component of B at the center (i.e., x=y=z=0) of the loop?
Express your answer in terms of I, r, and constants like μ0 and π.
Answer:
B=μ₀I/2r
Explanation:
Produced magnetic field due to an existing electric field through a coil or conductor can be explained by Biot-Savart Law. Formula for this law is:
dB=(μ₀I/4π.r²)dL
Here,
r=Radius of the loop
I and r are constants with respect to length L.
To convert linear displacement L into angular displacement Ф:
dL=r.dФ
So,
dB=(μ₀I/4π.r²)r.dФ
dB=(μ₀I/4π.r)dФ
Integrating both sides over the circle i.e. from 0 radians to 2π radians (360⁰), while the integration will apply only on dФ as all others are constants.
B=(μ₀I/4πr)(2π-0)
B=(μ₀I/2r)
The magnetic field produced due to current flowing through the coil is [tex]B = \frac{\mu_o I}{2r}[/tex].
The magnetic field produced due to current flowing through a coil given by Biot-Savart Law.
[tex]dB = \frac{\mu_o I}{4\pi r^2} dL[/tex]
where;
B is the magnetic field strengthr is the radius of the loopI is the current flowing in the coil[tex]dB = \frac{\mu_o I}{4\pi r^2} .(rd \phi)\\\\dB = \frac{\mu_o I}{4\pi r} \ d \phi\\\\B = \frac{\mu_o I}{4\pi r} [\phi ]^{2\pi} _{0}\\\\B = \frac{\mu_o I}{4\pi r} (2\pi)\\\\B = \frac{\mu_o I}{2 r}[/tex]
Thus, the magnetic field produced due to current flowing through the coil is [tex]B = \frac{\mu_o I}{2r}[/tex].
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If the water in a lake is everywhere at rest, what is the pressure as a function of distance from the surface? The air above the surface of the water is at standard sea-level atmospheric conditions. How far down must one go before the pressure is 1 atm greater than the pressure at the surface?
Answer:
Answer:
101325 + 10055.25h
//
h = 10.1 m
Explanation:
the pressure at sea level = 1 atm = 101325 Pa
density of sea water = 1025 kg/ m^(3)
pressure due to fluid height = pgh
Absolute pressure = 101325 + 1025*9.81*h
= 101325 + 10055.25h
where h= 0 at sea level at increases downwards
//
101325 = 1025* 9.81* h
h = 10.1 m
Explanation: