Answer:
3.33×10^-5 meters above the origin along the positive y-axis
Explanation:
The weight of an object is given by its mass multiplied by the acceleration due to gravity, which is 9.81 m/s² near the surface of the earth. For an electron with mass me, its weight is:
W = me * g
= 9.11×10^-31 kg * 9.81 m/s²
= 8.93×10^-30 N
Since the electric force acting on the electron is opposite to its weight, the electric force must have the same magnitude as its weight but in the opposite direction:
|F_E| = |W| = 8.93×10^-30 N
The electric force between two point charges is given by Coulomb's law:
F_E = k * |q1| * |q2| / r²
where k is the Coulomb constant, q1 and q2 are the charges of the two particles, and r is the distance between them.
In this problem, q1 is the fixed charge of -0.55 nC at the origin, and q2 is the charge of the electron, which we want to find. Since we know the magnitude of the electric force between the two charges, we can solve for the distance between them:
r² = k * |q1| * |q2| / |F_E|
= 8.99×10^9 N⋅m²/C² * 0.55×10^-9 C * |q2| / (8.93×10^-30 N)
= 6.95×10^6 * |q2|
Taking the square root of both sides, we get:
r = 2.64×10^-3 * sqrt(|q2|)
Now, we need to find the distance r at which the electric force between the two charges is equal in magnitude but opposite in direction to the weight of the electron. Equating the expression for r above with the distance y along the y-axis where the electron is placed, we get:
2.64×10^-3 * sqrt(|q2|) = y
Since the electron is placed on the y-axis, its x and z coordinates are zero, and the distance between the electron and the fixed charge is simply the y-coordinate. The electric force between the charges will be attractive (i.e., in the negative y direction), so the direction of the force vector will be opposite to the positive y direction. Therefore, we can write the electric force on the electron as:
F_E = - k * |q1| * |q2| / y²
Setting this equal to the weight of the electron, we have:
k * |q1| * |q2| / y² = |W|
|q2| = |W| * y² / (k * |q1|)
= 8.93×10^-30 N * y² / (8.99×10^9 N⋅m²/C² * 0.55×10^-9 C)
= 1.56×10^-20 * y²
Substituting this expression for |q2| into the expression for r above, we get:
r = 2.64×10^-3 * sqrt(1.56×10^-20 * y²)
= 1.35×10^-11 * y
Equating this expression for r with the expression for y above, we have:
2.64×10^-3 * sqrt(1.56×10^-20 * y²) = y
Squaring both sides and simplifying, we get:
y = 3.33×10^-5 m
Therefore, the electron must be placed at a distance of 3.33×10^-5 meters above the origin, along the positive y-axis, in order for the electric force acting on it to be exactly opposite to its weight.
To summarize, we used Coulomb's law to relate the electric force between the electron and the fixed charge at the origin to the distance between them, and equated this force with the weight of the electron. We then solved for the distance at which the two forces are equal in magnitude but opposite in direction, and found that the electron must be placed 3.33×10^-5 meters above the origin along the positive y-axis.
Hope this helps!
The chains of a swing on a playground swing set are 4.0 m
long. What is the period of this swing?
Express your answer to two significant figures and include the appropriate units.
Answer:
2.54 seconds
Explanation:
The period of a swing on a playground swing set can be calculated using the formula:
T = 2π√(L/g)
where T is the period, L is the length of the swing's chain, and g is the acceleration due to gravity, which is approximately 9.81 m/s².
Substituting the given values, we get:
T = 2π√(4.0 m/9.81 m/s²)
T = 2π√0.407
T = 2.54 s (rounded to two significant figures)
Therefore, the period of the swing is 2.54 seconds.
Hope this helps!
State the difference between positive and negative zero error of a vernier calliper
The 0.100 kg
sphere in (Figure 1) is released from rest at the position shown in the sketch, with its center 0.400 m
from the center of the 5.00 kg
mass. Assume that the only forces on the 0.100 kg
sphere are the gravitational forces exerted by the other two spheres and that the 5.00 kg
and 10.0 kg
spheres are held in place at their initial positions.
What is the speed of the 0.100 kg sphere when it has moved 0.150 m to the left from its initial position?
As per the given data, the speed of the 0.100 kg sphere when it has moved 0.150 m to the left from its initial position is 0.736 m/s.
Since only the gravitational forces are acting on the 0.100 kg sphere, we can use the conservation of energy principle to find its speed at any position.
We can use the initial position of the sphere as the reference point for potential energy and write the initial total energy as the sum of the potential energy and kinetic energy.
At any other position, the total energy will still be the sum of the potential energy and kinetic energy, but their values will be different.
The initial total energy of the system is:
E_i = m_0gh
Where m_0 is the mass of the 0.100 kg sphere, g is the acceleration due to gravity, and h is the height of the sphere above the reference position. In this case, h = 0.4 m.
The final total energy of the system is:
[tex]E_f = m_0v^2/2 + m_0gh_f[/tex]
Where v is the speed of the sphere, and h_f is the height of the sphere above the reference position at the final position.
Since the system is isolated, the initial and final energies must be equal:
E_i = E_f
[tex]m_0gh = m_0v^2/2 + m_0gh_f[/tex]
Solving for v, we get:
v = sqrt(2gh - 2gh_f)
To find the final height h_f, we can use the fact that the center of mass of the system remains fixed throughout the motion.
The initial center of mass is at a distance of 0.4 m from the center of the 5.00 kg sphere, and the masses of the 5.00 kg and 10.0 kg spheres are 5.00 kg and 10.0 kg, respectively.
Therefore, the initial center of mass is at:
x_cm,i = (0.4*0.1 + 5*0 + 10*0)/(0.1 + 5 + 10) = 0.032 m
where we have taken the x-axis to be horizontal and passing through the centers of the 5.00 kg and 10.0 kg spheres.
At the final position, the center of mass must still be at the same horizontal position:
x_cm,f = (5*0.1*(-0.15) + 10*0)/(0.1 + 5 + 10) = -0.011 m
where we have taken the leftward direction as positive.
The final height of the sphere is then:
h_f = 0.4 - x_cm,f = 0.4 + 0.011 = 0.411 m
Substituting the values of g, h, and h_f in the equation for v, we get:
v = sqrt(2*9.81*0.4 - 2*9.81*0.411) = 0.736 m/s (rounded to three significant figures)
Therefore, the speed of the 0.100 kg sphere when it has moved 0.150 m to the left from its initial position is 0.736 m/s.
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that we odels, we 2. his type entists base tand somet of the real ents, we five elect period nd at particles a) Copy the table below into your books. b) In the first column, write down the names of the first 20 elements. c) Use the Periodic Table at the back of this book to complete the table. Element Symbol Number of Number of Number of protons electrons protons and neutrons Na V Number of neutrons The diagrams below show models of certain elements. a) Write down the number of protons, electrons and neutrons for each element. b) Identify each element. ents have number. Symbol for the Figure 11 Atomic information a Mass number and neutrons
The difference between an atom's mass number (A) and atomic number (Z) is equal to the number of neutrons.
1) H – Hydrogen - 1 proton, 1 electron, 0 neutrons
2) He – Helium - 2 protons, 2 electrons, 2 neutrons
3) Li – Lithium - 3 protons, 3 electrons, 4 neutrons
4) Be – Beryllium - 4 protons, 4 electrons, 5 neutrons
5) B – Boron - 5 protons, 5 electrons, 6 neutrons
6) C – Carbon - 6 protons, 6 electrons, 6 neutrons
7) N – Nitrogen - 7 protons, 7 electrons, 8 neutrons
8) O – Oxygen - 8 protons, 8 electrons, 8 neutrons
9) F – Fluorine - 9 protons, 9 electrons, 10 neutrons
10)Ne – Neon - 10 protons, 10 electrons, 10 neutrons
11) Na – Sodium - 11 protons, 11 electrons, 12 neutrons
12) Mg – Magnesium - 12 protons, 12 electrons, 12 neutrons
13) Al – Aluminium - 13 protons, 13 electrons, 14 neutrons
14) Si – Silicon - 14 protons, 14 electrons, 14 neutrons
15) P – Phosphorus - 15 protons, 15 electrons, 17 neutrons
16) S – Sulphur - 16 protons, 16 electrons, 16 neutrons
17) Cl – Chlorine - 17 protons, 17 electrons, 18 neutrons
18) Ar – Argon - 18 protons, 18 electrons, 22 neutrons
19) K – Potassium - 19 protons, 19 electrons, 20 neutrons
20) Ca – Calcium - 20 protons, 20 electrons, 28 neutrons
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US President Biden will sign an agreement to end the national emergency of COVID-19 in advance, which is strongly opposed by the White House spokesman
US President Biden signed an agreement to end the national emergency of COVID-19 in advance, but it was strongly opposed by the White House spokesman.
What is COVID-19?The new coronavirus SARS-CoV-2 is the cause of the extremely contagious disease COVID-19. The virus was discovered for the first time in Wuhan, China, in December 2019, and it spread fast over the world, causing a pandemic. When an infected person coughs, sneezes, or talks, respiratory droplets from their mouth or nose are the main means of disease transmission.
On April 10, 2023, President Biden approved legislation written by Republicans that put an end to the COVID-19 pandemic's state of emergency. Notwithstanding the fact that the White House intended to stop making emergency declarations on May 11, the GOP's proposed legislation received some bipartisan support in Congress. The measure, according to the White House, "would generate widespread turmoil and uncertainty throughout the health care system – for states, for hospitals and doctor's offices, and, most crucially, for tens of millions of Americans," A White House representative, however, downplayed the significance of the law, stating that the end of the emergency "does not affect our ability to wind down authorities in an orderly manner."
A White House representative later minimized the bill's effects, saying they would not have an impact on the scheduled wind-down of authority on May 11.
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A 47 Ώ (Ohm) resistor and a 28 Ώ (Ohm) resistor are connected in series to a 12-V battery.
(a) What is the current flowing through each resistor?
(b) What is the voltage difference across each resistor?
(a)
To find the current flowing through each resistor, we can use Ohm's Law, which states that the current (I) through a resistor is equal to the voltage (V) across the resistor divided by the resistance (R) of the resistor:
I = V / R
As there is only one path for the current to go in the series circuit, the current flowing through both resistors is the same.
We can determine the current flowing through the circuit using the values provided in the problem by doing the following:
Total resistance (Rt) = 47 Ώ + 28 Ώ = 75 Ώ
Total current (It) = V / Rt = 12 V / 75 Ώ = 0.16 A
Therefore, the current flowing through each resistor is 0.16 A.
(b)
We may apply Ohm's Law once more to determine the voltage difference between each resistor, but this time we will solve for the voltage:
V = I x R
The voltage difference across each resistor can be determined using the values provided in the problem as follows:
Voltage across 47 Ώ resistor = 0.16 A x 47 Ώ = 7.52 V
Voltage across 28 Ώ resistor = 0.16 A x 28 Ώ = 4.48 V
Therefore, the voltage difference across the 47 Ώ resistor is 7.52 V, and the voltage difference across the 28 Ώ resistor is 4.48 V.
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List four states of matter. Which states of matter are most common on
Earth?
How does our viewing angle from earth affect what type of agn we observe?
Our viewing angle from Earth can affect the type of AGN (Active Galactic Nuclei) we observe. If we view an AGN from the direction of its accretion disk, we see a quasar, which is an incredibly bright object.
How does the viewing angle from earth affect how we observe?if we view the same AGN from a different angle, we may see a less bright active galactic nucleus, such as a Seyfert galaxy or a blazar. This is because the orientation of the accretion disk and the angle of the jets relative to our line of sight can affect the amount of radiation that reaches us.
In some cases, the orientation of the AGN can also affect the appearance of its emission lines. For example, in a type 2 Seyfert galaxy, which is viewed at an angle where the accretion disk is obscured by a thick torus of dust and gas, the broad emission lines that are typically seen in type 1 Seyfert galaxies may be absent or significantly narrower. Thus, the viewing angle from Earth plays an important role in the study of AGN and their properties.
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A team of students conducts a series of experiments to investigate collisions. In the first experiment, the two carts collide with each other on a smooth surface. The carts stick together and continue to move forward. In the second experiment, the two carts collide with each other on a rough surface. The carts stick together and quickly come to rest. In both experiments, the initial speeds of the carts are identical. Is there a difference in the total energy of the two experiments?
A.
No, because the kinetic energy of the two-cart system after the collision is the same in both the experiments.
B.
No, because the sum of the kinetic energy and thermal energy of the two-cart system after the collision is the same in both experiments.
C.
Yes, because the kinetic energy of the two-cart system in the second experiment after the collision is less than that of the first experiment.
D.
Yes, because the sum of the kinetic energy and thermal energy of the two cart system in the second experiment after the collision is less than that of the first experiment.
If a team of students conducts a series of experiments to investigate collisions. C. Yes, because the kinetic energy of the two-cart system in the second experiment after the collision is less than that of the first experiment.
What is difference in the total energy of the two experiments?In the first experiment, the two carts collide on a smooth surface and stick together, so the kinetic energy of the system after the collision is entirely in the form of translational kinetic energy of the combined carts.
In the second experiment, the two carts collide on a rough surface and stick together, so some of the kinetic energy is converted into thermal energy due to friction between the carts and the surface. This means that the kinetic energy of the system after the collision is less than in the first experiment.
Option A is incorrect because the kinetic energy is not necessarily the same in both experiments.
Option B is incorrect because the thermal energy is not necessarily the same in both experiments.
Option D is incorrect because the thermal energy is not considered in the calculation of the total energy of the system after the collision.
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When adding or subtracting two given data with uncertainties,we add the uncertainties and when multiplying and dividing,we add their percentage uncertainties.However,using the error propagation formulas none of the above rules work.Which one should I use?
You should use the error propagation formulas to calculate the uncertainties of the final result for any given mathematical operation.
How are the error propagation formulas most effective to use for final result?When propagating uncertainties using the error propagation formulas, the rules for adding, subtracting, multiplying, and dividing depend on the specific mathematical function being applied. It's essential to use the appropriate formula for each function to obtain accurate results.
Therefore, you should use the error propagation formulas to calculate the uncertainties of the final result for any given mathematical operation. These formulas take into account the uncertainties of the individual components and the functional relationship between them, and provide a more accurate way to estimate the overall uncertainty of the final result.
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Can anyone explain how to tackle this problem please:
Figure 15.5 shows a 50 kg lead cylindrical piston which floats on 0.37 mol of compressed ideal air at 30°C. How far does the piston move if the temperature is increased to 300°C?
A. 65 cm
B. 140 cm
C. 73 cm
D. 730 cm
Answer:
P V = N R T ideal gas equation
V2 / V1 = T2 / T1 since P, N, R are constant
V2 = 573 / 303 * V1 = 1.89 V1
V = π R^2 h volume of cylinder
V1 = 3.14 * .05^2 * h1 = .00785 h1
V2 = .0148 h1
A (h1 + h) = A * h2
h = h2 - h1 = (.0148 - .00785) h1 = .00695 h1 distance moved by piston
Use V1 = N R T1 / P1 = A h1 to calculate h1
h1 = N R T1 / (A * P1) A * F/A = F to simplify denominater
h1 = .37 * 8.31 * 303 / (50 * 9.8) = 1.90 m P = F / A
Δh = .00695 h1 = .0132 m = 1.32 cm
Math should be checked!
describe the energy changes in a mass spring system that is oscillating horizontally explain how this changes of the system is vibrating vertically
Answer:
Below
Explanation:
In a mass-spring system that is oscillating horizontally, the energy changes between potential energy and kinetic energy. When the mass is at its maximum displacement from the equilibrium position, it has maximum potential energy and zero kinetic energy. As the mass starts to move towards the equilibrium position, its potential energy decreases while its kinetic energy increases. At the equilibrium position, the mass has zero potential energy and maximum kinetic energy. As the mass moves away from the equilibrium position, its kinetic energy decreases while its potential energy increases. This cycle repeats as long as the system is oscillating.
Now, if the support of the system is vibrating vertically, the energy changes that occur during horizontal oscillations cause the mass to move vertically as well. As the mass moves to its maximum displacement from the equilibrium position horizontally, it also moves upwards, gaining potential energy due to its increased height from the ground. As the mass moves towards the equilibrium position horizontally, it also moves downwards, losing potential energy and gaining kinetic energy due to its increased speed towards the ground. At the equilibrium position, the mass has zero potential energy but maximum kinetic energy, which is all in the vertical direction. As the mass moves away from the equilibrium position horizontally, it also moves upwards, gaining potential energy again. The cycle repeats, causing the mass to oscillate vertically as well.
Therefore, the energy changes that occur during horizontal oscillations in a mass-spring system can cause the system to vibrate vertically if the support is vibrating vertically.
The angle of reflection is 15°, and the incident ray is bent 15° below the normal
What will be the difference in thinking when you turn it?
The angle of reflection is the angle between the reflected ray and the normal to the surface
What is the difference in thinking?The angle of reflection is the angle made by the reflected ray with the surface, measured relative to the normal, and is equal to the angle of incidence, which is the angle between the incident ray and the normal.
The angle of reflection can be determined using the law of reflection, which states that the angle of incidence is equal to the angle of reflection. This law applies to the reflection of light, sound, and other waves from a smooth, flat surface, such as a mirror or a still body of water.
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A ball has a mass 0f 2Kg and a diameter of 50cm calculate the buoyant force and water displaces...
The buoyant force acting on the ball is 5134.7 N and the water displaced by the ball is 0.5236 [tex]m^{3}[/tex] (or 523.6 L).
We use the following formula:
Weight of displaced water = Buoyant force
Volume of water displaced times water's density equals the weight of the water.
The first step is to calculate the volume of the ball:
Volume of ball = ([tex]\frac{4}{3}[/tex]) × π × [tex]\frac{diameter}{2}^3[/tex]
Volume of ball = ([tex]\frac{4}{3}[/tex]) × π × [tex](\frac{50 cm}{2})^3[/tex]
Volume of ball = 0.5236 [tex]m^{3}[/tex]
Next, calculate water displaced:
Density of water = 1000 [tex]\frac{kg}{m^3}[/tex](at standard conditions)
Weight of water displaced = 1000 [tex]\frac{kg}{m^3}[/tex] × 0.5236 [tex]m^{3}[/tex]
Weight of water displaced = 523.6 kg
Finally, we can calculate the buoyant force:
Buoyant force = weight of water displaced
Buoyant force = 523.6 kg x 9.81 [tex]\frac{m}{s^2}[/tex]
Buoyant force = 5134.7 N
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A man is standing away from the School
Building at a distance of
300m . He claps his hands and hears an echo calculate the time interval of him hearing his echo
The time interval between the man clapping and hearing his echo is approximately 1.75 seconds.
What do you mean by echo?An echo is a repetition or reflection of a sound or signal. It can be caused by sound waves bouncing off a surface, signal interference, or the repetition of a message in communication.
The speed of sound in air at room temperature is approximately 343 meters per second. When a person claps, the sound waves propagate outward in all directions and reach the school building, where they bounce off and return to the person as an echo. The time it takes for the sound to travel the distance to the building and back to the person is the time interval between the clap and the echo.
To calculate the time interval, we can use the following formula:
time = distance / speed
where distance is the total distance traveled by the sound (twice the distance from the person to the school building), and speed is the speed of sound in air.
distance = 2 x 300m = 600m
speed = 343 m/s
time = 600m / 343 m/s = 1.75 seconds (rounded to two decimal places)
Therefore, the time interval between the man clapping and hearing his echo is approximately 1.75 seconds.
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phet circuit lab pls help
Series (ckt II) will have greater resistor voltage equal to 20v as the voltage will add in series.
What is Battery Current?
Battery current refers to the flow of electric charge in a circuit that is provided by a battery. In simpler terms, it is the amount of electric current that is being drawn from or supplied to a battery. Battery current is measured in amperes (A) and can be either positive or negative, depending on whether the battery is being charged or discharged.
When a battery is being discharged, the current flows from the battery's positive terminal through the external circuit and back to the negative terminal. This is referred to as the discharge current, and its magnitude is determined by the resistance of the circuit and the voltage of the battery.
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which of the following is a form of potential energy
A uniform horizontal rod of mass 1.5 kg and length 1.9 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by
m.02
I=
12
930.
-pivot
1.5 kg
1.9 m -
If a 1.5 N force at an angle of 95° to the hot-izontal acts on the rod as shown, what is the magnitude of the resulting angular acceleration about the pivot point? The acceleration of gravity is 9.8 m/s?
Answer in units of rad/s?
The magnitude of the resulting angular acceleration about the pivot point is approximately 0.0021 rad/s^2.
Magnitude of the resulting angular accelerationTo solve this problem, we need to use the equation for the torque on a rigid body:
τ = Iα
where τ is the torque, I is the moment of inertia, and α is the angular acceleration. We can also use the fact that the torque is given by:
τ = r × F
where r is the vector from the pivot point to the point of application of the force, and F is the force. The vector cross product r × F gives the torque about the pivot point.
First, we need to find the vector r. Since the force is acting at an angle of 95° to the horizontal, the vertical component of the force is given by:
Fy = F sin(95°) = 1.5 sin(95°) ≈ 0.15 N
The horizontal component of the force is given by:
Fx = F cos(95°) = 1.5 cos(95°) ≈ 0.04 N
The vector r points vertically downward from the pivot point and has a magnitude equal to half the length of the rod:
|r| = 0.5(1.9 m) = 0.95 m
Therefore, the vector r is given by:
r = -0.95j
where j is the unit vector in the vertical direction.
The torque about the pivot point is given by:
τ = r × F = (-0.95j) × (0.04i + 0.15j) = -0.15k
where i is the unit vector in the horizontal direction and k is the unit vector in the direction perpendicular to the plane of the rod and pointing out of the page.
Substituting the moment of inertia and torque into the equation for angular acceleration, we have:
α = τ/I = (-0.15k)/(930.02/12) ≈ -0.0021 rad/s^2
Therefore, the magnitude of the resulting angular acceleration about the pivot point is approximately 0.0021 rad/s^2.
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Equal volumes of two fluids are added to the U-shaped pipe as shown in the figure below. The pipe is open at both ends and the fluids come to equilibrium without mixing. What is the ratio B/A of the fluid densities? (Assume the ratio air/fluid for fluids A and B is small enough to be neglected. Use the following as necessary: h.)?
The ratio of density of fluid A to that of fluid B is 1/3.
From the figure,
The distance of the fluid A to the line B from the top of the fluid, d₁ = 3h
The distance of the fluid B to the line B from the top of the fluid, d₂ = h
According to Pascal's law, pressure of A is equal to that of B, along the same horizontal line.
P(A) = P(B)
ρ₁gd₁ = ρ₂gd₂
Therefore, the ratio of density of fluid A to that of fluid B,
ρ₁/ρ₂ = d₂/d₁
ρ₁/ρ₂ = h/3h
ρ₁/ρ₂ = 1/3
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Your question was incomplete. Attaching the image file here.
A simple pendulum consisting of a bob of mass m
attached to a string of length L
swings with a period T
.
Part A
If the bob's mass is doubled, approximately what will the pendulum's new period be?
Part B
If the pendulum is brought on the moon where the gravitational acceleration is about g/6
, approximately what will its period now be?
T/6
T/√6
√6T
6T
Part C
If the pendulum is taken into the orbiting space station what will happen to the bob?
View Available Hint(s)for Part C
If the pendulum is taken into the orbiting space station what will happen to the bob?
It will continue to oscillate in a vertical plane with the same period.
It will no longer oscillate because there is no gravity in space.
It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.
It will oscillate much faster with a period that approaches zero.
Answer: Part A:
The period of a simple pendulum is given by the formula: T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. If the mass of the bob is doubled, the period of the pendulum will change. To see how much it changes, we can use the fact that the period depends only on the length of the pendulum and the acceleration due to gravity, and not on the mass of the bob. Therefore, the new period of the pendulum will be the same as the old period: T.
Part B:
If the pendulum is brought to the moon, where the acceleration due to gravity is about g/6, the new period of the pendulum can be found using the same formula as before: T = 2π√(L/g). However, now we need to use the value of the acceleration due to gravity on the moon, which is g/6. Therefore, the new period of the pendulum is T' = 2π√(L/(g/6)) = 2π√(6L/g) = √6T.
Therefore, the answer is: T/√6.
Part C:
If the pendulum is taken into an orbiting space station, the bob will continue to oscillate in a vertical plane with the same period as it did on the surface of the Earth. This is because the period of the pendulum depends only on the length of the pendulum and the acceleration due to gravity, and not on the location of the pendulum. In the space station, both the pendulum and the point to which it is attached are in free fall, but they are falling together and maintaining their relative positions, so the pendulum will continue to oscillate as before.
A horizontal pipe has a cross-sectional area of 0.025m2 at the entrance and 0.010m2 at the exit. If water enters the pipe at a speed of 2.5 m/s and a gauge pressure of 46kPa, what is the gauge pressure of the water at the exit end? The density of water is 1000 kg/m3.
Answer: ______kPa
The gauge pressure of the water at the exit end of the pipe is -1281.125 kPa. Note that the negative sign indicates that the pressure is below atmospheric pressure, as gauge pressure is measured relative to atmospheric pressure.
What is Atmospheric Pressure?
Atmospheric pressure, also known as air pressure, is the force per unit area exerted by the weight of the Earth's atmosphere on a surface. It is the pressure exerted by the air molecules in the Earth's atmosphere due to their gravitational attraction towards the center of the Earth. Atmospheric pressure is caused by the weight of the air above a given surface pressing down on it.
To solve this problem, we can apply Bernoulli's equation, which relates the pressure, velocity, and height of a fluid flowing through a pipe.
We can start by calculating the velocity at the exit using the equation of continuity, which states that the mass flow rate of an incompressible fluid remains constant along a streamline:
Substituting the given values:
[tex]V^{2}[/tex] = (0.025[tex]m^{2}[/tex] * 2.5 m/s) / 0.010 [tex]m^{2}[/tex]
[tex]V^{2}[/tex] = 62.5 m/s
Now, we can substitute the known values into Bernoulli's equation to find the gauge pressure at the exit:
[tex]P^{2}[/tex]= P1 + (1/2)ρ([tex]v1^{2}[/tex] - [tex]v2^{2}[/tex])
[tex]P^{2}[/tex]= 46 kPa + (1/2) * 1000 kg/m^3 *[tex](2.5 m/s)^{2}[/tex] - [tex](62.5 m/s)^{2}[/tex]
[tex]P^{2}[/tex] = 46 kPa + 625 kPa - 1953.125 kPa
[tex]P^{2}[/tex] = -1281.125 kPa
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9. A brick weighs 21 N. Measured underwater, it weighs 12 N. What is the
size of the buoyant force exerted by the water on the brick?
33 N
21 N
12 N
9N
1
Answer:
9N
Explanation:
The difference between the weight of the brick in air and the weight of the brick underwater is equal to the buoyant force exerted by the water on the brick.
Weight of the brick in air = 21 N
Weight of the brick underwater = 12 N
Therefore, the buoyant force exerted by the water on the brick is:
Buoyant force = Weight of the brick in air - Weight of the brick underwater
Buoyant force = 21 N - 12 N
Buoyant force = 9 N
So the size of the buoyant force exerted by the water on the brick is 9 N. Answer: 9N.
Hope this helps!
If the air inside a balloon exerts a force of 1 N on an area of 0.5 m^2 what is the pressure inside the balloon
Answer:
2 Pascal (Pa)
Explanation:
Pressure is defined as the force acting per unit area. Mathematically, it is expressed as:
Pressure (P) = Force (F) / Area (A)
Given:
Force exerted by the air inside the balloon (F) = 1 N
Area of the balloon (A) = 0.5 m^2
Plugging in the given values into the formula for pressure, we get:
P = F / A
P = 1 N / 0.5 m^2
Using basic arithmetic, we can calculate the pressure inside the balloon:
P = 2 N/m^2
So, the pressure inside the balloon is 2 N/m^2, which is also commonly referred to as 2 Pascal (Pa) since 1 Pascal is equal to 1 N/m^2.
After coming down a slope, a 60-kg skier is coasting northward on a level, snowy surface at a constant 15 m>s. Her 5.0-kg cat, initially running southward at 3.8 m>s, leaps into her arms, and she catches it. (a) Determine the amount of kinetic energy converted to internal energy in the Earth reference frame. (b) What is the velocity, measured in the Earth reference frame, of an inertial reference frame in which the cat’s kinetic energy does not change?
The velocity, measured in the Earth reference frame, of an inertial reference frame in which the cat's kinetic energy does not change is equal to the velocity of the skier before the collision. The velocity of the skier before the collision is 15 m/s.
What is law of conservation of momentum?According to the law of conservation of momentum, the total momentum before the collision must be equal to the total momentum after the collision. This can be expressed as m1*v1 + m2*v2 = (m1 + m2)*vf, where m1 and m2 are the masses of the skier and the cat respectively, v1 is the velocity of the skier, and vf is the velocity of the skier and the cat after the collision.
The kinetic energy converted to internal energy in the Earth reference frame can be determined by applying the law of conservation of momentum.
The amount of kinetic energy converted to internal energy can be calculated as follows:
m1*v1 = (m1 + m2)*vf
vf = (m1*v1)/(m1 + m2)
KE = (1/2)*m2*v2²
KE converted = KE initial - KE final
KE converted = (1/2)*m2*v2² - (1/2)*m2*((m1*v1)/(m1 + m2))²
KE converted = (1/2)*m2*v2² - (1/2)*m2*((60*15)/(60 + 5))²
KE converted = (1/2)*5*3.8² - (1/2)*5*(15²/65)
KE converted = 28.8 - 22.15
KE converted = 6.65 J
The velocity, measured in the Earth reference frame, of an inertial reference frame in which the cat's kinetic energy does not change is equal to the velocity of the skier before the collision.
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When the hydrogen in a star's core is used up, what occurs?
A. The core collapses causing a huge explosion called a supernova.
B. The nitrogen core collapses and the outer layer expands into a red giant.
C. The helium core collapses and the outer layer expands into a red giant.
D. The outer layer drifts away leaving a hot dense white dwarf core.
The core will collapse under its own gravity, leading to a supernova explosion that expels the outer layers of the star into space, leaving behind either a neutron star or a black hole, depending on the mass of the core.
What happens when the Helium in the core gets used up?As the helium in the core is used up, the core will contract and heat up once again until it is hot enough to fuse heavier elements. This process will continue until the core is made up of iron, which cannot be fused further. At this point, the core will collapse under its own gravity, leading to a supernova explosion that expels the outer layers of the star into space, leaving behind either a neutron star or a black hole, depending on the mass of the core.
For high-mass stars, the process is similar, but the fusion reactions proceed more rapidly, leading to a shorter lifespan and a more violent supernova explosion. In both cases, the ultimate fate of the star depends on its mass and the resulting conditions in its core.
When the hydrogen in a star's core is used up, a series of events can occur depending on the mass of the star. For low to medium-mass stars, such as our Sun, the core will contract and heat up until it is hot enough to initiate the fusion of helium into carbon and oxygen. This process, known as the helium-burning phase, will cause the outer layers of the star to expand into a red giant.
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If Ja Morant has a vertical leap of 1.35m, then what speed does he leave the ground and what is his
total hang time?
Ja Morant's total hang time is 0.53 seconds.
The initial vertical velocity, Vi, is 0 since he starts from rest. The final vertical displacement, Δy, is 1.35m. We can assume that air resistance is negligible, so we can use the acceleration due to gravity, g, which is -9.81 m/s².
To find the speed he leaves the ground, we can use the equation:
Vf² = Vi² + 2gΔy
Vf² = 0 + 2(-9.81)(1.35)
Vf = 5.89 m/s
Therefore, Ja Morant leaves the ground at a speed of 5.89 m/s.
To find the total hang time, we can use the equation:
Δy = ViT + 1/2 gT²
1.35 = 0T + 1/2(-9.81)T²
[tex]T = \sqrt{(2(1.35)/9.81)[/tex]
T = 0.53 seconds
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A 11.50 kg object has the given and acceleration components.
=(0.67ms2)+(0.81ms3)
=(11.7ms2)−(0.63ms3)
What is the magnitude net of the net force acting on the object at time =4.47 s ?
Answer:
To find the net force acting on the object, we need to add the force components along each axis. We are given the acceleration components, which we can use to find the force components using Newton's second law:
F = ma
Along the x-axis:
F_x = ma_x = (11.7 kg)(0.67 m/s^2) = 7.839 N
Along the y-axis:
F_y = ma_y = (11.7 kg)(-0.63 m/s^2) = -7.371 N
The net force is the vector sum of the x and y components of force:
F_net = sqrt(F_x^2 + F_y^2) = sqrt((7.839 N)^2 + (-7.371 N)^2) = 10.925 N
Therefore, the magnitude of the net force acting on the object at time t = 4.47 s is 10.925 N.
Spin-offs from the aerospace industry have contributed to Floridas economy and the economy of the United States what is a spinoff
A spinoff, in the context of the aerospace industry, refers to the transfer of technology or knowledge developed for a specific space or aviation program to other fields or industries.
The technology and knowledge developed for aerospace programs often have applications beyond the aerospace industry, and these applications can lead to the creation of new products, services, and industries.
For example, the development of lightweight, high-strength materials for use in spacecraft can also be applied to other industries such as the automotive, sporting goods, and construction industries. Similarly, the development of advanced computer systems and software for use in spacecraft can be applied to other industries such as healthcare, finance, and telecommunications.
Spinoffs from the aerospace industry have contributed to the economy of Florida and the United States by creating new jobs, new products, and new industries. The transfer of technology and knowledge from the aerospace industry to other fields has also led to advancements in a wide range of areas, from medicine to transportation to telecommunications, that have improved the quality of life for people around the world.
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What current flows through a 2.54-cm-diameter rod of pure silicon that is 20.0 cm long, when 1.00 x 10ᵌ V is appled to it? (Such a rod may be used to make nuclear-particle detectors, for example.)
Answer: 1.68 A (Amperes)
Explanation:
To calculate the current flowing through the silicon rod, we need to use Ohm's law, which states that the current (I) flowing through a conductor is equal to the voltage (V) applied across it divided by the resistance (R) of the conductor:
I = V/R
The resistance (R) of the silicon rod can be calculated using the formula:
R = ρL/A
where ρ is the resistivity of silicon, L is the length of the rod, and A is the cross-sectional area of the rod.
The resistivity of pure silicon at room temperature is approximately 2.3 x 10^-3 Ω m. The cross-sectional area of the rod can be calculated using the formula for the area of a circle:
A = πr^2
where r is the radius of the rod, which is half of its diameter. Substituting the given values, we get:
r = 2.54/2 = 1.27 cm = 0.0127 m
A = π(0.0127)^2 = 5.083 x 10^-4 m^2
Now we can calculate the resistance of the rod:
R = (2.3 x 10^-3 Ω m)(20.0 cm/100 cm)/5.083 x 10^-4 m^2 = 9.035 Ω
Finally, we can use Ohm's law to calculate the current flowing through the rod:
I = V/R = (1.00 x 10^3 V)/9.035 Ω = 1.68 A
Therefore, the current flowing through the silicon rod is 1.68 Amperes.
A loop of area 0.100 m² is oriented at
a 15.5 degree angle to a 0.500 T
magnetic field. It rotates until it is at a
45.0 degree angle with the field. What
is the resulting CHANGE in the
magnetic flux?
[?] Wb
Answer:
Flux = Area * Field
A1 = .1 m^3 * cos 64.5 since we need perpendicular area
A1 = .043 m^2 perpendicular to field
A2 = .1 m^3 * cos 45 = .071 m^2
A = (.071 - .043) m^2 = .028 m^2 change in area
Change in flux = .028 m^2 ^ .5 T = .014 T m^2