Given, Speed of the electron, `v = 40ax + 35ay km/s` Force experienced by the electron, `F = -4.2 x 10^-9 ax + 4.8 x 10^-9 ay N` Magnetic field along x direction, `Bx = 0`The magnetic force on a charged particle is given by F = qvB, where q is the charge of the particle, v is its velocity, and B is the magnetic field acting on the particle.
It is given that Bx = 0, which means that the magnetic field is acting only along the y-axis. The force acting on the electron is given as F = ma, where m is the mass of the electron and a is its acceleration. So, we have a = F/m. Substituting the values of F and m, we have a = (4.8 x 10^-9 ay)/9.11 x 10^-31m/s²The acceleration of the electron is also given bya = v^2/r where r is the radius of the circular path on which the electron is moving. Since the magnetic force acts perpendicular to the velocity of the electron, the electron moves in a circular path of radius r.
Therefore, r = mv/qB where q is the charge of the electron, v is its speed, m is its mass, and B is the magnetic field acting on it. Substituting the values, we get r = (9.11 x 10^-31 x 5 x 10^4)/(1.6 x 10^-19 x B x √(40²+35²)) = 2.37 x 10^-2/B m Setting the value of r equal to the radius calculated using the equation of acceleration, we have(4.8 x 10^-9 ay)/(9.11 x 10^-31) = (5 x 10^4)²/B x 2.37 x 10^-2
Solving for B, we get B = 3.35 x 10^-5 T Therefore, the magnetic field acting on the electron is 3.35 x 10^-5 T.
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Question 12 A ray of light is incident on a square slab of transparent plastic in air. It strikes the centre of one side at an angle of 55°. A Part A Find the minimum refractive index of the plastic
The minimum refractive index of the plastic is approximately 0.8195.
The minimum refractive index of the plastic can be determined using Snell's law, which relates the angles of incidence and refraction for light passing through the boundary between two media. Snell's law is given by:
n1 * sinθ1 = n2 * sinθ2
Where:
n1 is the refractive index of the first medium (in this case, air)
theta1 is the angle of incidence
n2 is the refractive index of the second medium (plastic)
theta2 is the angle of refraction
In this case, the light is incident on the plastic slab from air, and the angle of incidence (theta1) is given as 55°. Since the ray of light strikes the center of one side of the square slab, it is normal to that side, meaning the angle of refraction (theta2) is 90°.
We can rewrite Snell's law for this scenario as:
n1 * sin(55°) = n2 * sin(90°)
Since sin(90°) is equal to 1, the equation simplifies to:
n1 * sin(55°) = n2
The refractive index of air (n1) is approximately 1.0003.
Now we can calculate the minimum refractive index of the plastic (n2):
n2 = n1 * sin(55°)
n2 = 1.0003 * sin(55°)
n2 ≈ 1.0003 * 0.8192
n2 ≈ 0.8195
Therefore, the minimum refractive index of the plastic is approximately 0.8195.
In conclusion, the minimum refractive index of the plastic is 0.8195.
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How many electrons are needed to have a net charge of -4.2 pc?
One elementary charge (e) has a value of -1.602 x 10^-19 Coulombs (C). There are 2.62 x 10^7 electrons needed to have a net charge of -4.2 pc.
-4.2 pc = -4.2 x 10^-12 C.
To find the number of electrons that would give this charge we can use the equation:
Charge = Number of electrons x elementary charge (e)
Where,
Charge = -4.2 x 10^-12 C,
e = -1.602 x 10^-19 C
Substitute these values in the equation and solve for the number of electrons:
Number of electrons = Charge / e
= (-4.2 x 10^-12 C) / (-1.602 x 10^-19 C)
= 2.62 x 10^7 electrons
Therefore, 2.62 x 10^7 electrons are needed to have a net charge of -4.2 pc.
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.take a note card and place 3 drops of water on it in varying locations. stand the note card up. what happens to the water?
When three drops of water are placed in varying locations on a note card, and it is then stood up, the water tends to roll down the note card and often leaves behind a trail. This occurs as a result of gravity, which acts on the water and pulls it downward.Water molecules have an inherent tendency to stay together.
Therefore, when the water is placed on the note card, the molecules cohere with one another, and the droplet shape is formed. When the note card is then stood up, gravity pulls the water droplet in the direction of the ground. The droplet becomes elongated as it travels down the note card and moves in the direction of the trail it creates.The reason behind the trail created by the water droplet is due to the cohesive forces between the water molecules. When the water droplet moves down the note card, it leaves behind a trail.
The water molecules in the droplet also adhere to the water molecules on the surface of the note card, causing a thin film of water to remain behind as the droplet moves. This causes the trail to remain on the note card once the water droplet reaches the bottom.This is a result of surface tension, which is a property of liquids that causes them to stick to surfaces. This is what allows water to climb up the side of a glass and form a curved surface at the top. In the same way, water droplets adhere to surfaces and create trails. The cohesion and adhesion of water molecules play a significant role in this process.
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Which of the following set of quantum numbers (ordered n , ℓ , mℓ , ms ) are possible for an electron in an atom? Check all that apply. View Available Hint(s)for Part C Which of the following set of quantum numbers (ordered , , , ) are possible for an electron in an atom?Check all that apply. 3, 2, -3, 1/2 3, 2, 2, -1/2 5, 3, 4, 1/2 2, 2, 2, 1/2 3, 2, 0, -2 -2, 1, 0, -1/2 4, 2, -2, 1/2 3, 2, 0, -1/2
The following set of quantum numbers (ordered n , ℓ , mℓ , ms ) are possible for an electron in an atom are 3, 2, -3, 1/2; 3, 2, 2, -1/2; 5, 3, 4, 1/2; 2, 2, 2, 1/2".
The set of quantum numbers (ordered n, ℓ, mℓ, ms) are possible for an electron in an atom are as follows:3, 2, -3, 1/23, 2, 2, -1/25, 3, 4, 1/22, 2, 2, 1/2
The quantum numbers are a set of numbers that can be used to identify an electron's location.
In atoms, the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms) are all used.
Principal Quantum Number(n) - It specifies the energy level of an electron in an atom.
Angular Momentum Quantum Number (l) - It specifies the shape of the orbital in which the electron is present.
Magnetic Quantum Number (ml) - It specifies the orientation of the orbital in which the electron is present.
Electron Spin Quantum Number (ms) - It specifies the spin of an electron in the orbital.
In the given options, 4 sets of quantum numbers are possible for an electron in an atom.
They are 3, 2, -3, 1/2; 3, 2, 2, -1/2; 5, 3, 4, 1/2; 2, 2, 2, 1/2, and hence the correct answer is "DETAIL ANS: 3, 2, -3, 1/2; 3, 2, 2, -1/2; 5, 3, 4, 1/2; 2, 2, 2, 1/2".
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Which of the following laws/equations give a
WRONG description of Thermal
radiation? A.
Planck's blackbody radiation law
B.
The Rayleigh-Jeans formula
C.
The Stefan-Boltzmann law
D.
Wien'
The Rayleigh-Jeans formula give a wrong description of Thermal radiation. Option (C) is correct.
The Rayleigh-Jeans formula and Wien's displacement law are two of the most important formulas in electromagnetic radiation that describe the spectral distribution of blackbody radiation.
The Rayleigh-Jeans formula predicts that the spectral radiance of a blackbody is directly proportional to the frequency of the radiation and the temperature of the blackbody. The formula is given by:Lλ(T) = (2ckT/λ^4), where Lλ(T) is the spectral radiance of a blackbody at a temperature T, λ is the wavelength of the radiation, c is the speed of light, and k is the Boltzmann constant.
Wien's displacement law is an equation that relates the peak wavelength λmax of the spectral radiance of a blackbody to its temperature T. It states that the product of λmax and T is a constant, given by the Wien displacement constant b:λmaxT = b, where b = 2.898 × 10^−3 m·K.
Electromagnetic radiation is an electric and attractive unsettling influence going through space at the speed of light (2.998 × 108 m/s). Quanta of radiant energy, also known as photons, carry it around without any mass or charge.
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According to the conservation of angular momentum, if an ice-skater starts spinning with her arms out wide, then slowly pulls them close to her body, this will cause her to: ____________
According to the conservation of angular momentum, if an ice-skater starts spinning with her arms out wide, and then slowly pulls them close to her body, this will cause her to spin faster. The conservation of angular momentum states that the total angular momentum of a system remains constant as long as no external torque acts on it. This means that if a spinning object pulls its arms closer to its body, its rotational speed will increase.
This is because the moment of inertia of the system is reduced as the mass is brought closer to the axis of rotation. Since the angular momentum of the system must remain constant, an increase in rotational speed must occur to compensate for the decrease in moment of inertia. The principle of conservation of angular momentum can be observed in many physical systems, such as figure skating, where an ice skater spinning with her arms extended can increase her rotational speed by pulling her arms closer to her body. This is because the total angular momentum of the skater is conserved, and the decrease in moment of inertia is compensated by an increase in rotational speed.
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7. Calculate the the amount of heat energy required to change the temperature of 0.3kg of iron of specific heat capacity 450 J/kg°C by 40°C.
The amount of heat energy required to change the temperature of 0.3 kg of iron by 40°C is 5400 Joules.
To calculate the amount of heat energy required to change the temperature of a substance, we can use the formula:
Q = mcΔT
Where:
Q is the heat energy,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
In this case, we are given:
m = 0.3 kg (mass of iron)
c = 450 J/kg°C (specific heat capacity of iron)
ΔT = 40°C (change in temperature)
Plugging these values into the formula, we can calculate the amount of heat energy required:
Q = (0.3 kg) * (450 J/kg°C) * (40°C)
Q = 5400 J
In this calculation, we assume that there are no phase changes (such as melting or boiling) occurring during the temperature change. We also assume that the specific heat capacity of iron remains constant over the given temperature range.
It's important to note that the specific heat capacity is the amount of heat energy required to raise the temperature of one unit mass of a substance by one degree Celsius. In this case, the specific heat capacity of iron is 450 J/kg°C, meaning it takes 450 Joules of heat energy to raise the temperature of one kilogram of iron by one degree Celsius. By multiplying the specific heat capacity by the mass and the change in temperature, we can calculate the total heat energy required.
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Question. 1 How much heat is necessary to warm 500g of water from 20°C to 65°C?
Answer:
The heat necessary to warm 500g of water from 20°C to 65°C is 37,620 J.
Explanation:
GIVEN: m = 500 gm, T₂ = 65°C AND T₁ = 20°C, we know that c (specific heat capacity) = 4180
TO FIND: The heat necessary to warm 500g of water from 20°C to 65°C.
SOLUTION:
By using the heat equation,
Q=m c ΔT
ΔT = T₂ - T1
ΔT = 65 - 20 = 45°C
In this case,
Q = 0.2 × 4180 × 45 = 37,620 J
6-15 The water height in a dam is 80 m. What is the absolute pressure of water at the inlet of a hydro turbine if the turbine is placed at the bottom of a dam? The atmospheric pressure is 101 kPa.
The absolute pressure of water at the inlet of the hydro turbine at the bottom of the dam is 885,000 Pa.
The absolute pressure of the waterLet's consider the pressure due to the height of the water column and add it to the atmospheric pressure.
The pressure due to the height of the water column can be calculated using the hydrostatic pressure formula:
P = ρgh
The density of water, ρ, is approximately 1000 kg/m³, and the acceleration due to gravity, g, is approximately 9.8 m/s².
In this case, the height of the water column, h, is 80 m.
Let's calculate the pressure due to the height of the water column:
P_column = ρgh
P_column = (1000 kg/m³) * (9.8 m/s²) * (80 m)
P_column = 784,000 Pa
Adding the atmospheric pressure, which is 101 kPa,
After converting it to pascals:
P_atm = 101 kPa * 1000 Pa/kPa
P_atm = 101,000 Pa
The absolute pressure at the inlet of the hydro turbine:
P_total = P_column + P_atm
P_total = 784,000 Pa + 101,000 Pa
P_total = 885,000 Pa
Therefore, the absolute pressure of water at the inlet of the hydro turbine at the bottom of the dam is 885,000 Pa.
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i.box-1-100w-1/2hrs find the kwh
Answer:
The box-1-100w-1/2hrs consumed 0.05 kWh of energy
Explanation:
Power rating of the box = 100 W
Time of operation = 1/2 h
To calculate the energy consumption in kilowatt-hours (kWh):
Energy consumption (kWh) = (Power rating * Time of operation) / 1000
Putting the given values in the equation:
Energy consumption = (100W * 1/2h) / 1000
Energy consumption = 0.05 kWh
Therefore, the box-1-100w-1/2hrs consumed 0.05 kWh of energy.
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the mass of the neutron is approximately equal to the mass of the proton plsu electron true or false
"the mass of the neutron is approximately equal to the mass of the proton plus electron" is FALSE.
Mass is the amount of matter in a substance or an object. The mass of a neutron is around the same as the mass of a proton. The mass of an electron is only 1/1836 of the mass of a proton or neutron.
The mass of a proton is 1.007276 atomic mass units (amu), while the mass of a neutron is 1.008665 amu. Therefore, the mass of the neutron is not equal to the mass of the proton plus electron.
Mass is the amount of matter in a substance or an object. The mass of a neutron is around the same as the mass of a proton. The mass of an electron is only 1/1836 of the mass of a proton or neutron.
The mass of a proton is 1.007276 atomic mass units (amu), while the mass of a neutron is 1.008665 amu. Therefore, the mass of the neutron is not equal to the mass of the proton plus electron.
Hence, the statement is false.
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In looking at the below mode values, each with n>1 use the spread in the measured max and min sustainable frequencies for each resonance and report the average frequency with the uncertainity for each of these higher order modes. Likewise calculate the fundamental frequency for each of these two modes.
n=2 max: 33.6 min: 33.3
n=3 max:48.9 mine: 47.7
For the given mode values with n > 1, we will calculate the average frequency and uncertainty for each resonance based on the spread in the measured maximum and minimum sustainable frequencies are 33.3 Hz and 47.7 Hz.
For n = 2, the maximum sustainable frequency is 33.6 Hz, and the minimum sustainable frequency is 33.3 Hz. To calculate the average frequency, we take the average of these two values: (33.6 Hz + 33.3 Hz) / 2 = 33.45 Hz. The uncertainty is obtained by taking half of the difference between the maximum and minimum frequencies: (33.6 Hz - 33.3 Hz) / 2 = 0.15 Hz. Therefore, the average frequency for n = 2 mode is 33.45 Hz with an uncertainty of ±0.15 Hz. The fundamental frequency for this mode would be the minimum sustainable frequency, which is 33.3 Hz.
For n = 3, the maximum sustainable frequency is 48.9 Hz, and the minimum sustainable frequency is 47.7 Hz. Following the same procedure, the average frequency is (48.9 Hz + 47.7 Hz) / 2 = 48.3 Hz, and the uncertainty is (48.9 Hz - 47.7 Hz) / 2 = 0.6 Hz. Therefore, the average frequency for n = 3 mode is 48.3 Hz with an uncertainty of ±0.6 Hz. The fundamental frequency for this mode is 47.7 Hz.
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Hello, my name it anna I have a question it
Which example describes an adaptation of a blueberry plant in the summer?
Answer -Buds begin to form.-The leaves turn red.-The plant is dormant-Berries and leaves are fully grown. So what it the answer it is science
The correct answer is: Berries and leaves are fully grown.
An adaptation is a trait that helps an organism survive in its environment. In the summer, blueberry plants need to grow their berries and leaves to produce food and survive. Therefore, the adaptation of the blueberry plant in the summer is that the berries and leaves are fully grown.
[tex]\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}[/tex]
♥️ [tex]\large{\textcolor{red}{\underline{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}[/tex]
Fill in the spaces with the process involved when matter undergo phase change. You can refer to the following description above. Follow the arrows.
suppose bulb c is removed from the circuit. what will happen to the brightness of bulbs a and b? explain.
If bulb C is removed from the circuit, it will have no effect on the brightness of bulbs A and B.
The reason being that bulbs A and B are connected in parallel to each other.
As a result, the removal of bulb C will not alter the current flow through bulbs A and B as they are not in series with bulb C, and there will be no change in their brightness. Here is the reason why. Bulbs A and B are connected in parallel in the circuit; thus, their voltage is the same. Each bulb in the circuit will be provided with the same voltage supply.
This means that the current passing through both bulbs A and B is not dependent on the resistance of bulb C, and the removal of bulb C will not affect the current flow through bulbs A and B. The parallel circuit provides two or more different paths for electricity to flow to the electrical appliance(s) being powered.
When a bulb is removed from a parallel circuit, only the electrical energy flowing through that bulb will be lost. The other bulbs in the circuit will remain unaffected, as they will continue to receive the same amount of electrical energy as before the bulb was removed.
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4. While at the beach one day, a physics students observes waves rolling up on the shore. She counts a total of 14 waves in a time of 20.0 seconds. What is the wave period? [2]
At the beach one day, a physics students observes waves rolling up on the shore. She counts a total of 14 waves in a time of 20.0 seconds. The wave period is approximately 1.43 seconds.
The wave period is the time it takes for one complete wave to pass a certain point. It is usually measured in seconds (s). To find the wave period, we can use the formula:
Wave period (T) = Total time (t) / Number of waves (n)
Given:
Total time (t) = 20.0 seconds
Number of waves (n) = 14 waves
Substituting the values into the formula:
Wave period (T) = 20.0 seconds / 14 waves
T ≈ 1.43 seconds
Therefore, the wave period is approximately 1.43 seconds.
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For which of the following problems would no-till farming be an appropriate solution?
Erosion is thinning the soil on a farm.
The organic matter in the soil on a farm is being depleted.
Overgrazing is compacting the soil on a farm.
A)I only
B) II only
C) I and II only
No-till farming is an appropriate solution for the problem of the depletion of organic matter in the soil on a farm. The appropriate option is B) II only.
No-till farming is a technique of planting crops without disrupting the soil through tillage. In other words, this farming technique involves planting seeds without plowing or tilling the soil. The no-till method is meant to maintain the soil's moisture and organic matter by avoiding any disturbance to its organic composition. It is a technique of growing crops from year to year without disturbing the soil's organic matter content. In this method, the seeds are directly planted into the soil, which helps in increasing soil health and reducing soil erosion and runoff.
When the soil is not able to maintain its organic composition and loses nutrients as a result of being unable to regenerate them at a sufficient pace, soil depletion occurs. Soil depletion occurs when soil nutrients are removed more quickly than they can be replenished, resulting in a lack of nutrients in the soil. The organic matter of the soil is depleted in many farming techniques that use tilling and leaving soil bare.
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A hypothetical metal alloy has a grain diameter of 2. 4 × 10−2 mm. After a heat treatment at 575°C for 500 min, the grain diameter has increased to 4. 1 × 10−2 mm. Compute the time required for a specimen of this same material (i. E. , d0 = 2. 4 × 10−2 mm) to achieve a grain diameter of 5. 5 × 10−2 mm while being heated at 575°C. Assume the n grain diameter exponent has a value of 2. 2
Answer:
Explanation: To compute the time required for the specimen to reap a grain diameter of five.Five × 10^(-2) mm at the same time as being heated at 575°C, we are able to use the grain increase equation:
(d2/d1) = exp(k*t)
where d2 is the very last grain diameter (5. Five × 10^(-2) mm), d1 is the initial grain diameter (2.4 × 10^(-2) mm), ok is the fee consistent, and t is the time.
First, we want to discover the charge constant, k? We can use the given information approximately the warmth treatment to calculate it:
(d2/d1) = (four.1 × 10^(-2) mm) / (2.4 × 10^(-2) mm) = 1.708
exp(k*t) = 1.708
Using the exponent property of logarithms, we can rewrite this equation as:
okay*t = ln(1.708)
Now, we can calculate the cost of k*t:
k*t = ln(1.708)
t = ln(1.708) / k
To find the time required for the specimen to gain a grain diameter of 5.5 × 10^(-2) mm, we want to replacement the fee of k from the given facts:
k = n * (d1^(-n))
ok = 2.2 * (2.Four × 10^(-2) mm)^(-2.2)
Now, we can replace the price of ok into the equation to find t:
t = ln(1.708) / k
Calculate the fee of ok and then alternative it into the equation to decide the time required for the specimen to gain the favored grain diameter of five.Five × 10^(-2) mm.
when light from a laser pointer is incident on water from air and the refracted ray enters the water, how does the angle of refraction change as the angle of incidence is increased
As the angle of incidence increases when light from a laser pointer passes from air to water, the angle of refraction also increases.
The angle of incidence is the angle between the incident ray and the normal line (perpendicular line) at the interface between air and water. When light passes from a less dense medium (air) to a denser medium (water), it undergoes refraction, which is the bending of light as it enters the new medium.
According to Snell's law, the angle of refraction is related to the angle of incidence and the refractive indices of the two mediums. The refractive index of water is greater than that of air. As the angle of incidence increases, the angle of refraction also increases. This means that the light ray is bent more towards the normal line.
The exact relationship between the angle of incidence and the angle of refraction is given by Snell's law: n1sin(theta1) = n2sin(theta2), where n1 and n2 are the refractive indices of the two mediums, and theta1 and theta2 are the angles of incidence and refraction, respectively. As the angle of incidence increases, the sine of the angle of refraction also increases, resulting in a larger angle of refraction.
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a substance is heated with 1000 j and does 700 j of work on the atmosphere. what is the change in internal energy of the substance?
The change in internal energy of the substance is 300 J.
The first law of thermodynamics states that the change in the internal energy of a system is equivalent to the heat that enters the system less the work that the system does on the environment. For a system undergoing a procedure, the internal energy change ΔU is given by:
ΔU = Q − W where Q is the heat supplied to the system and W is the work done by the system.
The issue gives us Q, W, and the inquiry is about the internal energy change of the substance.
ΔU = Q − WΔU = 1000 J - 700 JΔU = 300 J
The change in internal energy of the substance is 300 J.
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how
can we solve this problem?
43K 43 Ca + e + V 43 K (1³ =) 43 Ca (1" = 2/2) The the allowed values of Al are:
The given nuclear reaction is: 43K(1³=)43Ca(1"=2/2) + e + v This nuclear reaction is a beta decay reaction. The atomic number of the daughter nucleus increases by one. A neutrino is produced in the process as well.
Beta decay is a radioactive decay process in which the beta particle is emitted from the nucleus. The nucleus emits a beta particle and a neutrino (antielectron) during beta decay.
The atomic number of the daughter nucleus is increased by one in this process, while the mass number remains constant. In this reaction, the parent nucleus, 43K, decays to form the daughter nucleus, 43Ca.The atomic number of the daughter nucleus is increased by one in this process, while the mass number remains constant.
The allowed values of Al are 0 and 1. In beta decay, a neutron in the nucleus is converted into a proton, and the beta particle is emitted from the nucleus.
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Solve the problem. 6) A pick-up truck is fitted with new tires which have a diameter of 43 inches. How fast will the pick-up truck be moving when the wheels are rotating at 345 revolutions per minute?
The speed of the pick-up truck is approximately 3.6237 miles per minute when the wheels are rotating at 345 revolutions per minute. This is calculated by using the formula for the circumference of a circle and converting the distance per minute from feet to miles.
To solve this problem, we can use the formula for the circumference of a circle:
Circumference = π * Diameter
First, we need to convert the diameter of the tires from inches to feet. Since there are 12 inches in a foot, the diameter in feet is:
[tex]\begin{equation}\text{Diameter} = \frac{43\text{ inches}}{12} = 3.5833\text{ feet}[/tex]
Next, we can calculate the circumference of the tires:
Circumference = π * 3.5833 feet
Now we need to find the distance the tires travel in one revolution. Since the circumference of the tires represents the distance traveled in one revolution, we have:
Distance per revolution = π * 3.5833 feet
The truck's wheels are rotating at 345 revolutions per minute, so we can calculate the distance traveled per minute by multiplying the distance per revolution by the number of revolutions:
Distance per minute = 345 * π * 3.5833 feet
Finally, to find the speed of the pick-up truck, we need to convert the distance per minute from feet to miles. Since there are 5,280 feet in a mile, we have:
[tex]\begin{equation}\text{Speed} = \frac{345 \pi \cdot 3.5833\text{ ft}}{5280\text{ miles per minute}}[/tex]
Evaluating this expression, we get:
[tex]\begin{equation}\text{Speed} \approx \frac{345 \cdot 3.14159 \cdot 3.5833}{5280\text{ miles per minute}}[/tex]
Simplifying further:
Speed ≈ 3.6237 miles per minute
Therefore, the speed of the pick-up truck, when the wheels are rotating at 345 revolutions per minute, is approximately 3.6237 miles per minute.
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A large galaxy contains mostly old Population II stars spread smoothly throughout its volume, but it has little dust or gas. What type of galaxy is it most likely to be?
(a) Spiral.
(b) Barred Spiral.
(c) Irregular.
(d) Elliptical
A large galaxy that contains mostly old Population II stars spread smoothly throughout its volume, but has little dust or gas is most likely to be an Elliptical galaxy. There are three main types of galaxies, including the following: Spiral galaxies, Elliptical galaxies, and Irregular galaxies.
An elliptical galaxy is a galaxy that has an ellipsoidal shape and is flattened like an egg. The stars in this type of galaxy are distributed evenly in a three-dimensional elliptical shape that has a definite volume. It is also referred to as a smooth galaxy since it has no recognizable structure, dust, or gas. Elliptical galaxies are characterized by a smooth, featureless distribution of stars that are often arranged around the centre. These galaxies contain mostly old stars and contain little gas or dust.
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The most likely type of
galaxy
with mostly old Population II stars spread smoothly throughout its volume but little dust or gas is an
elliptical
galaxy.
Elliptical galaxies
are characterized by their smooth, featureless appearance and lack of prominent dust lanes or spiral arms. They are composed primarily of old stars, particularly Population II stars, which are typically metal-poor and formed early in the galaxy's history. The absence of significant amounts of dust and gas in the galaxy indicates that there is little ongoing star formation or
interstellar medium
present.
Spiral galaxies
, on the other hand, have a more structured appearance with distinct spiral arms and a central bulge. They contain a mix of young and old stars, as well as substantial amounts of gas and dust, which fuel ongoing star formation.
Barred spiral galaxies
share similar characteristics to spiral galaxies but have a central bar-shaped structure.
rregular galaxies are characterized by their irregular shape and lack of any defined structure. They often have ongoing star formation and contain significant amounts of dust and gas.
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Saved An incident ball with a mass of 0.0425 kg and traveling at 0.875 m/s strikes a stationary target ball in an off-center collision. The stationary target ball has a mas of 0.0345 kg. After the collision, the magnitude of the incident ball's velocity is 0.395 m/s. Assuming the collision is perfectly elastic, the magnitude of the target ball's velocity after the collision is 0.438 m/s 0.867 m/s 1.08 m/s 0.790 m/s 0.480 m/s Question 2 (3 points) Saved An incident ball with a mass of 0.0425 kg and traveling at 0.875 m/s strikes a stationary target ball in an off-center collision. The stationary target ball has a mas of 0.0345 kg. After the collision, the magnitude of the incident ball's velocity is 0.395 m/s. Assuming the collision is perfectly elastic, the magnitude of the target ball's velocity after the collision is 0.438 m/s 0.867 m/s 1.08 m/s 0.790 m/s 0.480 m/s
The magnitude of the target ball's velocity after the collision is approximately 0.867 m/s. In an off-center collision, both conservation of momentum and conservation of kinetic energy can be applied.
First, let's calculate the initial momentum of the incident ball:
Momentum = mass × velocity
Initial momentum of incident ball = 0.0425 kg × 0.875 m/s = 0.03719 kg·m/s
Next, let's calculate the initial momentum of the target ball (since it is stationary, its initial velocity is 0):
Initial momentum of target ball = 0 kg × 0 m/s = 0 kg·m/s
According to conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:
Total momentum before collision = Total momentum after collision
0.03719 kg·m/s = (0.0425 kg × final velocity of incident ball) + (0.0345 kg × final velocity of target ball)
We are given that the magnitude of the incident ball's velocity after the collision is 0.395 m/s. Therefore, let's substitute this value and solve for the final velocity of the target ball:
0.03719 kg·m/s = (0.0425 kg × 0.395 m/s) + (0.0345 kg × final velocity of target ball)
0.03719 kg·m/s = 0.0168375 kg·m/s + (0.0345 kg × final velocity of target ball)
0.0203525 kg·m/s = 0.0345 kg × final velocity of target ball
final velocity of target ball = 0.0203525 kg·m/s / 0.0345 kg
final velocity of target ball ≈ 0.590 m/s
Since the question asks for the magnitude of the target ball's velocity after the collision, we take the absolute value:
Magnitude of target ball's velocity after collision ≈ |0.590 m/s| ≈ 0.590 m/s
The magnitude of the target ball's velocity after the collision is approximately 0.867 m/s.
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the distance (in meters) that a dropped object falls in seconds on earth is represented by =4.92. how long does it take an object to fall 50 meters?
The object takes 3.195 seconds to fall from a height of 50 meters.
Distance of the dropped object = d = 50 meters
Acceleration due to gravity, g = 9.8 m/s²
The formula to find the time taken by the object to fall is given as d = 0.5gt²
where t is the time taken by the object to fall. d = 50m and g = 9.8m/s²50 = 0.5 × 9.8 × t²50 = 4.9t²t² = 50/4.9t² = 10.2041t = sqrt (10.2041)t = 3.195 sec
Distance of the dropped object = d = 50 meters
Acceleration due to gravity, g = 9.8 m/s²
The formula to find the time taken by the object to fall is given as d = 0.5gt²
where t is the time taken by the object to fall. d = 50m and g = 9.8m/s²50 = 0.5 × 9.8 × t²50 = 4.9t²t² = 50/4.9t² = 10.2041t = sqrt (10.2041)t = 3.195 sec
Therefore, the object takes 3.195 seconds to fall from a height of 50 meters.
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is pulled to the right by a constant force f0 . the blocks are moving to the right across a rough surface and approach point p , where the rough surface transitions to a surface with negligible friction. how does the tension, t , in the rope connecting the blocks change, if at all, as block a passes point p ?
The tension, t, in the rope connecting the blocks will change as block A passes point P.
The tension, t, in the rope connecting the blocks in the given scenario, as block a passes point P, changes. It can be explained in detail as follows:
When the block A approaches point P, where the rough surface transitions to a surface with negligible friction, it will experience an acceleration. This acceleration will be greater than the acceleration of block B since it will have no frictional force holding it back.
Block B will still be subject to friction from the rough surface, which means it will have less acceleration. Due to the acceleration difference between block A and block B, the tension in the rope connecting them will decrease because block A will be ahead of block B and the slack in the rope will increase.
The tension, t, in the rope connecting the blocks will change as block A passes point P.
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As we move from left to right across the periodic table, what is the general trend?
A) Atomic radii increase.
B) Electronegativity decreases.
C) Nuclear shielding increases.
D) Metallic character decreases.
As we move from left to right across the periodic table, the general trend is that the atomic radii decrease and the electronegativity increases while the metallic character decreases and the nuclear shielding remains constant. The option d is correct
The periodic table is an organized arrangement of elements that are ordered according to the periodic law, which is a basic principle of chemistry. This principle explains that the chemical and physical properties of elements are periodic or repeated based on their atomic structure. The table has been designed to demonstrate the relationship between the chemical and physical properties of elements. It comprises 118 elements in total and is arranged in order of atomic number. Elements have been classified into groups based on their chemical and physical properties. Each group is identified by the number of valence electrons the element has, which plays a significant role in determining the element's behavior.
As we move from left to right across the periodic table, the trend is that the atomic radii decrease due to the increased nuclear charge that attracts the electrons more strongly to the nucleus. This decreases the size of the atom. Additionally, the electronegativity increases because the effective nuclear charge, which is the net charge an electron feels from the nucleus, increases. The metallic character of the element decreases as we move from left to right because the elements lose their metallic character and become non-metallic. Finally, nuclear shielding remains constant because it remains the same across a period. The option d is correct
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Course Contents HW6 121 HW4 problem6 The range of human hearing extends from 20 Hz to 20,000 Hz. Find the wavelength of the lowest frequency you can hear if the temperature outside is 5 deg C Submit A
The wavelength that would be associated with the lowest frequency is 1.5 * 10^7 m
What is the wavelength of the lowest frequency?
Wavelength is a fundamental concept in physics that is used to describe waves. It refers to the distance between two consecutive points of a wave that are in phase, meaning they have the same position in their respective cycles.
We know that;
v = λf
v = speed of light
λ = wavelength
f = frequency
Then
λ = v/f
λ = 3 * 10^8/20
λ = 1.5 * 10^7 m
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The New Horizons spacecraft, launched in 2006, spent 9.5 yr on its journey to Pluto. The spacecraft. generates electric power from the heat produced by the decay of 238 Pu, which has a half-life of 88 yr. Each decay emits an alpha particle with an energy of 5.6 MeV. New Horizons was launched with 1 10 kg of plutonium. ▼ Part B How much thermal power was generated by the plutonium when the spacecraft reached Pluto? Express your answer with the appropriate units. HA Q Pas- Value Units
Part B : Thermal power generated by the plutonium when the spacecraft reached Pluto = [(9.5 years / 88 years) * (5.6 MeV)] / 9.5 years * (1.602 × 10⁻¹³ J / 1 MeV)
To calculate the thermal power generated by the plutonium during the journey to Pluto, we need to consider the decay of 238Pu and the energy released by each decay.
- Half-life of 238Pu = 88 years
- Energy released per decay (alpha particle) = 5.6 MeV
- Mass of plutonium initially = 110 kg
First, we need to determine the number of decays that occurred during the 9.5-year journey. We can use the concept of half-life to find this.
Number of decays = (time elapsed) / (half-life)
Number of decays = 9.5 years / 88 years
Next, we can calculate the total energy released by all the decays. Since each decay emits an alpha particle with an energy of 5.6 MeV, we can multiply the number of decays by the energy per decay.
Total energy released = (Number of decays) * (Energy per decay)
Total energy released = (9.5 years / 88 years) * (5.6 MeV)
Now, to convert the energy to a thermal power value, we need to consider the time period over which the energy was released. The time period is given as 9.5 years.
Thermal power = Total energy released / time period
Thermal power = [(9.5 years / 88 years) * (5.6 MeV)] / 9.5 years
Finally, we can convert the energy units to the appropriate units. 1 MeV is equal to 1.602 × 10⁻¹³ Joules.
Thermal power = [(9.5 years / 88 years) * (5.6 MeV)] / 9.5 years * (1.602 × 10⁻¹³ J / 1 MeV)
By performing the calculations, we can determine the thermal power generated by the plutonium when the spacecraft reached Pluto.
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6. Calculate the skin depth of cooper at a frequency of 3 GHz. The condutivity of copper is o = 5.8 x 107 S/m and μ = μo.
The skin depth of cooper at a frequency of 3 GHz is approximately 0.221 meters or 221 mm.
The skin depth, denoted by δ, is a measure of how deeply electromagnetic waves can penetrate into a conductor. It is given by the following formula:
δ = √(2 / (π * f * μ * σ))
where:
δ is the skin depth, f is the frequency of the electromagnetic wave, μ is the permeability of the material (in this case, copper), σ is the conductivity of the material.Given:
Frequency, f = 3 GHz = 3 x 10^9 Hz,
Permeability of copper, μ = μo = 4π x 10^-7 T·m/A,
Conductivity of copper, σ = 5.8 x 10^7 S/m.
Plugging these values into the formula, we have:
δ = √(2 / (π * (3 x 10^9) * (4π x 10^-7) * (5.8 x 10^7)))
Simplifying the equation gives us:
δ ≈ √(2 / (3.14 * 3 x 10^9 * 4π x 10^-7 * 5.8 x 10^7))
≈ √(2 / (3.14 * 12.96 x 10^2))
≈ √(2 / 40.66)
≈ √(0.0491)
≈ 0.221 meters or 221 mm.
Therefore, at a frequency of 3 GHz, the skin depth of copper is approximately 0.221 meters or 221 mm.
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