Write the symbol for the nucleus that completes each nuclear equation
The symbol for the nucleus (X) that completes the nuclear equation is 224/79Au.
The nuclear equation for the alpha decay of Thallium-228 can be represented as follows:
228/81Tl -> 4/2He + X
In this equation, Thallium-228 (symbolized as 228/81Tl) undergoes alpha decay, resulting in the emission of an alpha particle (symbolized as 4/2He). The symbol "X" represents the unknown nucleus formed after the alpha decay.
To determine the identity of the unknown nucleus (X), we need to consider the conservation of both mass number and atomic number in nuclear reactions.
For the mass number:
The mass number of Thallium-228 is 228, which is equal to the sum of the mass numbers of the alpha particle (4) and the unknown nucleus (X). Therefore, we can write the equation:
228 = 4 + A, where A represents the mass number of the unknown nucleus.
By rearranging the equation, we find:
A = 228 - 4
A = 224
For the atomic number:
The atomic number of Thallium-228 is 81, which is equal to the sum of the atomic number of the alpha particle (2) and the atomic number of the unknown nucleus (Z). Therefore, we can write the equation:
81 = 2 + Z, where Z represents the atomic number of the unknown nucleus.
By rearranging the equation, we find:
Z = 81 - 2
Z = 79
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4
What is a mechanical wave? Give an example.
A mechanical wave propagates through a medium by physical motion. Example: sound wave or seismic wave.
A mechanical wave is a kind of wave that multiplies through a medium, similar to a solid, liquid, or gas, through real development of the particles of the medium. The development of the particles makes a disturbance in the medium, which then, makes some separation from the wellspring of the exacerbation as a wave. A delineation of a mechanical wave is a sound wave, which is made by the vibration of air particles. Right when a sound is conveyed, it causes a disrupting impact in the enveloping air, making areas of stuffed and slim air that development away from the wellspring of the sound as a wave. Another representation of a mechanical wave is a seismic wave, which is made by a tremor and goes through the ground.
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when is mechanical energy conserved
Mechanical energy is conserved in a closed system when there are no external forces acting upon it.
According to the principle of conservation of mechanical energy, the total amount of mechanical energy, which is the sum of kinetic energy and potential energy, remains constant as long as there is no work done by non-conservative forces like friction or air resistance.
In the absence of external forces, the total mechanical energy of the system remains unchanged throughout its motion. For example, in the case of a pendulum swinging back and forth, neglecting air resistance, the mechanical energy is conserved as the pendulum oscillates between its highest and lowest points.
However, it's important to note that mechanical energy conservation is an idealization and may not hold true in all real-world scenarios due to factors like friction, air resistance, and energy losses. In practical situations, mechanical energy conservation is often a useful approximation but may not be strictly maintained.
THerefore, mechanical energy is conserved in a closed system when there are no external forces acting upon it.
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A student rides a bicycle in a circle at a constant speed and constant radius. A force diagram for the student-bicycle system is shown in the figure above. The value for each force is shown in the figure. What is the acceleration of the student-bicycle system?
Fnormal = 500N
Ffriction to the right = 250N
Fgravity = 500N
So the acceleration of the student-bicycle system is [tex]a=5m/s^{2}[/tex].
The gravitational force that acts on the bicycle system is
[tex]F_{g}=500N[/tex]
Now the force, that is the gravitational force is related to mass of the system and the acceleration due to gravity of the system, 'm' and 'g' respectively.
Therefore, we can write
[tex]F_{g}=mg[/tex]
500 = m x 10 (since , g = 10 m/s-s)
∴ m = 50 kg
Now the net vertical force acting on the student bicycle system is 0. And the vertical acceleration of system is also 0. The total horizontal force acts to the right of the system. So by Newton's 2nd law of motion, we can write
[tex]F_{f}=ma[/tex]
[tex]a=\frac{F_{f}}{m}[/tex]
[tex]=\frac{250}{50}[/tex]
Therefore [tex]a=5m/s^{2}[/tex]
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The complete question is
A student rides a bicycle in a circle at a constant speed and constant radius. A force diagram for the student-bicycle system is shown in the figure above. The value for each force is shown in the figure. What is the acceleration of the student-bicycle system?
Shoshauna and Tamir are students who have
been investigating whether a material reflects blue
light. Read and compare their arguments, then
answer the questions below.
Shoshauna's Argument
The material does not reflect blue light. The
material appears green, not blue. This is evidence
that it does not reflect blue light.
Tamir's Argument
The material does not reflect blue light. A material
appears a certain color when that color of visible
light reflects off the material and into someone's
eyes. This material appears green and not blue, so
it must reflect green light, not blue light.
Which argument is more convincing? (highlight
one)
Shoshauna's argument
Tamir's argument
What makes one argument more convincing than
the other?
Tamir's argument is more convincing than Shoshauna's argument. This is because Tamir's argument provides a more thorough explanation for why the material appears green instead of blue.
What argument is more convincing?Tamir explains that the material appears green because it is reflecting green light into someone's eyes, not blue light.
This shows a clear understanding of the physics of light and how it interacts with materials. In contrast, Shoshauna's argument only states that the material appears green, without providing any explanation or reasoning for why this is the case.
Therefore, Tamir's argument is more convincing as it provides a logical explanation for why the material appears green instead of blue.
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A particle executes linear simple harmonic motion of amplitude A .what is the ratio of the kinetic energy and total energy when the displacement is half of the amplitude
The ratio of kinetic energy to total energy at x = A/2 is 4/√m k.
The total mechanical energy of a particle executing linear simple harmonic motion is given by:
E = 1/2 k A²
where k is the spring constant and A is the amplitude of the motion.
The kinetic energy of the particle is given by:
K = 1/2 m v²
where m is the mass of the particle and v is its velocity.
When the displacement of the particle is half of the amplitude (x = A/2), the velocity of the particle can be found using the equation of motion:
x = A/2 = A cos(ωt)
v = -Aω sin(ωt)
where ω is the angular frequency of the
At x = A/2, the velocity of the particle is:
v = -Aω sin(π/2) = -Aω
The kinetic energy of the particle is then:
K = 1/2 m (-Aω)² = 1/2 m A²ω²
The total energy of the particle at x = A/2 is:
E = 1/2 k (A/2)² = 1/8 k A²
The ratio of kinetic energy to total energy at x = A/2 is:
K/E = (1/2 m A²ω²) / (1/8 k A²) = 4 m ω² / k
We can also express ω in terms of k/m:
ω = √(k/m)
Therefore:
K/E = 4 m (k/m) / k = 4/√m k
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The kinematic viscosity of oxygen at 40 °C and a pressure of 160 kPa is 0.104 stokes. Determine the dynamic viscosity of the oxygen at this temperature and pressure. (Ro₂ = 0.2598 kPa.m³/kg.K)
The dynamic viscosity of oxygen at 40°C and 160 kPa is 64.17 × 10⁻⁶ Pa.s.
The dynamic viscosity of a fluid is equal to its kinematic viscosity multiplied by its density.
Given:
Kinematic viscosity of oxygen at 40°C and 160 kPa = 0.104 stokes
Density of oxygen at 40°C and 160 kPa = (160000 Pa / 0.2598 kPa.m³) = 616.55 kg/m³ (using ideal gas law)
Using the formula:
Dynamic viscosity = Kinematic viscosity * Density
We get:
Dynamic viscosity of oxygen = 0.104 stokes * 616.55 kg/m³ = 64.17 × 10⁻⁶ Pa.s
Therefore, the dynamic viscosity of the oxygen at 40°C and 160 kPa is 64.17 × 10⁻⁶ Pa.s.
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Robert Delaunay's Homage to Blériot (1914) was inspired by
O the invention of stroboscopic photography
the construction of the Eiffel Tower
his wife's new dress designs
the first flight across the English channel
The correct option is (d) the first flight across the English channel. Robert Delaunay's Homage to Blériot (1914) was the first flight across the English channel.
This painting pays homage to French aviator 'Louis Blériot', who made history by becoming the first person to cross English Channel in an airplane in 1909. Delaunay was fascinated by dynamism and speed of modern technology, and his painting captures the sense of movement and energy associated with achievements Blériot.
This painting also demonstrate use of Delaunay of vivid color and geometric shapes, which drew inspiration from early Cubism and the art movements of the early 20th century.
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the question:
The radio waves of a particular AM radio station broadcasts with a
frequency of 720 kHz. What is the frequency of a radio station that
transmits a radio wave with half the wavelength? (give answer in
standard AM radio station format)
and I have to answer in kHz
I get that 3x10^8/720,000Hz =416.66m
I'm just unsure what to do next. any help I get would be appreciated. Thank you!
The answer is "frequency of radio station that transmits a radio wave with half the wavelength of the 720 kHz radio station is 1440 kHz".
The frequency (f) and wavelength (λ) of a radio wave are related by the equation: c = f λ
Where, c is the speed of light (approximately 3.00 x [tex]10^8[/tex] m/s) in vacuum.
As for the first radio station, [tex]f_{1}[/tex] = 720 kHz.
the wavelength [tex]\lambda_{1}[/tex] of its radio waves can be calculated by using the above equation, i.e.
c = [tex]f_{1}[/tex][tex]\lambda_{1}[/tex]
[tex]\lambda_{1}[/tex] = c / [tex]f_{1}[/tex]
= (3.00 x [tex]10^8[/tex] m/s) / (720 x [tex]10^3[/tex] Hz)
[tex]\lambda_{1}[/tex]= 416.67 m
And for the second radio station, as given, its radio waves have half the wavelength of the first radio station. So, the wavelength [tex]\lambda_{2}[/tex] of its radio waves is [tex]\lambda_{1}[/tex]/2 = 208.33 m.
We can find the frequency [tex]f_{2}[/tex] of the second radio station using the same equation:
c = [tex]f_{2}[/tex].[tex]\lambda_{2}[/tex]
[tex]f_{2}[/tex] = c / [tex]\lambda_{2}[/tex]
= (3.00 x [tex]10^8[/tex] m/s) / (208.33 m)
[tex]f_{2}[/tex] = 1440 kHz
Therefore, frequency of radio station that transmits a radio wave with half the wavelength of the 720 kHz radio station is 1440 kHz.
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A beetle that has an inertia of 3.4 × 10 −6 kg sits on the floor. It jumps by using its muscles to push against the floor and raise its center of mass. Part A If its center of mass rises 0.75 mm while it is pushing against the floor and then continues to travel up to a height of 290 mm above the floor, what is the magnitude of the force exerted by the floor on the beetle
The magnitude of the force exerted by the floor on the beetle is approximately 1.33 Newtons.
To determine the magnitude of the force exerted by the floor on the beetle, we need to consider the conservation of energy. As the beetle pushes against the floor and raises its center of mass, the work done by the force exerted by the floor is equal to the change in potential energy of the beetle.
First, let's calculate the change in potential energy. The height change from the initial position on the floor to the final height above the floor is 290 mm - 0.75 mm = 289.25 mm, which we need to convert to meters: 289.25 mm = 0.28925 m.
The change in potential energy can be calculated using the formula: ΔPE = mgh, where m is the mass of the beetle and g is the acceleration due to gravity. Since the mass of the beetle is given as 3.4 × 10^(-6) kg, and g is approximately 9.8 m/s², we have:
ΔPE = (3.4 × 10^(-6) kg) × (9.8 m/s²) × (0.28925 m) = 1.0 × 10^(-6) J.
According to the work-energy principle, the work done by the force exerted by the floor is equal to ΔPE. Therefore, the magnitude of the force exerted by the floor on the beetle is:
Force = ΔPE / distance = (1.0 × 10^(-6) J) / (0.75 mm) = 1.33 N.
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5. A 0.25-kg skeet (clay target) is fired at an angle of 30° to the horizon with a speed of 25.0 m/s. When it reaches the maximum height, it is hit from below by a 15-g pellet traveling vertically upward at a speed of 200 m/s. The pellet is embedded in the skeet. (Air resistance can be ignored throughout.)
a. How much higher ddoes the skeet rise?
b. How much extra distance will the skeet travel as a result of the collision?
To solve this problem, we can use the principles of projectile motion and conservation of momentum.
Given:
Mass of the skeet, m_skeet = 0.25 kg
Angle of projection, θ = 30°
Initial speed of the skeet, v_skeet = 25.0 m/s
a. To find how much higher the skeet rises, we need to determine the change in height of the skeet from its initial position to the maximum height.
First, let's find the initial vertical velocity (v_y_initial) of the skeet. Since the skeet is fired at an angle of 30° to the horizon, we can calculate it as follows:
v_y_initial = v_skeet * sin(θ)
v_y_initial = 25.0 m/s * sin(30°)
Calculating the value:
v_y_initial ≈ 12.5 m/s
At the maximum height, the vertical velocity (v_y) becomes zero. We can use this information to find the time it takes for the skeet to reach the maximum height.
Using the equation for vertical motion under constant acceleration:
v_y = v_y_initial - g * t
Since v_y becomes zero at the maximum height:
0 = v_y_initial - g * t
Solving for t:
t = v_y_initial / g
Substituting the values:
t = 12.5 m/s / 9.8 m/s^2
Calculating the value:
t ≈ 1.28 s
Now, we can find the change in height (Δh) by using the equation for vertical motion:
Δh = v_y_initial * t - 0.5 * g * t^2
Substituting the values:
Δh = 12.5 m/s * 1.28 s - 0.5 * 9.8 m/s^2 * (1.28 s)^2
Calculating the value:
Δh ≈ 8.02 m
Therefore, the skeet rises approximately 8.02 meters higher.
b. To find the extra distance the skeet will travel as a result of the collision, we need to determine the horizontal component of the velocity added by the pellet.
The horizontal component of the velocity of the pellet (v_pellet_horizontal) is zero because it is traveling vertically upward.
The horizontal component of the velocity of the skeet (v_skeet_horizontal) remains unchanged throughout the motion because no external horizontal forces act on the system.
Since the horizontal component of the velocity is conserved, the extra distance traveled by the skeet (Δx) is equal to the horizontal component of the velocity of the pellet (v_pellet_horizontal). Therefore:
Δx = v_pellet_horizontal
Calculating the value:
Δx = 200 m/s * cos(90°)
Since cos(90°) = 0:
Δx = 0
Therefore, the skeet does not travel any extra distance as a result of the collision.
Who was MOST known for country blues?
Taj Mahal
Robert Johnson
Muddy Waters
Bessie Smith
Answer: Robbert Jhonson
Explanation:Robert Johnson (May 8, 1911 – August 16, 1938). Delta blues singer, songwriter, guitarist, and harmonica player. Despite his short life and limited recordings, he is perhaps the best known blues musician of the pre-war era, sometimes called the "King of the Delta Blues".
The writings of Transcendentalists had the greatest influence on which of the following movements?
The movement known as American Romanticism was most influenced by the writings of Transcendentalists.
In the United States, transcendentalism was a philosophical and literary movement that began in the 19th century. The intrinsic goodness of people and nature was emphasized, along with the value of intuition and personal experience and the rejection of societal norms and institutions.
Ralph Waldo Emerson and Henry David Thoreau, among other transcendentalist philosophers, promoted independence, spiritual inquiry, and a close relationship with nature.
American Romanticism evolved under the strong impact of Transcendentalism's ideas and principles. Often integrating themes of nature, the sublime, and the search for spiritual truth, this literary and artistic movement praised individualism, imagination, and emotional expression.
The Transcendentalists' writings shaped the philosophical and artistic underpinnings of the Romantic movement, serving as a cornerstone for it.
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Lester wants to fly his helicopter 537 meters, from one side of a field to the other. His helicopter has enough battery power remaining to fly for 3,600 seconds. If he flies the helicopter at a constant velocity of 10.0 meters per second to the west, how long will it take the helicopter to cross the field?
The time taken by the helicopter to cross the field is 53.7 seconds.
Distance to be travelled by the helicopter, d = 537 m
Velocity with which the helicopter is flying, v = 10 m/s
Velocity of a moving object is defined as the amount of distance covered by the object in unit time. It is the time rate of change of the displacement travelled.
So,
v = d/t
Therefore, the time taken by the helicopter to cross the field,
t = d/v
t = 537/10
t = 53.7 s
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A stanza of 12-bar blues is typically four phrases long.
True
False
Answer:
False
Explanation:
Nearly all blues music is played to a 4/4 time signature, which means that there are four beats in every measure or bar and each quarter note is equal to one beat. A 12-bar blues is divided into three four-bar segments.
An electron is located within an interval of 0.115 nm in the north-south direction. What is the minimum uncertainty delta v. in the electron's velocity in that direction?
The Heisenberg uncertainty relation is given different forms in different textbooks. Use the form employing greater than or equal to h/(4pi).
Deltav=_____m/s
The minimum uncertainty in the electron's velocity in the north-south direction is Δv = 4.38 × 10⁵ m/s.
The Heisenberg uncertainty principle states that the product of the uncertainties in position and momentum of a particle is greater than or equal to h/(4π), where h is the Planck's constant. Mathematically, it can be written as:
Δx.Δv ≥ h/(4π)
Here, we are given the uncertainty in the position of the electron as:
Δx = 0.115 nm = 1.15 × 10⁻¹⁰ m
We need to find the minimum uncertainty in the velocity of the electron in the north-south direction. Let's assume this uncertainty as Δv.
Plugging in the given values, we get:
Δx .Δv ≥ h/(4π)
1.15 × 10⁻¹⁰ m * Δv ≥ (6.626 × 10⁻³⁴J.s)/(4π)
Δv ≥ (6.626 × 10⁻³⁴ J.s)/(4π * 1.15 × 10⁻³⁴ m)
Δv ≥ 4.38 × 10⁵ m/s
Therefore, the minimum uncertainty in the electron's velocity in the north-south direction would be Δv = 4.38 × 10⁵ m/s.
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Which of the following statements is NOT true?
The blues is often in a 12-bar format
The roots of the blues were in New York City.
The blues influenced music that came later.
Lyrics are important to the blues.
Answer:
The roots of the blues were in new York City
Explanation:
The origins are South American
(c) The angular velocity of a disc of mass 3 kg and radius 0.6 m changes from 10 rev/s to 15 rev/s in 0.07 s. Calculate the impulse and the torque on the disc.
Explain, in terms of the air molecules, why the pressure at the top of a mountain is less
than at sea level.
For 3 marks
If a 0.5kg basketball is flying through the air at 10 m/s then what is its kinetic energy.
The equation KE = 0.5*m*v2 may be used to determine the kinetic energy of a 0.5 kilogramme basketball travelling through the air at 10 m/s.
Given that the ball's mass is 0.5 kg and its speed is 10 m/s, its kinetic energy equals 25 J (0.5 * 0.5 * 10 * 10). The energy an item has as a result of motion is known as kinetic energy.
It is the energy an object has as a result of motion, and it is equivalent to the effort necessary to propel the item from rest to its present velocity. Since kinetic energy is inversely related to velocity, it will increase by a factor of four if the ball's velocity is twice.
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Calculate the energy in electron volts of (a) an electron that has de Broglie wavelength 400 nm and (b) a photon that has wave. length 400 nm
Answer:
(a) To calculate the energy in electron volts of an electron with a de Broglie wavelength of 400 nm, we can use the de Broglie wavelength equation:
λ = h / p
where λ is the de Broglie wavelength, h is Planck's constant (6.626 x 10^-34 J s), and p is the momentum of the particle. We can rearrange this equation to solve for the momentum:
p = h / λ
Plugging in the given de Broglie wavelength, we get:
p = (6.626 x 10^-34 J s) / (400 x 10^-9 m)
= 1.6565 x 10^-24 kg m/s
To calculate the kinetic energy of the electron, we can use the formula:
KE = p^2 / (2m)
where m is the mass of the electron (9.109 x 10^-31 kg). Plugging in the momentum we just calculated, we get:
KE = (1.6565 x 10^-24 kg m/s)^2 / (2 x 9.109 x 10^-31 kg)
= 1.423 x 10^-17 J
Finally, we can convert this energy from joules to electron volts (eV) using the conversion factor 1 eV = 1.602 x 10^-19 J:
KE = 1.423 x 10^-17 J / (1.602 x 10^-19 J/eV)
= 88.8 eV
Therefore, the energy in electron volts of an electron with a de Broglie wavelength of 400 nm is 88.8 eV.
(b) To calculate the energy in electron volts of a photon with a wavelength of 400 nm, we can use the formula:
E = hc / λ
where E is the energy of the photon, h is Planck's constant, c is the speed of light (299,792,458 m/s), and λ is the wavelength of the photon. Plugging in the given wavelength, we get:
E = (6.626 x 10^-34 J s) x (299,792,458 m/s) / (400 x 10^-9 m)
= 4.965 x 10^-19 J
Finally, we can convert this energy from joules to electron volts using the conversion factor 1 eV = 1.602 x 10^-19 J:
E = 4.965 x 10^-19 J / (1.602 x 10^-19 J/eV)
= 3.10 eV
Therefore, the energy in electron volts of a photon with a wavelength of 400 nm is 3.10 eV.
Explanation:
If a cosmic ray proton approaches the Earth from outer space along a line toward the center of the Earth that lies in the plane of the equator, in what direction will it be deflected by the Earth’s magnetic field? What about an electron? A neutron?
A cosmic ray proton approaching the Earth's equator along a line towards the center of the Earth will be deflected by the Earth's magnetic field in a direction perpendicular to both the direction of the proton's motion and the Earth's magnetic field lines.
Which function does direction depend on?The direction of deflection will depend on the orientation of the Earth's magnetic field at the point of entry. An electron would also be deflected in the same direction as the proton because the deflection of a charged particle in a magnetic field is determined by its charge and velocity, and the electron has a negative charge that would cause it to deflect in the opposite direction to the proton.
A neutron, on the other hand, has no charge and would not be deflected by the Earth's magnetic field.
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A 250 kg beam is raised through 25 m at a constant velocity by a crane.
b) Determine the work done by the crane on the beam.
c) Determine the work done by gravity on the beam.
Complete this sentence: If the human population surpasses Earth's global capacity, there will be increased _________ of resources.
abundance
amount
scarcity
sources
Which definition refers to electrical power?
OA. The energy of an electric charge minus the resistance in a circuit
B. The rate at which electrical energy is transferred in a circuit
C. The energy of an electric charge due to its position in an electric
field
D. The rate at which an electric charge flows past a point in a circuit
The definition that refers to electrical power is B. The pace at which electrical energy in a circuit is transmitted.
The pace at which electrical energy is transported through a circuit is referred to as its power. It is measured in watts and is calculated by multiplying the voltage by the current in the circuit.
Power is the amount of energy transferred per unit time, and it represents the rate at which work is done. In an electrical circuit, power is the product of the voltage and the current.
The unit of power is the watt, which is equivalent to one joule of energy transferred per second. As a result, B is the proper response to the question.
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Particles q₁ = +18.1 µC, q2 = -11.2 μC, and 93 = +5.67 μC are in a line. Particles q₁ and q2 are separated by 0.280 m and particles q2 and q3 are separated by 0.350 m. What is the net force on particle q₂?
Answer:
[tex]\vec F_{net \ on \ q_2}=15.9637N}} \ \text{at 180\textdegree (or to the left/negative x-axis)}[/tex]
Explanation:
Using Coulomb's law to answer this question.
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Coloumb's Law:}}\\\\\vec F_e=\frac{k_eq_1q_2}{r^2} \cdot \hat r \end{array}\right}[/tex]
[tex]k_e[/tex] is Coulomb's constant ([tex]8.99 \times10^9\frac{Nm^2}{C^2}[/tex])[tex]\hat r[/tex] is a direction vector that points towards the charge you are calculating the force on~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:
[tex]q_1=18.1 \ \mu C \rightarrow 18.1 \times 10 ^{-6} \ C\\r_{2_{1}}=0.280 \ m\\q_2=-11.2 \ \mu C \rightarrow -11.2 \times 10 ^{-6} \ C\\q_3=5.67 \ \mu C \rightarrow 5.67 \times 10 ^{-6}\ C\\r_{2_{3}}=0.350 \ m[/tex]
Find:
[tex]|| \vec F_{net \ on \ q_{2}}||= \ ?? \ N[/tex]
**Assuming q_2 is the center of a coordinate system.
(1) - Find the force on q_2 exerted by q_1
[tex]\vec F_{2_{1}}=\frac{k_eq_2q_1}{(r_{2_{1}})^2} \cdot \hat r_{2_{1}}\\\\\Longrightarrow \vec F_{2_{1}}=\frac{(8.99 \times 10^{9})(-11.2 \times 10^{-6})(18.1 \times 10^{-6})}{(0.280)^2} \cdot \frac{ < 0.280,0 > }{\sqrt{(0.280)^2+(0)^2} } \\\\\Longrightarrow \vec F_{2_{1}}=-23.2456 \cdot < 1,0 > \\\\\therefore \boxed{\vec F_{2_{1}}= < -23.2456,0 > N}[/tex]
(2) - Find the force on q_2 exerted by q_3
[tex]\vec F_{2_{1}}=\frac{k_eq_2q_3}{(r_{2_{3}})^2} \cdot \hat r_{2_{3}}\\\\\Longrightarrow \vec F_{2_{3}}=\frac{(8.99 \times 10^{9})(-11.2 \times 10^{-6})(5.67 \times 10^{-6})}{(0.280)^2} \cdot \frac{ < - 0.350,0 > }{\sqrt{(-0.350)^2+(0)^2} } \\\\\Longrightarrow \vec F_{2_{3}}=-7.2819 \cdot < -1,0 > \\\\\therefore \boxed{\vec F_{2_{3}}= < 7.2819,0 > N}[/tex]
(3) - Find the net charge on q_2
[tex]\vec F_{net \ on \ q_2}=\vec F_{2_{1}}+\vec F_{2_{3}}\\\\\text{Recall that} \ \vec F_{2_{1}}= < -23.2456,0 > N \ \text{and} \ \vec F_{2_{3}}= < 7.2819,0 > N\\\\\Longrightarrow \vec F_{net \ on \ q_2}= < -23.2456,0 > + < 7.2819,0 > \\\\\therefore \boxed{\vec F_{net \ on \ q_2}= < -15.9637,0 > N}\\\\\Longrightarrow ||\vec F_{net \ on \ q_2}||=\sqrt{(-15.9637)^2+(0)^2} \\\\\therefore \boxed{\boxed{||\vec F_{net \ on \ q_2}||=15.9637N}} \ \text{at 180\textdegree (or to the left/negative x-axis)}[/tex]
1. A car has a total mass of 1200 kg and is traveling at 100 km per hour when the driver experience is a brake failure and collides with the barrels. Calculate the change in momentum it will experience whilst coming to a standstill.2. According to GSU's HyperPhysics Project this crash would have been fatal for an average 80 kg person. The safety zone in terms of momentum, ranges from 0 to 1,000 kg per metre per second. Determine the minimum velocity the car can slow down to during a collision with the barrels without the crash being fatal.
Answer:
1. To calculate the change in momentum of the car, we need to use the formula:
Δp = m * Δv
where Δp is the change in momentum, m is the mass of the car, and Δv is the change in velocity.
At the moment of collision, the car is traveling at 100 km/h, which is 27.78 m/s. When the car comes to a standstill, its velocity is 0 m/s. So the change in velocity is:
Δv = 0 - 27.78 = -27.78 m/s
The mass of the car is 1200 kg. So the change in momentum is:
Δp = m * Δv = 1200 kg * (-27.78 m/s) = -33,336 kg m/s
Therefore, the change in momentum of the car is -33,336 kg m/s.
2. To determine the minimum velocity the car can slow down to during a collision with the barrels without the crash being fatal, we need to use the formula:
p = mv
where p is momentum, m is mass, and v is velocity.
The safety zone in terms of momentum is from 0 to 1000 kg/m/s. We know that the mass of an average person is 80 kg. So we can calculate the maximum momentum that an 80 kg person can safely withstand:
p_max = 80 kg * 1000 kg/m/s = 80,000 kg m/s
Now we can rearrange the formula to solve for the minimum velocity:
v_min = p_max / m
v_min = 80,000 kg m/s / 1200 kg
v_min = 66.67 m/s
Therefore, the minimum velocity that the car can slow down to during a collision with the barrels without the crash being fatal is 66.67 m/s.
Complete the sentence. _________ is a health impact related to light pollution.
Low blood pressure
Disrupted circadian rhythms
Burning of the skin
Digital macular degeneration
Disrupted circadian rhythms are a health impact related to light pollution. The correct option is B.
Light pollution can indeed have an impact on human health, and disrupted circadian rhythms are one of the well-documented consequences. When exposed to excessive artificial light, especially during nighttime, the natural circadian rhythms that regulate our sleep-wake cycle and other physiological processes can be disrupted.
Option A, low blood pressure, is not directly related to light pollution. While light pollution may indirectly affect blood pressure through its influence on sleep quality and overall health, it is not a direct health impact of light pollution itself.
Option C, burning of the skin, is not associated with light pollution. Burning of the skin typically occurs due to excessive exposure to ultraviolet (UV) radiation from the sun or artificial sources like tanning beds, but it is not a consequence of light pollution.
Option D, digital macular degeneration, is also not specifically linked to light pollution. Macular degeneration is a progressive eye condition that affects the central part of the retina (the macula). While prolonged exposure to screens or certain types of artificial light may have implications for eye health, the term "digital macular degeneration" is not a recognized medical condition.
Therefore, disrupted circadian rhythms have been widely studied and recognized as a health impact associated with light pollution, making option B the most appropriate choice.
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An automobile, whose speed is increasing at a rate of 0.6m / (s ^ 2) , travels along a circular road of radius 20m. What will be automobile's angular acceleration when its instantaneous speed of the automobile is 4 m/s?
Sphere A is located at the origin and has a charge of +2.0x10^-6 C. Sphere B is located at +0.60 m on the x-axis and has a charge of -3.6x10^-6 C. Sphere C is located at +0.80 m on the x-axis and has a charge of +4.0x10^-6 C. Determine the net force on sphere B.
Answer:
[tex]\vec F_{net}=0.067425 \ N \ \text{at} \ 0 \textdegree \ \text{(or to the right/positive x-axis)}[/tex]
Explanation:
Refer to the attached image(s).