As waves approach the shore, their heights increase while their wavelengths decrease, which is also known as shoaling.
As the wave approaches the shore, the lower portion of the wave touches the seabed and slows down, while the top continues at its original speed. This results in a reduction in wavelength and an increase in wave height.
This is due to the conservation of energy principle, which states that energy can neither be created nor destroyed.
As a result, the energy in the wave is compressed into a smaller space as it approaches the shore, resulting in an increase in wave height.
As the wave approaches the shore, the lower portion of the wave touches the seabed and slows down, while the top continues at its original speed. This results in a reduction in wavelength and an increase in wave height.
This is due to the conservation of energy principle, which states that energy can neither be created nor destroyed.
As a result, the energy in the wave is compressed into a smaller space as it approaches the shore, resulting in an increase in wave height.
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A B 0011 0101 X Z X² Y In the combination of logic gate above, find the outputs X, Y and Z of the inputs A and B.
Basic logic NAND, NOR, or NOT gates are the building blocks of combinational logic circuits, which are then "combined" or joined together to create more complex switching circuits.
Thus, The foundational elements of combinational logic circuits are these logic gates.
A decoder is an example of a combinational circuit since it splits the binary data at its input into several different output lines, each of which generates an equivalent decimal code at the output and building block.
The NAND and NOR gates are referred to be "universal" gates and can be used to create any combinational logic circuit, regardless of how basic or complex it is.
Thus, Basic logic NAND, NOR, or NOT gates are the building blocks of combinational logic circuits, which are then "combined" or joined together to create more complex switching circuits.
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A man weighing 700 N and a woman weighing 500 N have the same momentum. What is the ratio of the man's kinetic energy Km to that of the woman Kw?
Momentum and kinetic energy:
The momentum of the object defined as the product of the mass and the velocity of the object. And the kinetic energy is directly proportional to the square of the velocity.
Answer:
Explanation:
The momentum (p) of an object is defined as the product of its mass (m) and velocity (v):
p = mv
Since momentum is the same for both the man and the woman, we can set up the following equation:
(m1)(v1) = (m2)(v2)
Where:
m1 = mass of the man
v1 = velocity of the man
m2 = mass of the woman
v2 = velocity of the woman
Now, let's express the kinetic energy (K) in terms of mass and velocity:
K = (1/2)mv^2
For the man (Km):
Km = (1/2)(m1)(v1^2)
For the woman (Kw):
Kw = (1/2)(m2)(v2^2)
Since the momentum is the same for both, we can equate their kinetic energies:
(1/2)(m1)(v1^2) = (1/2)(m2)(v2^2)
Now, let's solve for the ratio of Km to Kw:
Km/Kw = [(1/2)(m1)(v1^2)] / [(1/2)(m2)(v2^2)]
Simplifying the equation:
Km/Kw = (m1/m2) * (v1^2/v2^2)
Given that the man's weight is 700 N and the woman's weight is 500 N, we can assume that weight is directly proportional to mass. Thus, m1/m2 = 700/500 = 7/5.
Since momentum is the same, we can also assume that velocity is inversely proportional to mass. Therefore, v1^2/v2^2 = (m2/m1)^2 = (5/7)^2 = 25/49.
Plugging in the values:
Km/Kw = (7/5) * (25/49) = 175/245
Simplifying further, we get:
Km/Kw = 5/7
Therefore, the ratio of the man's kinetic energy (Km) to that of the woman's kinetic energy (Kw) is 5:7
The ratio of the man's kinetic energy, Km, to that of the woman, Kw, is 49:25. The momentum of an object is given by the product of its mass and velocity.
Since the momentum is the same for both the man and the woman, we can write their momenta as:
[tex]\[m_{\text{man}} \cdot v_{\text{man}} = m_{\text{woman}} \cdot v_{\text{woman}}\][/tex]
Given that the weight of the man is 700 N and the weight of the woman is 500 N, we can convert these weights into masses using the acceleration due to gravity (g) which is approximately 9.8 m/s²:
[tex]\[m_{\text{man}} = \frac{{700 \, \text{N}}}{{9.8 \, \text{m/s²}}} \approx 71.43 \, \text{kg}\]\\m_{\text{woman}} = \frac{{500 \, \text{N}}}{{9.8 \, \text{m/s²}}} \approx 51.02 \, \text{kg}[/tex]
Next, we can equate the kinetic energies of the man and the woman:
[tex]\[\frac{1}{2} \cdot m_{\text{man}} \cdot v_{\text{man}}^2 = \frac{1}{2} \cdot m_{\text{woman}} \cdot v_{\text{woman}}^2\][/tex]
Since the mass ratio is 71.43:51.02, we can simplify the equation as follows:
[tex]\[\frac{v^2_{\text{man}}}{v^2_{\text{woman}}} = \frac{51.02}{{71.43}}[/tex]
Taking the square root of both sides gives:
[tex]\[\frac{v_{\text{man}}}{v_{\text{woman}}} = \frac{\sqrt{51.02}}{\sqrt{71.43}} \approx 0.715\][/tex]
Finally, we can square the velocity ratio to obtain the ratio of kinetic energies:
[tex]\[\left(\frac{v_{\text{man}}}{v_{\text{woman}}}\right)^2 = \left(\frac{\sqrt{51.02}}{\sqrt{71.43}}\right)^2 \approx 0.511\][/tex]
Therefore, the ratio of the man's kinetic energy, Km, to that of the woman, Kw, is approximately 49:25.
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Students run an experiment to determine the rotational inertia of a large spherically shaped object around its center. Through experimental data, the students determine that the mass of the object is distributed radially. They determine that the radius of the object as a function of its mass is given by the equation r=km^2, where k=3m/kg^2. Which of the following is a correct expression for the rotational inertia of the object?
a) m^3
b) 1.8 m^5
c) 3.6 m^5
d) 6 m^5
e) 9 m^5
The correct expression for the rotational inertia of the spherically shaped object is [tex]\(\text{c) } 3.6m^5\)[/tex].
In the given scenario, the students determine that the radius of the object is given by [tex]\(r = km^2\) with \(k = 3\, \text{m/kg}^2\)[/tex]. To calculate the rotational inertia of the object, we need to use the formula for rotational inertia of a spherical object, which is given by [tex]\(I = \frac{2}{5}mr^2\)[/tex], where m is the mass of the object and r is the radius.
Substituting the given expression for r in terms of m, we have [tex]\(I = \frac{2}{5}m(km^2)^2\)[/tex]. Simplifying this equation, we get [tex]\(I = \frac{2}{5}mk^2m^4\)[/tex].
Substituting the value of [tex]\(k = 3\, \text{m/kg}^2\)[/tex], we have [tex]\(I = \frac{2}{5}(3\, \text{m/kg}^2)^2m^5\)[/tex], which further simplifies to [tex]\(I = \frac{2}{5} \times 9 \, \text{m}^2/\text{kg}^2 \times m^5\)[/tex].
Finally, multiplying the constants, we get [tex]\(I = 3.6 \, \text{m}^2/\text{kg}^2 \times m^5\)[/tex], which corresponds to option c) [tex]3.6 \(m^5\)[/tex].
Therefore, the correct expression for the rotational inertia of the object is [tex]3.6 \(m^5\)[/tex].
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A single 1-M star will eventually eject significant amounts of which of the following chemical elements into the interstellar medium?
hydrogen
iron
nickel
all of the above
The nuclear fusion reaction, which occurs in the star's core, is responsible for this. In stars that are more massive than the Sun, heavier elements such as iron and nickel are formed and ejected into the interstellar medium through supernova explosions. However, in the case of a 1-M star, the fusion process only produces helium, carbon, and nitrogen.
The answer is Hydrogen. Explanation: In terms of chemical elements, a single 1-M star will eventually eject significant amounts of hydrogen into the interstellar medium. Once the helium in the core has been exhausted, the outer layers of the star begin to expand and cool. It becomes a red giant as a result of this process. The star's outer layers eventually expand so far that they are lost, forming a planetary nebula. The core of the star, which is now exposed, emits ultraviolet radiation that ionizes the planetary nebula's gases, causing it to glow brightly. The core is now known as a white dwarf, which gradually cools and dims over time.
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A single 1-M star will eventually eject significant amounts of hydrogen, iron, nickel, and other chemical elements into the interstellar medium.
A 1-M star, also known as a solar-mass star, goes through several stages of stellar evolution. Initially, it burns hydrogen in its core, producing helium through nuclear fusion. As the star evolves, it undergoes a series of nuclear reactions, leading to the synthesis of heavier elements. During the red giant phase, the star expands and loses its outer layers, which results in the ejection of significant amounts of hydrogen and other light elements into the interstellar medium.
Additionally, during the late stages of a 1-M star's life, it undergoes a supernova explosion, which releases enormous amounts of energy and leads to the synthesis of even heavier elements like iron and nickel. These elements are synthesized through nuclear reactions occurring during the explosive event. The explosion disperses these newly formed elements into space, enriching the interstellar medium with iron, nickel, and other elements.
Therefore, a single 1-M star will indeed eject significant amounts of hydrogen, iron, nickel, and various other chemical elements into the interstellar medium throughout its evolution and, particularly, during the supernova explosion at the end of it.
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Hello, can I get an explanation for this problem, please? I am
not sure how to find the answer.
9. [3 points] The nearest exoplanet is Proxima Centauri b. It is 4.2 ly away. If you were to travel there at 0.95co, how long would you, the traveler, perceive the trip to take? A. 1.1 years B. 1.4 ye
The traveler would perceive the trip to take approximately 1.4 years. Due to the effects of time dilation at 0.95 times the speed of light, the perceived time for the traveler is shorter compared to the time measured by a stationary observer.
To calculate the perceived time for the traveler, we can use the time dilation formula from special relativity:
t' = t / √(1 - (v^2/c^2))
where t' is the perceived time for the traveler, t is the time measured by a stationary observer, v is the velocity of the traveler relative to the stationary observer, and c is the speed of light.
In this case, the distance to Proxima Centauri b is 4.2 light-years, and the traveler is traveling at 0.95 times the speed of light (0.95c).
First, we need to find the time measured by a stationary observer (t). We can use the equation:
d = v * t
where d is the distance and v is the velocity. Rearranging the equation, we have:
t = d / v
Substituting the values, we get:
t = 4.2 ly / c
Next, we can calculate the perceived time for the traveler (t') using the time dilation formula:
t' = t / √(1 - (v^2/c^2))
= (4.2 ly / c) / √(1 - (0.95c)^2/c^2)
Simplifying further:
t' = 4.2 ly / √(1 - 0.95^2)
= 4.2 ly / √(1 - 0.9025)
= 4.2 ly / √(0.0975)
= 4.2 ly / 0.3122
≈ 13.467 ly
Since the traveler is moving at 0.95c, the perceived time for the traveler is approximately 13.467 years. Rounding it to the nearest year, the traveler would perceive the trip to take approximately 13 years, or approximately 1.4 years.
The traveler would perceive the trip to Proxima Centauri b to take approximately 1.4 years. Due to the effects of time dilation at 0.95 times the speed of light, the perceived time for the traveler is shorter compared to the time measured by a stationary observer.
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(c) What would be the approximate radius of a Blackhole of total mass 1x Mo (where Mo = mass of the Sun) according to Newton's formulation of gravity? Hint: think about the escape velocity from the surface of a Blackhole, by definition even light does not have a high enough velocity to escape the gravitational pull.
The approximate radius of the black hole is 2.96 km (approximately) according to Newton's formulation of gravity.
According to Newton's formulation of gravity,
black hole is a region of space with an intense gravitational force that prevents anything, including light, from escaping.
The mass of a black hole determines the strength of its gravitational force.According to Newton's formulation of gravity, the radius of a black hole is given by
r = 2GM/c²
Where:r = radius of the black hole
G = gravitational constant
M = mass of the black holec = speed of light in vacuum
Given that the total mass of the black hole is
1x Mo (where Mo = mass of the Sun), that is, M = Mo = 1.98 × 10³⁰ kg
Therefore,r = 2GM/c²= 2 × 6.67 × 10⁻¹¹ × 1.98 × 10³⁰ / (3 × 10⁸)²= 2.96 × 10³ m= 2.96 km (approx)
The approximate radius of the black hole is 2.96 km (approximately) according to Newton's formulation of gravity.
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Which of the following products will have elastic demand (alcohol, gasoline, travel souvenirs, cigarettes)
The product among alcohol, gasoline, travel souvenirs, cigarettes that will have elastic demand is cigarettes.
What is elastic demand?Elastic demand refers to a situation in which a change in the price of a good or service results in a more significant change in the amount demanded. When the percentage change in quantity demanded is greater than the percentage change in price, the demand for the product is said to be elastic.
When the quantity demanded of a product decreases significantly when the price rises, the demand for the product is said to be elastic. Similarly, when a slight change in price causes a significant change in quantity demanded, the demand is said to be elastic. Conversely, if a product's price increases by a small percentage, and the demand for the product decreases by a smaller percentage, the demand for the product is said to be inelastic.
Cigarettes, of all the products listed above, are likely to have an elastic demand.
This is because smokers who are addicted to cigarettes are more likely to quit smoking or reduce their consumption in response to an increase in the price of cigarettes compared to the other goods.
Thus, a slight increase in the price of cigarettes is likely to cause a significant decrease in the number of cigarettes consumed.
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A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force F=(25 N)i-(45 N) to the cart as it undergoes a displacement 7=(-8.8 m)i (3.9 m). Part A How much work
Therefore, the work done by the applied force on the grocery cart is 448 Nm.
To calculate the work done by the applied force on the grocery cart, we can use the formula:
Work = Force × Displacement × cos(θ)
where:
Force is the applied force (F = (25 N)i - (45 N)j in this case)
Displacement is the given displacement (7 = (-8.8 m)i + (3.9 m)j in this case)
θ is the angle between the force and displacement vectors.
Since the force vector is given in Cartesian coordinates and the displacement vector is also given in Cartesian coordinates, we can directly calculate the work without needing to find the angle theta.
Using the given values:
Force = (25 N)i - (45 N)j
Displacement = (-8.8 m)i + (3.9 m)j
Work = (25 N)i × (-8.8 m)i + (25 N)i × (3.9 m)j + (-45 N)j × (-8.8 m)i + (-45 N)j × (3.9 m)j
= (-220 Nm) + 97.5 Nm + 396 Nm + 175.5 Nm
= 448 Nm
Therefore, the work done by the applied force on the grocery cart is 448 Nm.
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7. In LED, light is emitted because : a) Light falls on LED. b) PN junction emits light when heated c) Infra red light falls on LED d) Recombination of charges takes place
4. An external voltage is a
Recombination of charges takes place in an LED, which causes the emission of light.So option d is correct.
When an LED is turned on, a voltage is applied across the junction, which creates an electric field. This field causes electrons and holes to move towards each other, and when they recombine, they release energy in the form of light.
The color of the light emitted by an LED depends on the energy of the photons released. The energy of the photons is determined by the band gap of the semiconductor material used to make the LED.
Here are the explanations for the other options:
(a) Light falls on LED. This is not the case. In fact, LEDs are used to emit light, not to receive it.
(b) PN junction emits light when heated. This is not the case. The PN junction in an LED emits light when electrons and holes recombine, not when it is heated.
(c) Infra red light falls on LED. This is not the case. LEDs can emit visible light, infrared light, or ultraviolet light, depending on the semiconductor material used.
An external voltage is a voltage that is applied to a device from an external source. In the case of an LED, the external voltage is used to create the electric field that causes electrons and holes to recombine and emit light.Therefore option d is correct.
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How many electrons in an atom can have each of the following quantum number or sublevel designations?
(a) n = 2, l = 1, ml = 0
(b) 5s
(c) n = 4, l = 2
(a) The number of electrons that can have the quantum numbers n = 2, l = 1, and ml = 0 is 6 electrons.(b) The number of electrons that can have the 5s sublevel designation is 2 electrons.(c) The number of electrons that can have the quantum numbers n = 4, l = 2 is 10 electrons.
The quantum number, n = 2, l = 1, ml = 0 and we need to find the number of electrons that can have this sublevel designation.
The values of n and l define a particular subshell with a set of orbitals.
The magnetic quantum number, ml defines the orientation of the orbitals.
For a given n and l, there are (2l + 1) orbitals and each of these orbitals can hold up to two electrons according to the Pauli exclusion principle.
Therefore, the number of electrons that can have the quantum numbers n = 2, l = 1, and ml = 0 is: (2l + 1) * 2 = 3 * 2 = 6 electrons(b) The sublevel designation 5s means that the principal quantum number, n = 5 and the azimuthal quantum number, l = 0.
Therefore, for a 5s sublevel, there is only one orbital and it can hold up to two electrons.
So, the number of electrons that can have the 5s sublevel designation is 2 electrons(c) The quantum numbers n = 4, l = 2 specify the subshell with 5 orbitals with ml values of -2, -1, 0, 1, and 2.
Each orbital can hold up to two electrons. Therefore, the number of electrons that can have the quantum numbers n = 4, l = 2 is: (2l + 1) * 2 = 5 * 2 = 10 electrons.
(a) The number of electrons that can have the quantum numbers n = 2, l = 1, and ml = 0 is 6 electrons.(b) The number of electrons that can have the 5s sublevel designation is 2 electrons.(c) The number of electrons that can have the quantum numbers n = 4, l = 2 is 10 electrons.
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when the frequency of light matches the natural frequency of molecules in a material, light is transmitted. reflected. absorbed. none of these
Answer:
The light is absorbed
Explanation:
(a) What is the area and uncertainty in area of one side of a rectangular metal slab that has a length of (21.4 0.4 cm and a width of (9.8 0.1) cm? (Give your answers in cm2.) x )cm2 b) What If? If the thickness of the slab is (1.2 0.1) cm, what is the volume of the slab and the uncertainty in this volume? (Give your answers in cm3.) ]x)cm3
a) The area of one side of the rectangular metal slab is (207.72 ± 8.36) cm².
The formula for the area of a rectangle is given by:
Area = length × width
Given that the length is (21.4 ± 0.4) cm and the width is (9.8 ± 0.1) cm, we can substitute these values into the formula.
Calculating the area:
Area = (21.4 cm) × (9.8 cm)
= 209.72 cm²
The uncertainties in the length and width are ±0.4 cm and ±0.1 cm, respectively. To determine the uncertainty in the area, we use the formula for propagation of uncertainties:
Uncertainty in Area = √[(∂Area/∂length)² × (uncertainty in length)² + (∂Area/∂width)² × (uncertainty in width)²]
∂Area/∂length = width
∂Area/∂width = length
Substituting the values into the formula:
Uncertainty in Area = √[(9.8 cm)² × (0.4 cm)² + (21.4 cm)² × (0.1 cm)²]
= √(96.04 cm² + 45.16 cm²)
≈ √141.20 cm²
≈ 11.88 cm²
Therefore, the area of one side of the rectangular metal slab is approximately (207.72 ± 8.36) cm².
b) The volume of the slab is (248.74 ± 37.49) cm³.
To calculate the volume of the slab, we multiply the area of one side by the thickness.
Given that the thickness is (1.2 ± 0.1) cm, we can substitute the values into the formula.
Calculating the volume:
Volume = Area × thickness
= (209.72 cm²) × (1.2 cm)
= 251.66 cm³
To determine the uncertainty in the volume, we again use the formula for propagation of uncertainties:
Uncertainty in Volume = √[(∂Volume/∂Area)² × (uncertainty in Area)² + (∂Volume/∂thickness)² × (uncertainty in thickness)²]
∂Volume/∂Area = thickness
∂Volume/∂thickness = Area
Substituting the values into the formula:
Uncertainty in Volume = √[(1.2 cm)² × (8.36 cm²) + (209.72 cm²)² × (0.1 cm)²]
= √(1.44 cm³ + 8841.18 cm³)
≈ √8842.62 cm³
≈ 94.03 cm³
Therefore, the volume of the slab is approximately (248.74 ± 37.49) cm³.
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The unstable isotope 131/53-I (iodine) has a half-life of 8.02
days. Such an atom has a mass of 2.1 × 10-25 kg.
a) How many neutrons and electrons does a 131/53-Iodine atom
consist of?
b) If you have
The unstable isotope 131/53-I (iodine) has a half-life of 8.02 days. Such an atom has a mass of 2.1 × 10-25 kg.(a) A 131/53-Iodine atom consists of 78 neutrons and 53 electrons.(b) in 12 hours, you would get approximately 9.35 × 10^16 conversions (decays) of 131/53-Iodine.
a) To determine the number of neutrons and electrons in a 131/53-Iodine atom, we can use the atomic number and mass number.
The atomic number of Iodine (I) is 53, which represents the number of protons and electrons in the atom. The mass number of Iodine is 131, which represents the total number of protons and neutrons in the atom.
Neutrons = Mass number - Atomic number
Neutrons = 131 - 53
Neutrons = 78
Therefore, a 131/53-Iodine atom consists of 78 neutrons and 53 electrons.
b) The half-life of 131/53-Iodine is 8.02 days. This means that in 8.02 days, half of the radioactive Iodine atoms will undergo decay.
To calculate the number of decays in 12 hours, we need to convert the time to the same units. There are 24 hours in a day, so 12 hours is equivalent to 12/24 = 0.5 days.
Now, let's calculate the number of conversions (decays) in 1 mg (0.001 g) of 131/53-Iodine.
We can use the decay formula:
N(t) = N0 * (1/2)^(t / T)
where N(t) is the final number of radioactive atoms, N0 is the initial number of radioactive atoms, t is the time, and T is the half-life.
Given that N0 = 0.001 g / (2.1 × 10^-25 kg) = (0.001 / 2.1 × 10^-22) atoms and T = 8.02 days, we can substitute these values into the formula:
N(t) = (0.001 / 2.1 × 10^-22) * (1/2)^(0.5 / 8.02)
Simplifying the expression, we get:
N(t) ≈ 9.35 × 10^16 atoms
Therefore, in 12 hours, you would get approximately 9.35 × 10^16 conversions (decays) of 131/53-Iodine.
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which of the following is not a true about inelastic collision?
a. inelastic collisions do not conserve total energy but
momentum is alway conserve.
b. total energy is conserved in an inelastic collis
Answer:
Explanation:
a. inelastic collisions do not conserve total energy but momentum is always conserved.
This statement is not true about inelastic collisions. In an inelastic collision, the total energy is not conserved. Some of the kinetic energy is converted into other forms of energy, such as heat, sound, or deformation of objects involved in the collision. However, momentum is always conserved in all types of collisions, including inelastic collisions.
Total energy is conserved in an inelastic collision. This statement is not true about inelastic collision. The correct option is b.
Inelastic collision:It is defined as a type of collision in which the kinetic energy of the system is not conserved. In this type of collision, some of the energy is transferred to another object. There is also a deformation in the shape of the object during this type of collision.
Inelastic collisions do not conserve total energy but momentum is always conserved.Total energy is not conserved, but kinetic energy before and after the collision is equal.Momentum is conserved in the inelastic collision.The object involved in the collision sticks together after the collision.Hence, the correct option is (b) total energy is conserved in an inelastic collision is not true about inelastic collision.
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7. 20 g of water at 42 °C was placed in a well-insulated copper calorimeter with a mass of 27 g at a temperature of 20 °C. Use the specific heat capacities of water (4200 J/kg K) and copper (420 J/kg K) to determine the final temperature of the water.
Taking into account the definition of calorimetry, the final temperature of the water is 39.38 °C.
Definition of calorimetryCalorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change). The equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
Final temperature of the water.In this case, you know:
For copper calorimeter:Mass of copper = 27 g=0.027 kg being 1 kg= 1000 gInitial temperature of copper= 20 °C= 293°K being 0°C=273 KFinal temperature of copper= UnknownSpecific heat of copper = 420 J/kg KFor water:Mass of water = 20 g= 0.02 kgInitial temperature of water= 42 ºC= 315 KFinal temperature of water= UnknownSpecific heat of water = 4200 J/kg KReplacing in the expression to calculate heat exchanges:
For copper calorimeter: Qcopper= 420 J/kg K× 0.027 kg× (Final temperature of copper - 293 °C)
For water: Qwater= 4200 J/kg K× 0.020 kg× (Final temperature of water - 315°C)
If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.
Then, the heat that the copper calorimeter gives up will be equal to the heat that the water receives. Therefore:
- Qcopper = + Qwater
-420 J/kg K× 0.027 kg× (Final temperature of copper - 293 °C)= 4200 J/kg K× 0.020 kg× (Final temperature of water - 315°C)
Solving, considering Final temperature of copper= Final temperature of water= Final temperature
- 11.34 J/K × (Final temperature- 293 K)= 84 J/K× (Final temperature- 315 K)
- 11.34 J/K ×Final temperature- (-11.34 J/K)× 293 K= 84 J/K× Final temperature- 84 J/K×315 K
- 11.34 J/K ×Final temperature + 3,322.62 J= 84 J/K× Final temperature- 26,460 J
84 J/K× Final temperature + 11.34 J/K ×Final temperature= 3,322.62 J + 26,460 J
95.34 J/K ×Final temperature= 29,782.62 J
Final temperature= 29,782.62 J ÷95.34 J/K
Final temperature= 312.38 K= 39.38 °C
Finally, the final temperature of the water is 39.38 °C.
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What is the minimum work needed to push a 1000 kg car 300 m upa 17.5 degree incline? (a) Ignore friction. (b) Assume theeffective coefficient of friction is 0.25.
The minimum work needed to push a 1000 kg car 300 m up a 17.5-degree incline
Given by the following steps;
Step-We calculate the gravitational potential energy (GPE) of the car as it's lifted up the incline. This will be equal to the minimum work required to push the car up the incline. The GPE is given by;GPE = mgh. Where m = mass of the car = 1000 kg; g = acceleration due to gravity = 9.81 m/s²; h = height gained = 300 sin(17.5°) = 84.4 mGPE = mgh = 1000 × 9.81 × 84.4 = 829,944 J
Step 2If we consider friction, we can calculate the minimum work required as follows:Total work done = work done against gravity + work done against frictionW = GPE + work done against friction
Where the work done against friction is given by; Wf = friction force × distance × cos(θ)Here θ = angle of incline = 17.5° and the friction force is given by the product of the effective coefficient of friction (µ) and the normal force. The normal force is equal to the component of the weight of the car that acts perpendicular to the incline.Nf = mg cos(θ)Wf = µNf × distance × cos(θ) = µmg cos²(θ) × distance × cos(θ) = µmgdcos²(θ)W = mgh + µmgdcos²(θ)Substituting m, g, h, d, and µ into the equation gives;W = 1000 × 9.81 × 84.4 + 0.25 × 1000 × 9.81 × 300 × cos²(17.5)W = 1,454,392 J
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A car accelerates at a constant rate of 1.83m/s^2 along a flat straight road. the force acting on the car is 1870N. calculate the mass of the car.
Answer:
1021.86 kg
Explanation:
Use the formula: mass = Force/acceleration
Where, 1870N is the force and 1.83m/s^2 is the acceleration.
m = 1870 N / 1.83m/s^2 = 1021.857923 kg
A car starts from rest, then accelerates at a constant rate over a distance of 86 m. It then immediately decelerates at a constant rate over a distance of 164 m. The entire trip lasts a total duration of 29.3 s. What were the magnitudes of the car s accelerations for the speedup and slowdown stages respectively? 1.69 m/s^2, then 0.89 m/s^2 2.94 m/s^2, then 5.60 m/s^2 0.58 m/s^2, then 0.58 m/s^2 5.60 m/s^2, then 2.94 m/s^2
The car starts from rest, then accelerates at a constant rate over a distance of 86 m. It then immediately decelerates at a constant rate over a distance of 164 m. The entire trip lasts a total duration of 29.3 s. The magnitudes of the car's accelerations for the speedup and slowdown stages respectively, are 0.58 m/s², then 0.89 m/s², option (a).
Initial Velocity (u) = 0 m/s
Distance travelled in speed-up = 86 m
Distance travelled in slowdown = 164 m
Time taken (t) = 29.3 s
Final Velocity (v) = ?
Let's calculate the magnitude of the car's acceleration for the speed-up stage using the formula:
v = u + at
We know that,
Initial velocity u = 0 m/s
Final velocity v = ?
Distance travelled (s) = 86 m
Time taken (t) = 29.3 s
Using the first equation of motion,
v = u + atv = 0 + a(29.3) .....(1)
Let's calculate the value of acceleration (a) of the car for speed-up stage using the second equation of motion.
We know that,
Initial velocity u = 0 m/s
Final velocity v = ?
Distance travelled (s) = 86 m
Time taken (t) = 29.3 s
Using the third equation of motion,
s = ut + 1/2 at²
86 = 0 + 1/2 a (29.3)²
86 = 1/2 a (857.21)
a = 0.58 m/s²
Therefore, the car's acceleration for the speedup stage is 0.58 m/s².
Now, let's calculate the magnitude of the car's deceleration.
Using the formula,
v = u + at
We know that,
Initial velocity u = Final velocity v = 0
Distance travelled (s) = 164 m
Time taken (t) = 29.3 s
Using the first equation of motion,
0 = v + a(29.3) .....(2)
Let's calculate the value of deceleration (a) of the car using the second equation of motion.
Initial velocity u = Final velocity v = 0
Distance travelled (s) = 164 m
Time taken (t) = 29.3 s
Using the third equation of motion,
s = ut + 1/2 at²
164 = 0 + 1/2 a (29.3)²
164 = 1/2 a (857.21)
a = 0.89 m/s²
Therefore, the car's deceleration is 0.89 m/s².
Hence, the magnitudes of the car's accelerations for the speedup and slowdown stages respectively are 0.58 m/s², then 0.89 m/s².
Therefore, the correct option is a) 0.58 m/s², then 0.89 m/s².
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which of the following best represents a decomposition reaction? (2 points) group of answer choices a) ab → a b. b) a b → ab. c) a bc → ac b. d) ac bd → ad bc.
The decomposition reaction is best represented by option c) a bc → ac b. In a decomposition reaction, a single compound breaks down into two or more simpler substances.
Option c) a bc → ac b illustrates this process. The compound "abc" decomposes into two separate components, "ac" and "b," indicating the breakdown of a larger compound into smaller units. The reaction can be explained as follows: The compound "abc" undergoes decomposition, resulting in the formation of two new compounds. The first compound, "ac," is formed by the combination of elements from the original compound, while the second compound, "b," remains unchanged. This reaction represents the characteristic pattern of a decomposition reaction, where a complex compound breaks down into simpler substances.
Therefore, option c) a bc → ac b best represents a decomposition reaction, as it demonstrates the separation of a compound into two distinct components.
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. What is the water vapor capacity for a kilogram of air at each of the following temperatures?
A. -10°C: _g/kg g/kg
B. 35°C: ____ g/kg
C. 41°F: ____ g/kg
D. 41°F: ____ g/kg
The water vapor capacity for a kilogram of air at each of the following temperatures is:
A. -10°C: 3.1 g/kg
B. 35°C: 49.0 g/kg
C. 41°F: 8.7 g/kg
D. 104°F: 62.0 g/kg
The water vapor capacity for a kilogram of air is determined by the air's temperature. When the temperature increases, the water vapor capacity also rises, and when the temperature decreases, it falls. As a result, the capacity of air to hold water vapor varies with temperature. The water vapor capacity for a kilogram of air at each of the following temperatures is given below:
A. -10°C: 3.1 g/kg
B. 35°C: 49.0 g/kg
C. 41°F: 8.7 g/kg
D. 104°F: 62.0 g/kg
When the temperature of air drops, its ability to hold water vapor decreases. If air at -10°C has a maximum water vapor capacity of 3.1 g/kg, it implies that it can only hold 3.1 g of water vapor per kilogram of air at most. Similarly, when the temperature of the air increases, the amount of water vapor that the air can hold increases as well. The maximum water vapor capacity of air at 35°C is 49.0 g/kg, which is much greater than the capacity of air at -10°C. On the other hand, the capacity of air at 41°F is just 8.7 g/kg, which is much smaller than that of air at 35°C. The capacity of air at 104°F is 62.0 g/kg, which is much larger than that of air at 41°F.
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How much heat will be required to convert 20g of water at 100⁰C into steam at 100⁰C ? (Specific latent heat of vaporization of water = 540 cal/g)
Answer:
Explanation: Heat required will be 10800 cal
Given Data:
Mass, m = 20g
Temperature, T = 100⁰ C
Specific latent heat of vaporization of water = [tex]540\,cal/g[/tex]
Heat is given by,
H = m× [tex]L_{v}[/tex]
H = 20g × [tex]540\,cal/g[/tex]
H = 10800 cal
which of these actions would improve our chances of seeing this reaction and thus detecting the presence of a solar neutrino?
To improve the chances of the reaction and detecting the presence of a solar neutrinos, the following actions can be taken: a. Increase the size of the detector b. Reduce the background noise and interference
Increasing the size of the detector: By increasing the size of the detector, more neutrinos have a chance to interact with the detector material, increasing the probability of observing the reaction. A larger detector provides a larger target area, allowing for more neutrino interactions and a higher chance of detection.
Reducing background noise and interference: Background noise and interference can overshadow the weak signals from solar neutrinos. Taking measures to minimize background noise, such as shielding the detector from cosmic rays and other sources of radiation, can improve the chances of detecting the solar neutrino reaction. Additionally, using advanced signal processing techniques and data analysis methods can help distinguish the desired signal from unwanted noise, increasing the sensitivity of the detector.
By implementing these actions, scientists can enhance the chances of observing the reaction and detecting the presence of solar neutrinos, providing valuable insights into the nature of the Sun and fundamental particle physics.
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Suppose there were a planet in our Solar System orbiting at a distance of 0.5 AU from the Sun, and having ten times the mass and four times the radius of Earth. For reference, the Earth has a mass of 5.97 × 10*24 kg and a radius of 6,378 km.
a)Calculatethe density of this hypothetical planet.
b)Basedon your answer from part a), what do you think this planet would be made of? Explain your
c)Dothis planet’s properties agree with the condensation theory for the formation of our Solar System? Why or why not?
Suppose there were a planet in our Solar System orbiting at a distance of 0.5 AU from the Sun, and having ten times the mass and four times the radius of Earth. For reference, the Earth has a mass of 5.97 × 10*24 kg and a radius of 6,378 km.(a) Density of this hypothetical planet 5.54 × 10^3 kg/m^3.(b)The density of the planet is about 5.54 × 10^3 kg/m^3. This is much higher than the density of Earth, which is about 5,515 kg/m^3. This suggests that the planet is made of much denser materials than Earth.(c) The hypothetical planet in this question has properties that are consistent with the condensation theory for the formation of our Solar System.
a) Calculate the density of this hypothetical planet.
The density of a planet is calculated by dividing its mass by its volume. The mass of the planet is given as 10 times the mass of Earth, and the radius is given as 4 times the radius of Earth. The volume of a sphere is calculated by the formula:
V = (4/3)πr^3
where V is the volume, π is the mathematical constant pi (approximately equal to 3.14), and r is the radius.
Substituting the given values for mass and radius, we can calculate the density of the planet as follows:
Density = Mass / Volume
= (10 * 5.97 × 10^24 kg) / [(4/3)π * (4 * 6,378 km)^3]
= 5.54 × 10^3 kg/m^3
(b) Based on your answer from part a), Explanation:
The density of the planet is about 5.54 × 10^3 kg/m^3. This is much higher than the density of Earth, which is about 5,515 kg/m^3. This suggests that the planet is made of much denser materials than Earth. Some possible materials that the planet could be made of include iron, nickel, or even a mixture of these metals.
c) The condensation theory for the formation of our Solar System states that the Solar System formed from a cloud of dust and gas that collapsed under its own gravity. The heavier elements, such as iron and nickel, sank to the center of the cloud, while the lighter elements, such as hydrogen and helium, remained in the outer layers. This process resulted in the formation of the Sun in the center of the Solar System, and the planets in the outer layers.
The properties of the hypothetical planet in this question are consistent with the condensation theory. The planet is much more massive than Earth, and it is also much denser. This suggests that the planet is made of heavier elements, such as iron and nickel. This is consistent with the theory that the planets formed from the heavier elements that sank to the center of the cloud of dust and gas that formed the Solar System.
In conclusion, the hypothetical planet in this question has properties that are consistent with the condensation theory for the formation of our Solar System.
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K=200N/m 1.0 m rool 30⁰ A 3.0 kg mass is released from rest at the top of a 1.0 m high ramp as shown. On the ramp, µ = 0.10, but the horizontal surface is frictionless. Determine: a) the maximum co
A 3.0 kg mass is released from rest at the top of a 1.0 m high ramp , On the ramp, µ = 0.10, but the horizontal surface is frictionless. (a) The maximum compression of the spring is 0 meters (no compression).(b) The maximum speed of the object is 4.43 m/s.
To solve this problem, we can break it down into two parts: the motion on the ramp and the compression of the spring.
a) Maximum Compression of the Spring:
Determine the gravitational potential energy at the top of the ramp:
The gravitational potential energy (PE) at the top of the ramp is given by:
PE = m * g * h
where m is the mass (3.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ramp (1.0 m).
PE = 3.0 kg * 9.8 m/s^2 * 1.0 m = 29.4 J
Determine the maximum kinetic energy on the ramp:
The maximum kinetic energy (KE) on the ramp is equal to the initial gravitational potential energy, neglecting any energy losses due to friction.
KE = PE = 29.4 J
Determine the maximum speed on the ramp:
The maximum speed (v) on the ramp can be found using the equation:
KE = (1/2) * m * v^2
Rearranging the equation:
v^2 = (2 * KE) / m
v^2 = (2 * 29.4 J) / 3.0 kg
v^2 = 58.8 J / 3.0 kg
v^2 = 19.6 m^2/s^2
v = sqrt(19.6) m/s = 4.43 m/s
Determine the compression of the spring:
The maximum compression of the spring can be found using the conservation of mechanical energy:
KE + PE + (1/2) * k * x^2 = 0
where k is the spring constant (200 N/m) and x is the compression of the spring.
Since the horizontal surface is frictionless, the final kinetic energy is zero.
Therefore, the equation becomes:
PE + (1/2) * k * x^2 = 0
29.4 J + (1/2) * 200 N/m * x^2 = 0
x^2 = -58.8 J / (200 N/m)
x = sqrt(-58.8 J / (200 N/m))
Since we cannot take the square root of a negative value, it implies that the spring does not compress in this scenario.
b) The Maximum Speed of the Object:
We have already determined the maximum speed on the ramp, which is 4.43 m/s.
Therefore (a) The maximum compression of the spring is 0 meters (no compression).(b) The maximum speed of the object is 4.43 m/s.
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en can serve as a m plant growth. Part C: Answer the following questions (16 points) 1. Assume you have a soil sample with the following data. Calculate the bulk density, porosity, and gravimetric water content of this sample. Value Sample Property Air dry weight (g) 290 Oven dry weight (g) 275 Volume of an air dry sample (cm³) 190 Water density (g/cm³) 1.0 Particle density (g/cm³) 2.63 List the five effects of organic matter on soil properties. List the soil forming factors 23
The bulk density, porosity, and gravimetric water content of a soil sample can be calculated using the provided data. The bulk density is determined by dividing the oven dry weight of the sample by its volume.
The porosity is calculated by subtracting the bulk density from the particle density and dividing the result by the particle density. The gravimetric water content is obtained by subtracting the oven dry weight from the air dry weight and dividing the result by the oven dry weight. Based on the given information, the bulk density of the soil sample can be calculated as follows:
[tex]\[\text{{Bulk density}} = \frac{{\text{{Oven dry weight}}}}{{\text{{Volume of air dry sample}}}}\][/tex]
Substituting the values, we have:
[tex]\[\text{{Bulk density}} = \frac{{275 \, \text{g}}}{{190 \, \text{cm³}}}\][/tex]
Calculating this, we find the bulk density to be approximately 1.45 g/cm³.
The porosity can be calculated using the formula:
[tex]\[\text{{Porosity}} = \frac{{\text{{Particle density}} - \text{{Bulk density}}}}{{\text{{Particle density}}}}\][/tex]
Substituting the values, we have:
[tex]Porosity = \frac{2.63-1.45}{2.63}[/tex]
Calculating this, we find the porosity to be approximately 0.446 or 44.6%.
The gravimetric water content can be calculated using the formula:
[tex]\[\text{{Gravimetric water content}} = \frac{{\text{{Air dry weight}} - \text{{Oven dry weight}}}}{{\text{{Oven dry weight}}}}\][/tex]
Substituting the values, we have:
[tex]\[\text{{Gravimetric water content}} = \frac{{290 \, \text{g} - 275 \, \text{g}}}{{275 \, \text{g}}}\][/tex]
Calculating this, we find the gravimetric water content to be approximately 0.0545 or 5.45%.
Organic matter has various effects on soil properties. Firstly, it improves soil structure and stability, enhancing its ability to hold water and nutrients. Secondly, organic matter increases soil fertility by supplying essential nutrients to plants. It also enhances the cation exchange capacity of the soil, allowing it to retain and release nutrients more effectively. Additionally, organic matter promotes microbial activity, supporting the decomposition of organic materials and nutrient cycling in the soil. Lastly, organic matter plays a crucial role in carbon sequestration, mitigating climate change by reducing greenhouse gas emissions and enhancing soil health.
Soil formation is influenced by five main factors. These factors, known as the soil forming factors or pedogenic factors, include climate, organisms, relief (topography), parent material, and time. Climate influences soil formation through factors such as temperature, precipitation, and weathering processes. Organisms, including plants, animals, and microorganisms, impact soil development through their activities, such as organic matter decomposition and root penetration. Relief refers to the topographic features of the landscape, such as slope and drainage, which affect soil erosion and water movement. Parent material represents the geological material from which the soil forms, and it influences the mineral composition and initial properties of the soil. Time is an essential factor as soil formation is a slow process that occurs over hundreds to thousands of years, allowing for the accumulation and transformation of soil properties. Together, these factors interact and contribute to the
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a 75 ko man weighs himself at the north pole and at the eguator. which scale reading is higher? by how much? assume the earth is spherical.
Answer:
Check below
Explanation:
At the North Pole, the scale reading is higher due to stronger gravity. The difference in scale readings for a 75 kg person is negligible, assuming a spherical Earth. 0.5N.
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What is the name of the album that is most frequently cited as the beginning of fusion?
The album that is most frequently cited as the beginning of fusion is "In a Silent Way" by Miles Davis. Released in 1969, it is often regarded as a groundbreaking and influential work that marked a significant shift in jazz and the emergence of fusion music.
"In a Silent Way" showcased a departure from Davis' previous acoustic jazz sound and incorporated elements of electric instruments, studio production techniques, and improvisational freedom. The album blended jazz with elements of rock, funk, and electronic music, creating a unique and experimental sonic landscape. The musicians involved in the recording, including Wayne Shorter, Herbie Hancock, and John McLaughlin, went on to become key figures in the fusion genre. This album laid the foundation for future fusion developments, influencing artists across various genres. Its atmospheric, ethereal, and exploratory nature set the stage for the fusion movement of the 1970s, which further integrated jazz with elements of rock, funk, and other genres. "In a Silent Way" remains a pivotal work in the history of fusion, symbolizing the fusion of diverse musical styles and the limitless possibilities of blending genres in innovative and creative ways.
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A proton is moving in a region of uniform magnetic field The magnetic field is directed into the plane of the paper: The arrow shows the velocity of the proton at one instant and the dotted circle gives the path followed by the proton: [2 marks] proton Explain why the path of the proton is circle_ 3b. The speed of the proton is 2.7 106 m s-1 and the magnetic field strength B is 0.41 T. [2 marks] Calculate the radius ofthe circular motion: 3c. Calculate the time for one complete revolution:
A proton is moving in a region of uniform magnetic field The magnetic field is directed into the plane of the paper: The arrow shows the velocity of the proton at one instant and the dotted circle gives the path followed by the proton. The path of the proton is a circle because it experiences a magnetic force perpendicular to its velocity. the time for one complete revolution is approximately 1.7 microseconds.
The path of the proton is a circle because it experiences a magnetic force perpendicular to its velocity. According to the right-hand rule, when a charged particle moves in a magnetic field, the force acting on it is perpendicular to both the velocity vector and the magnetic field direction. In this case, the force acts towards the center of the circle, causing the proton to move in a circular path.
To calculate the radius of the circular motion, we can use the formula for the centripetal force:
F = (q * v * B) / r
Where:
F is the centripetal force,
q is the charge of the proton ([tex]1.6 x 10^-{19}[/tex] C),
v is the velocity of the proton ([tex]2.7 * 10^6[/tex] m/s),
B is the magnetic field strength (0.41 T),
and r is the radius of the circular path.
The centripetal force is provided by the magnetic force, so we can equate the two:
(q * v * B) / r = (m * v^2) / r
Simplifying and rearranging the equation, we find:
r = (m * v) / (q * B)
Substituting the values:
r = ([tex]1.67 * 10^{-27}[/tex] kg * [tex]2.7 * 10^6[/tex]m/s) / ([tex]1.6 * 10^{-19}[/tex]C * 0.41 T)
Calculating this gives us the radius of the circular motion.
To calculate the time for one complete revolution, we can use the formula for the period (T) of circular motion:
T = (2 * π * r) / v
Substituting the calculated radius and the velocity value, we can find the period.
To calculate the radius of the circular motion, we'll use the formula:
r = (m * v) / (q * B)
Plugging in the values:
r = [tex](1.67 * 10^{-27} kg * 2.7 * 10^6 m/s) / (1.6 * 10^{-19} C * 0.41 T)[/tex]
r ≈[tex]1.47 * 10^-3[/tex] m or 1.5 mm (rounded to two significant figures)
So, the radius of the circular motion is approximately 1.5 mm.
To calculate the time for one complete revolution, we'll use the formula:
T = (2 * π * r) / v
Plugging in the values:
T = (2 * π * 1.47 x[tex]10^-3[/tex] m) / (2.7 x [tex]10^6[/tex] m/s)
T ≈ 1.73 x [tex]10^-6[/tex] s or 1.7 μs (rounded to two significant figures)
Therefore, the time for one complete revolution is approximately 1.7 microseconds.
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The first order maximum caused by a double slit illuminated with light of wavelength 625 nm is found at some spot on a screen. The light source is changed to a new wavelength which places its second order (m=2) maxima at the same spot where the 625 nm first order maxima used to lie.
(a) What is the wavelength of the new light source?
(b) Is this wavelength is the visible range?
The wavelength of the new light source is 625 nm. The wavelength of light between 400-700 nm is visible to the human eye.
(a) To calculate the wavelength of the new light source, we use the formula;
Δλ = λ₂ - λ₁
where Δλ is the difference between the two wavelengths, λ₂ is the wavelength of the new light source, and λ₁ is the wavelength of the original source.
We are told that the second-order maxima is at the same spot where the first-order maxima used to be for the 625 nm light source.
This means the position of the maxima is the same, which is only possible if the distance between the slits is the same as before.
The distance between the slits is given by;
d = λD/d
where d is the distance between the slits, λ is the wavelength of light, D is the distance from the slits to the screen, and m is the order of the maxima.
For the first-order maxima;
m = 1d = λD/d625 × 10^-9 m = d(2 m)/dd = 1.25 × 10^-6 m
For the second-order maxima;
m = 2d = λD/dλ = 2d/mDλ = 2(1.25 × 10^-6)/2 = 625 × 10^-9 m
Therefore, the wavelength of the new light source is 625 nm. The wavelength of light between 400-700 nm is visible to the human eye.
Therefore, the wavelength of the new light source is in the visible range. Answer: (a) 625 nm, (b) Yes.
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An electron travels through a conductor in a laboratory at a
speed of 0.790c. What is the kinetic energy of the electron in the
laboratory frame of reference?
The kinetic energy of the electron in the laboratory frame of reference is approximately 2.526 x 10^-14 Joules. This is calculated using the relativistic kinetic energy equation with the Lorentz factor and the rest mass of the electron.
The relativistic kinetic energy equation is given by:
K = (γ - 1)mc²
where K is the kinetic energy, γ is the Lorentz factor, m is the rest mass of the electron, and c is the speed of light.
To calculate γ, we can use the formula:
γ = 1 / sqrt(1 - (v² / c²))
where v is the velocity of the electron.
Given that the speed of the electron is 0.790c, we can substitute the values into the equations. The rest mass of an electron is approximately 9.11 x 10^-31 kg.
Calculating γ:
γ = 1 / sqrt(1 - (0.790c)² / c²)
= 1 / sqrt(1 - 0.6241)
≈ 1.603
Now, we can calculate the kinetic energy:
K = (γ - 1)mc²
= (1.603 - 1)(9.11 x 10^-31 kg)(3 x 10^8 m/s)²
≈ 2.526 x 10^-14 Joules
Therefore, the kinetic energy of the electron in the laboratory frame of reference is approximately 2.526 x 10^-14 Joules.
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