The reaction to the action force of the impact of the ball against the player's glove is option B): "the force the glove exerts on the ball".
This is known as Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. So, as the ball exerts a force on the glove, the glove exerts an equal and opposite force on the ball, allowing the player to catch the ball. The other options listed (A, C, and D) are not directly related to the action-reaction force pair involved in the catch.
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a slingshot fires a pebble from the top of a building at a speed of 15.0 m/s. the building is 33.0 m tall. ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired in the following directions.
The required speed with which the pebble strikes the ground when the pebble is fired from the slingshot with specified initial velocity is calculated to be 29.53 m/s.
The initial speed of the pebble with which it is fired is 15 m/s.
The height of the building is given as 33 m.
Acceleration due to gravity g = 9.8 m/s²
Now, let us calculate the velocity of the pebble with which it strikes the ground.
The suitable equation of motion is, v² - u² = 2 a s
where,
v is final velocity
u is initial velocity
a is acceleration
s is distance
Entering the values into the above equation by making v as subject, we have,
v² - u² = 2 a s
v² = u² + 2 a s
v = √(u² + 2 a s) = √[15² + 2(9.8)(33)] = √(225 + 646.8) = √871.8 = 29.53 m/s
Thus, the final velocity is calculated to be 29.53 m/s.
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8. (a) The mass m of an object is proportional to its volume V: m = kV. What are the SI units of k? (b) If the distance d an object moves in time t is given by the equation d = At-, find the SI units of A. (c) If the speed v of an object depends on time t according to the equation v = A + Bt + Ct4, what are the SI units of A, B, and C?
density is the mass of a specific material per unit volume. d = M/V, in which d is densities, M is mass, as well as V is volume, is the formula for density. Common units for expressing density are grams per cubic inch.
What is the mass unit?
There are several ways to measure mass, including kilograms, grams, pounds, and pounds, but the SI unit de mass is the "kilogram" or kg. Using the right conversion formula, any unit of weight can be changed into another unit without changing the significance or meaning of the quantity being measured.
Describe mass.
Definition, Units, Formula, and Examples of Mass The quantity of matter that makes up every object or body is the greatest way to understand mass. Everything that we can see has mass. Examples of objects with mass include a table, an chair, your mattress, a ball, a tumbler, and even air. The mass of a thing determines whether it is light or heavy.
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When tuning a piano, a technician strikes a tuning fork for the A above middle C and sets up a wave motion that can be approximated by y = 0.001 sin 880πt, where t is the time (in seconds).(a) What is the period of the function?(b) The frequency f is given by f=1p. What is the frequency of the note?
The required period of the function is calculated to be 1/440 sec and frequency is calculated to be 440 hertz.
The equation of the sinusoidal wave is given as, y = 0.001 sin 880πt
Comparing the given equation with that of the general equation of the wave in simple harmonic motion, we have, y = a sin ωt
where,
y is displacement
a is amplitude
t is time
ω is angular frequency
So, Angular frequency ω = 880π
The period of the function is given by the formula, p = 2π/ω = 2π/880π = 1/440 sec
We know the equation for frequency as, f = 1/p = 1/(1/440) = 440 hertz
Thus, the period of the function is 1/440 sec and frequency is calculated to be 440 hertz.
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Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and 16 x 106 branch instructions. The CPI for each type of instruction is 1 1, 4, and 2, respectively. Assume that the processor has a 2 GHz clock rate 1.141 [10] <$1.10 By how much must we improve the CPI of FP instructions if we want the program to run two times faster?
To make the program run two times faster, we need to improve the CPI of FP instructions from 1 to 5.859375.
To determine the execution time of the program, we can use the following formula:
Execution Time = (Number of Instructions x CPI) / Clock Rate
We can calculate the total number of instructions as follows:
Total Instructions = 50 * 10⁶ FP instructions + 110 * 10⁶ INT instructions + 80 * 10⁶ L/S instructions + 16 * 10⁶ branch instructions
Total Instructions = 256 x 10⁶ instructions
We can calculate the current CPI for the program as follows:
CPI = (50 * 1 + 110 * 1 + 80 * 4 + 16 * 2) / 256
CPI = 1.8125
We can calculate the current execution time of the program as follows:
Execution Time = (256 * 10⁶ * 1.8125) / (2 * 10⁹)
Execution Time = 0.29296875 seconds
To make the program run two times faster, we need to reduce the execution time to 0.146484375 seconds. We can use the same formula to calculate the new CPI for the FP instructions:
New CPI = (Execution Time x Clock Rate) / (Number of Instructions x 2)
We can rearrange the formula to solve for the new CPI:
New CPI = (Execution Time x Clock Rate) / (Number of Instructions x 2)
New CPI = (0.146484375 * 2 * 10⁹) / (50 * 10⁶ * 1)
New CPI = 5.859375
To make the program run two times faster, we need to improve the CPI of FP instructions from 1 to 5.859375. This can be achieved through various optimizations, such as using SIMD instructions, reducing data dependencies, and optimizing memory access.
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An idealized object that does not reflect or scatter any radiation that hits it, but simply absorbs every bit of radiation that falls on it is called:
A blackbody is a hypothetical object that absorbs all incident radiation without reflecting or scattering any of it.
What exactly is a blackbody?A blackbody is a hypothetical object that emits all wavelengths of radiation while simultaneously absorbs all electromagnetic waves that strike it.
Another name for it is a perfect emitter or absorber.
Blackbody radiation is the specific spectral distribution of radiation produced by a blackbody that is only temperature dependant.
Understanding the behaviour of thermal radiation and the properties of stars, galaxies, as well as other celestial objects in physics and astronomy is dependent on this concept.
Planck's law, which gives the spectral density of a radiation emitted by the a blackbody at a temperature t as a function of the wavelength of the radiation, describes the radiation released by a blackbody as being solely dependent on its temperature.
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which of the following are responsible for the repulsive force between two identically charged atoms?
Electrostatic force is responsible for the repulsive force between two atoms charged identically. The electrostatic force is an attractive force as well as a repulsive force induced by the electric charge particles.
The following that are responsible for the repulsive force between two identically charged atoms are electrostatic interactions.
A push or pull on an object that is caused by the interaction of that thing with another object is what we refer to as a force. When two things interact with one another, a force is applied to both of those items. Any force that causes two items to move away from each other is referred to as a repulsive force.
Electrostatic attraction is another type of attraction that can take place between molecules or atoms. On the other hand, it does not create a bond. They interact with one another not through full charges but rather by half charges or dipoles. Electrostatic interactions can either be attractive or repulsive. When the charges have the same positive or negative sign, the electrostatic force acts in a repelling manner.
Your question is incomplete, but most probably your full question was:
Which of the following are responsible for the repulsive force between two identically charged atoms?
van der Waals interactions
covalent bonds
hydrogen bonds
electrostatic interactions
hydrophobic interactions
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Two blocks connected by a string are pulled across a rough horizontal surface by a force applied to one of the blocks, as shown. The acceleration of gravity is 9.8
m
/
s
2
. If each block has an acceleration of 5.2
m
/
s
2
to the right, what is the magnitude of the applied force? Answer in units of N
.
The force imparted to one of the blocks is 14.59 N in magnitude.
To solve this problem, we can use Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration.
Let's denote the mass of the left block as m1 and the mass of the right block as m2. We can then write the following equations:
m1 * a = F - f
m2 * a = f
where a is the acceleration of both blocks, F is the magnitude of the applied force, and f is the frictional force acting on the blocks. Note that the acceleration is the same for both blocks because they are connected by a string.
The frictional force can be expressed as f = μ * N, where μ is the coefficient of friction and N is the normal force. The normal force acting on the blocks is equal to their weight, which can be expressed as m1 * g and m2 * g, respectively.
Substituting the expression for the frictional force into the first equation and solving for F, we get:
F = m1 * a + μ * m1 * g
Substituting the given values, we get:
F = (m1 + m2) * a + μ * (m1 + m2) * g
F = (m1 + m2) * 5.2 + 0.4 * (m1 + m2) * 9.8
F = 5.2 * (m1 + m2) + 3.92 * (m1 + m2)
F = 9.12 * (m1 + m2)
Now, we need to find the value of m1 + m2. We are not given their individual masses, but we know that the ratio of their masses is 2:3. Let's denote the smaller mass as m and the larger mass as 1.5m. Then we have:
m + 1.5m = 2.5m
1.5m/m = 3/2
So the masses are m = 2/5 and 1.5m = 6/5.
Substituting these values into the expression for F, we get:
F = 9.12 * (2/5 + 6/5)
F = 9.12 * 8/
F = 14.59 N
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What is the name for the glue that holds a fitness team together and keeps them focused on their purpose?
A. competition
B. cohesion
C. strategy
D. psychology
Answer:
Explanation:
cohesion
A satellite is in a circular orbit about the earth [tex]M_{E} = 5.9 x 10^{24} kg).[/tex] The period of the satellite is [tex]1.38 x 10^{4}[/tex] What is the speed at which the satellite travels? V = [Blank] [Units]
The gravitational force acting on satellite at an altitude R causes the centripetal acceleration it required.
Thus, (2R) 2 GMm = 2Rmv 2
What is the speed of a circularly orbiting earth satellite?
The required speed to do this—known as the circular satellite velocity—is around 8 kph, or 17,500 mph in more commonly used quantities. See how gravity affects the orbits of the Earth, Moon, and International Space Station in this additional PhET interactive.
While elliptically circling the planet, a satellite will accelerate as its height (or distance from the world) decreases and decelerate as its height (or distance from the earth) increases.
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The electric potential at points in an xy plane is given V-(2.0 V/m2)x2-(3.0 V/m)y2. In unit-vector notation, what is the electric field at the point (3.0 m, 2.0 m)? What is the angle that the field there makes with the positive x direction?.
The electric field at the point (3.0 m, 2.0 m) is (4.0 V/m^2) i + (6.0 V/m^2) j, and the angle that the field makes with the positive x direction is 56.3°.
Steps
To find the electric field at a point in an xy plane, we can take the negative gradient of the electric potential at that point:
E = -∇V
where ∇ is the gradient operator. In two dimensions, the gradient operator is:
∇ = i(∂/∂x) + j(∂/∂y)
where i and j are the unit vectors in the x and y directions, respectively.
Taking the partial derivatives of V with respect to x and y, we get:
(∂V/∂x) = -4.0 V/m^2 x
(∂V/∂y) = -6.0 V/m^2 y
Substituting these into the expression for the electric field, we get:
E = -i(∂V/∂x) - j(∂V/∂y)
E = 4.0 V/m^2 i + 6.0 V/m^2 j
Now we can substitute the given values of x and y to find the electric field at the point (3.0 m, 2.0 m):
E = 4.0 V/m^2 i + 6.0 V/m^2 j
= (4.0 V/m^2)(1.0 i) + (6.0 V/m^2)(1.0 j)
= (4.0 V/m^2)(cosθ i + sinθ j)
where θ is the angle that the electric field makes with the positive x direction. To find θ, we can take the arctangent of the y component divided by the x component:
θ = tan^-1(6.0/4.0)
= 56.3°
Therefore, the electric field at the point (3.0 m, 2.0 m) is (4.0 V/m^2) i + (6.0 V/m^2) j, and the angle that the field makes with the positive x direction is 56.3°.
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Compare dispersing light into colors with a prism and then into a grating. The 30-60-90 prism is made of glass with indices n {red}=1.52 and n {blue}=1.53. The grating has 500 slits per millimeter. a. What is the angular separation between red and blue light leaving the prism in degrees? b. What is the angular separation between red (650 nm) and blue (450 nm) light leaving the grating in degrees? Explain
Both a prism and a diffraction grating can be used to disperse white light into its component colors, which are a range of wavelengths of visible light.
What is diffraction?Due to the differing refractive indices of the prism material for various wavelengths of light, a prism operates by refracting the various hues of light as they pass through at slightly different angles. This causes the colours to disperse into a spectrum that resembles a rainbow. The prism equation provides the angle of departure for the various hues.θ = sin^(-1) (-1)
[n avg] = [(n blue - n red) sin()]
where n red and n blue are the prism's respective refractive indices for red and blue light, and n avg is the average of the two refractive indices, and is the angle of deviation, is the angle of incidence of the light on the prism, and so forth.Assuming an angle and the values given as n red = 1.52, n blue = 1.53, and n avg = (n red + n blue)/2 = 1.525
of incidence of 45 degrees, we can calculate the angle of deviation to be:
θ = sin^(-1)[(1.53 - 1.52) sin(45°) / 1.525] = 0.34°
Therefore, the angular separation between red and blue light leaving the prism is 0.34 degrees.
A diffraction grating works by diffracting light through a series of closely spaced parallel slits or lines, which act like a series of tiny prisms that diffract the light at different angles depending on its wavelength. The angle of diffraction of a particular color is given by the grating equation:nλ = d(sin θ_i + sin θ_d)
where n is the order of the diffraction (usually n=1), λ is the wavelength of the light, d is the spacing between the grating lines, θ_i is the angle of incidence of the light on the grating, and θ_d is the angle of diffraction.
Using the given values of d=500 nm (i.e., 500 slits per mm), λ_red = 650 nm and λ_blue = 450 nm, and assuming an angle of incidence of 0 degrees (i.e., the light is perpendicular to the grating), we can calculate the angles of diffraction for red and blue light to be:θ_red = sin^(-1)[(nλ_red / d) - sin(θ_i)] = sin^(-1)[(1650 nm / 500 nm) - sin(0°)] = 0.837°
θ_blue = sin^(-1)[(nλ_blue / d) - sin(θ_i)] = sin^(-1)[(1450 nm / 500 nm) - sin(0°)] = 0.579°
Therefore, the angular separation between red and blue light leaving the grating is:
θ_separation = θ_red - θ_blue = 0.837° - 0.579° = 0.258°
This is smaller than the angular separation obtained from the prism, and is due to the fact that a grating has a higher dispersion than a prism, meaning that it can separate colors more effectively. However, a grating also has narrower angular acceptance and higher order diffraction peaks that can interfere with the desired signal.To know more about diffraction , check out :
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Suppose that two stars in a binary star system are separated by a distance of 100 million kilometers and are located at a distance of 200 light-years from Earth.
Part A
What is the angular separation of the two stars? Give your answer in degrees.
Express your answer using two significant figures.
Part B
What is the angular separation of the two stars? Give your answer in arcseconds.
Express your answer using two significant figures.
Part C
Can the Hubble Space Telescope resolve the two stars
Part A
The angular separation of the two stars is 0.03 degrees
Part B
The angular separation of the two stars is 108 arcseconds.
Part C
Yes, the Hubble Space Telescope can resolve the two stars.
What is angular seperation?Angular separation is the angular distance between two objects as seen from a particular point in space. This is usually expressed in degrees, arcminutes, or arcseconds. It is a measure of the angular separation between two objects, such as stars, planets, galaxies, or other celestial objects, as seen from Earth. The angular separation of two objects is the angle between them as seen from the observer's point of view.
For example, the Moon and the Sun have an angular separation of 0.5 degrees when they are in opposition.
Part A:
To find the angular separation of the two stars, we can use the formula:
θ = arctan(d/D)
where θ is the angular separation, d is the physical distance between the two stars, and D is the distance to the stars from Earth.
Substituting the given values, we get:
θ = arctan(100,000,000 km / (200 x 9.461 trillion km))
θ = arctan(100,000,000 km / 1.8922 x 10^15 km)
θ = 0.03 degrees
Therefore, the angular separation of the two stars is 0.03 degrees.
Part B:
To convert degrees to arcseconds, we can use the formula:
1 degree = 3600 arcseconds
Substituting the value we found in part A, we get:
θ = 0.03 degrees x 3600 arcseconds/degree
θ = 108 arcseconds
Therefore, the angular separation of the two stars is 108 arcseconds.
Part C:
To determine if the Hubble Space Telescope (HST) can resolve the two stars, we need to compare the angular separation of the stars to the angular resolution of the HST. The angular resolution of a telescope is the smallest angle between two point sources that the telescope can distinguish.
The angular resolution of the HST is about 0.05 arcseconds. Since the angular separation of the two stars is 108 arcseconds, the HST can easily resolve the two stars as separate objects.
Therefore, the HST can resolve the two stars in the binary star system.
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A the cannon recoils from firing a cannonball. The speed of the cannon's recoil is small because a. cannon has more mass than the ball b. momentum of the cannon is smaller. c. force against the cannon is smaller than against the ball. d. momentum is mainly concentrated in the ball. e. none of these
The cannon has more mass than the ball.
option A is the correct answer.
Why is the speed of the cannon's recoil small?
The recoil of a cannon after firing a cannonball is a consequence of the conservation of momentum. According to this principle, the total momentum of a closed system is conserved, which means that the momentum of the cannon and cannonball before the firing must be equal to the momentum of the cannon and the cannonball after the firing.
Thus, the cannon has more mass than the ball, which means that it has a greater inertia and will be more resistant to acceleration. This results in a smaller speed of recoil compared to the cannonball.
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Which motion might be represented by this position-versus-time graph? Х t t 0 - 0 O A car drives toward the left at first slowly, then faster. O A car drives toward the left at first quickly, then slower. O A car drives toward the right at first slowly, then faster. O A car drives toward the right at first quickly, then slower. O A ball rolls down a ramp that is at first not too steep, then steeper. O An elevator descends at first slowly, then faster.
The motion that could be represented by the given position-versus-time graph is "A car drives toward the right at first slowly, then faster."
We can see that the graph has a positive slope, which indicates a motion in the positive direction. The fact that the slope is increasing over time means that the object is accelerating, which is consistent with the car starting off slowly and then getting faster.
phase 1 : object moves towards its mean position, velocity must be slow.
phase 2 : object moved to its initial position taking much less time compared to phase 1 but in the same direction to reach its mean position.
Therefore, the correct option is "A car drives toward the right at first slowly, then faster."
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Four point charges q 1
,q 2
,q 3
and q 4
are placed at the corners of the square of side a, as shown in figure. The potential at the centre of the square is: (Given : q 1
=1×10 −8
C,q 2
=−2×10 −8
C,q 3
=3×10 −8
C,q 4
=2×10 −8
C,a=1m).
The potential at the center of the square is 507 V. Option (a) is correct answer.
The electric potential at a distance r from a point charge q is given by:
V = kq/r
where k is the Coulomb constant.
In this case, the charges q₂ and q₄ are negative, so the potentials due to these charges will be negative. The charges q₁ and q₃ are positive, so the potentials due to these charges will be positive. The potential at the center of the square is given by:
[tex]V = \dfrac{kq_1}{a} + \dfrac{kq_2}{\sqrt{2}a} + \dfrac{kq_3}{a} + \dfrac{kq_4}{\sqrt{2}a}[/tex]
Plugging in the values of the charges and the distance a, we get:
[tex]V = \dfrac{9 \times 10^9 \times 1 \times 10^{-8}}{1} + \dfrac{9 \times 10^9\times -2\times 10^{-8}}{\sqrt{2}\times 1} + \dfrac{9 \times 10^9\times 3 \times 10^{-8} }{1} + \dfrac{9 \times 10^9 \times 2 \times 10^{-8} }{\sqrt{2} \times 1}[/tex]
Simplifying this expression,
V = 507 V
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--The complete question is, Four point charges q1 ,q2 ,q3 and q4 are placed at the corners of the square of side a, as shown in figure. The potential at the centre of the square is: (Given: q1 =1×10^−8 C , q2 =−2×10^−8 C, q3 =3×10^−8 C, q4 =2×10^−8C, a=1m).
a. 507 V
b. 607 V
c. 550 V
d. 650 V--
A worker wants to load some bricks in to his van. there are 1000 bricks and when stacked neatly they measure 2m by 1m by 1m. If the vans maximum load is 1000kg, how many bricks can he load?
The worker can load {500/ρ} number of bricks into the van.
What is mass?Mass is the total amount of matter present inside a body.
Given is that a worker wants to load some bricks in to his van. There are 1000 bricks and when stacked neatly they measure 2m by 1m by 1m.
Assume the density of the brick to be {ρ}. Let's assume that he can load {n} bricks inside the van. So, we can write -
n x mass of 1 brick = maximum load
n x ρ x V = L{max}
nρV = L{max}
n = L{max}/ρV
n = 1000/(2 x 1 x 1 x ρ)
n = 500/ρ
Therefore, the worker can load {500/ρ} number of bricks into the van.
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If a sample that has a density of 3.3 g/mL is taken to the moon, how does the value of the density change?
If a sample that has a density of 3.3 g/mL is taken to the moon, the value of the density does not change.
The simple reason for this reason is density depends on mass and volume. The whole reason for this is that gravity does not depend on mass and volume.
Density can be explained as a measure of mass per unit volume. It is a very intensive property, which means that its value does not change depending on the size of the object. In physics, density can be explained is the ratio obtained from the mass of an object to its volume. It is often also defined as mass per unit volume.
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The density of a sample with a density of 3.3 g/mL would not change if taken to the moon.
The density of a material depends on its mass and volume, both of which are affected by the gravitational force acting on the material. On the moon, the gravitational force is about 1/6th of that on Earth, so the mass and weight of the sample would change.
However, the density of the sample would remain the same, regardless of its location. Density is defined as mass per unit volume, so as long as the mass and volume of the sample do not change, its density will remain constant. The value of 3.3 g/mL is a property of the material itself and is not affected by the gravitational force or location. Therefore, the density of a sample with a density of 3.3 g/mL would not change if taken to the moon.
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A rail road car of mass 1500kg rolls to the right at 4 m/s and collides with another railroad car of mass 3000kg that is rolling to the left at 3 m/s. The cars stick together. Their speed immediately after the collision. 2/3m/s 1 m/s 5/3 m/s 10/3 m/s 7 m/s
The speed of the cars right after the collision is 1/3 m/s.
To solve this problem, we can apply the law of conservation of momentum, which states that the total momentum of a system remains constant if there are no external forces acting on it. In this case, the two railroad cars form a closed system, and there are no external forces acting on them during the collision. Therefore, we can write:
(m1 * v1) + (m2 * v2) = (m1 + m2) * v
where m1 and v1 are the mass and velocity of the first railroad car, m2 and v2 are the mass and velocity of the second railroad car, m1 + m2 is the total mass of the system after the collision, and v is the velocity of the system after the collision.
Plugging in the given values, we get:
(1500 kg * 4 m/s) + (3000 kg * (-3 m/s)) = (1500 kg + 3000 kg) * v
Simplifying and solving for v, we get:
v = (1500 kg * 4 m/s - 3000 kg * (-3 m/s)) / (1500 kg + 3000 kg)
v = (6000 kgm/s + 9000 kgm/s) / 4500 kg
v = 15/45 m/s = 1/3 m/s
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Communication cannot be reversed
Communication cannot be reversed. This statement is clearly true. Because, we cannot take the words back and undo all the initial effects during communication.
What is reversibility ?An action that can be reversed back to its initial state is called reversible otherwise it is called irreversibility. Communication is irreversible and can be defined as a principle of interpersonal interaction in which we cannot take or retrieve information we say or pass to another party, whether it is what we intended to say or not.
We can wish we hadn't said something, regret it, and apologies later, but we can't take it back because once it's out, it's out.
We're only talking to make up for the unintended consequences of previous communication errors. We believe that taking more care with what we say in the first place can break this seemingly never-ending cycle.
Once a word, phrase, or comment is spoken, or an impulsive text message or e-mail is sent, it cannot be erased from the memory of others. Because communication is irreversible, one should always be mindful of what they communicate to others.
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Your question is incomplete. But your complete question is as follows:
Communication cannot be reversed. True/false ?
Consider a roller coaster loop with a constant radius of 9m. Find the centripetal acceleration and force factor at the bottom if the tangential velocity is 10m/s. Find the centripetal acceleration and force factor at the top if the tangential velocity is 7.5.
Answer: F = ma = 1000 * 6.25 = 6,250 N.
Explanation:
The centripetal acceleration (a) of an object moving in a circular path is given by:
a = v^2 / r
where v is the tangential velocity and r is the radius of the loop.
The force factor, or centripetal force (F), required to keep an object moving in a circular path is given by:
F = ma
where m is the mass of the object.
At the bottom of the loop, where the tangential velocity is 10 m/s, the centripetal acceleration is:
a = v^2 / r = 10^2 / 9 = 100 / 9 m/s^2
The force factor required to maintain this acceleration depends on the mass of the object. For example, if the mass of the roller coaster car is 1000 kg, then the force factor required would be:
F = ma = 1000 * (100 / 9) = 11,111 N
At the top of the loop, where the tangential velocity is 7.5 m/s, the centripetal acceleration is:
a = v^2 / r = 7.5^2 / 9 = 6.25 m/s^2
Using the same mass of 1000 kg, the force factor required would be:
F = ma = 1000 * 6.25 = 6,250 N.
Charge +3 nC is in a hollow cavity inside a large chunk of metal that is electrically neutral. The total charge on the exterior surface of the metal is A. 0 nC. B. +3 nC. C. -3 nC. D. Can't say without knowing the shape and location of the hollow cavity.
The correct answer is C. -3 nC.
When a charge is placed inside a hollow cavity of metal, the metal will rearrange its charges to cancel out the charge inside the cavity. This means that the exterior surface of the metal will have a charge opposite to the charge inside the cavity.
In this case, since the charge inside the cavity is +3 nC, the charge on the exterior surface of the metal will be -3 nC. This is known as the principle of charge neutrality.
Therefore, when a charge of +3nC is placed in a hollow cavity inside a large chunk of metal that is electrically neutral, the total charge on the exterior surface of the metal is C. -3 nC.
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drops his phone of mass 650 g from his hand at a height
of 1.50 m above the ground. As the phone is dropped, gravity
does 8.80 J of work.
What is the speed of the phone just before touching the ground?
The speed of the phone just before touching the ground is approximately 3.92 m/s.
What is conservation of energy?We can use the conservation of energy to find the speed of the phone just before touching the ground. Initially, the phone has potential energy due to its height above the ground, which is converted into kinetic energy just before the phone touches the ground.
The potential energy of the phone just before it is dropped is:
[tex]PE = mgh[/tex]
where m is the mass of the phone, g is the acceleration due to gravity (9.81 m/s^2), and h is the height of the phone above the ground (1.50 m). Thus, the initial potential energy of the phone is:
[tex]PE = (0.65 kg)(9.81 m/s^2)(1.50 m) = 9.57 J[/tex]
As the phone is dropped, gravity does 8.80 J of work on the phone. This work is converted into kinetic energy just before the phone touches the ground. Thus, the final kinetic energy of the phone just before it touches the ground is:
[tex]KE = W = 8.80 J[/tex]
The kinetic energy of an object is given by:
[tex]KE = (1/2)mv^2[/tex]
where v is the speed of the object.
Thus, we can solve for the speed of the phone just before it touches the ground:
[tex]v = sqrt(2KE/m) = sqrt(2(8.80 J)/(0.65 kg)) = 3.92 m/s[/tex]
Therefore, the speed of the phone just before touching the ground is approximately 3.92 m/s.
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Question is attached
(a) The elevator is be moving upwards and accelerating upwards.
(b) The acceleration of the student is 0.5 m/s².
What is the net force of the elevator?
The net force acting on the elevator is determined by applying Newton's second law of motion as shown below.
F (net) = Ws + We ( when the elevator is moving upwards )
F (net) = Ws - We ( when the elevator is moving downwards)
where;
Ws is the weight of the studentWe is the weight of the elevatorSince the resultant force of the elevator is greater than the weight of the student, the elevator must be moving upwards.
The mass of the student = 600 N / 9.8 m/s² = 61.22 kg
The acceleration of the student = ( 630 N - 600 N ) / ( 61.22 kg )
= 0.5 m/s²
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Water is poured into a container that has a small leak. The mass m of the water is given as a function of time t by m=5.00 t 0.8
−3.00t+20.00, with t≥0,m in grams, and t in seconds.
At what time is the water mass greatest.
Answer:
Explanation:
when gravity pulls the water molecules downwards, they will fall. But if they're in a container, the container will keep them from spreading out completely. This is what's called 'taking the shape of the container'
A small object of mass 3.80g and charge −18.0μC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What are the magnitude and direction of the electric field?
The electric field will have a magnitude of 2.07 x [tex]10^{3}[/tex] N/C and the electric field will be directed downwards.
How do you determine the size of an electric field?The electromagnetic force exerted on the object by the field must be directed upward and have a strength equivalent to the object's weight for the object to appear to "float" in it.
Thus, [tex]F_{e} = qE = mg[/tex]
The magnitude of the electric field is,
E = mg/ IqI = (3.80 x [tex]10^{-3} Kg[/tex])(9.80 m/s²)/1.80 x [tex]10^{-6}[/tex] C = 2.07 x [tex]10^{3}[/tex] N/C.
How do you tell which way the electric field is going?A negatively charged object experiences an electric force that is directed in the opposite direction as the electric field. Because it must be directed downward, the electric force.
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The purpose of using epicycles and deferents to explain the motion of the planets in the night sky was to account for a. prograde motion b. Mercury and Venus' limited angular distance from the Sun. c. retrograde motion d. non-uniform speed of the planets in their orbits. e. precession of the equinoxes.
Epicycles and deferents were used to account for the planets' irregular orbital speeds in order to explain how they moved in the night sky.
What is meant by an epicycle?A planet moves in a circle that has a core that is also rotated simultaneously on the rim of a wider circle, according to an early theory of astronomy.
Ptolemy was forced to develop a model of planetary motion that utilised epicycles in order to maintain the geocentric cosmology of the time and account for Mars' retrograde motion. An epicycle is essentially a small "wheel" that revolves around a larger wheel.
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Rigid rods of negligible mass lying along the y axis connect three particles. The system rotates about the x axis with an angular speed of 1.00 rad/s. It is known that m1 = 3.00 kg, m2 = 2.00 kg, m3 = 4.00 kg, and y1 = -4.00 m, y2 = -2.00 m, and y3 = 3.00 m.(a) Find the moment of inertia about the x axis.(b) Find the total rotational kinetic energy evaluated from 1/2 Iω2(c) Find the tangential speed of each particle.
The following values are given: m1 = € kg, resistor = y1 Means 3.00 m, sq.m = 1.75 kg, r2 / y2. = 1.75 m, m3 = 3.00 metric tons, r3 = restrictions that limit. = 4.00 m, and = 1.75 rad/s about x-axis = m1r. 2.
How do you calculate the total kinetic energy of rotation?The combined translational and rotational kinetic energies of an object in motion determine its total kinetic energy. 12 mvCM2 is the translational kinetic energy. Rotational kinetic energy is equal to 12 I2. Total kinetic energy is 12mvCM2 plus 12I2.
What is the equation for the case of rotating kinetic energy?E s = x 2 x I ω 3 . . This equation shows that its kinetic energy of such a rotating solid body is inversely related to the cube of the angular acceleration and the moment of inertia. Flywheel energy-storage systems, which are intended to, take advantage.
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let f (t) be the velocity (in miles/hour) of a runner running along a north-south path, where f (t) > 0 means running in the north direction, and where t
The velocity of the runner at t=π/2 seconds is 15 miles/hour.
The directional speed of an item in motion, as measured by a specific unit of time and viewed from a certain point of reference, is what is referred to as velocity.
The function is as follows:
f(t)=-10cos(t) + 15
The time in this case is shown as t.
We must change t=/2 in the given function to obtain the runner's speed at t=/2 seconds:
f(π/2) = -10cos(π/2) + 15
f(π/2) = -10(0) + 15
f(π/2) = 15
As a result, the runner's speed at t=/2 seconds is 15 miles per hour.
We may deduce that the runner is moving towards the north at a speed of 15 miles per hour at t=/2 seconds since f(t) > 0 denotes moving in that direction.
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Two wires AC and BC are tied at C of a small sphere of mass 5 kg, which revolves at a constant speed v in the horizontal plane with the speed v of radius 1.6 m. Find the minimum value of v.
he minimum value of v is determined by the forces acting on the sphere. The forces acting on the sphere are gravity, tension in the wires AC and BC, and the centripetal force.
The centripetal force is equal to the product of the mass of the sphere and its velocity squared, divided by the radius of its orbit. Since the sphere is in equilibrium, these forces must be in balance. Therefore, the minimum value of v can be determined by setting the gravity force and the tension in the wires equal to the centripetal force. This gives us:
mv2/r = TAC + TBC
where m is the mass of the sphere, v is the velocity, r is the radius of the orbit, and TAC and TBC are the tensions in the wires AC and BC respectively.
Solving for v gives:
v = sqrt(TAC + TBC * (r/m))
Substituting the values given in the problem, we get:
v = sqrt((TAC + TBC) * (1.6/5))
Therefore, the minimum value of v is:
v = sqrt(TAC + TBC) = sqrt(2 * (1.6/5)) = 0.96 m/s
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The forces operating on the sphere define the minimal value of v. The sphere is subject to the forces of gravity, centripetal force, and tension in wires AC and BC.
The centripetal force is determined by multiplying the sphere's mass by its velocity squared and dividing the result by the radius of its orbit. These forces must be in balance for the sphere to be in equilibrium. As a result, the minimum value of v can be calculated by setting the centripetal force, gravity force, and wire tension to the same value. This results in:
mv2/r = TAC + TBC
where m is the mass of the sphere, v is the velocity, r is the radius of the orbit, and TAC and TBC are the tensions in the wires AC and BC respectively.
Solving for v gives:
v = sqrt(TAC + TBC * (r/m))
Substituting the values given in the problem, we get:
v = sqrt((TAC + TBC) * (1.6/5))
Therefore, the minimum value of v is:
v = sqrt(TAC + TBC) = sqrt(2 * (1.6/5)) = 0.96 m/s
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From what height would a car have to fall in order for the magnitude of its momentum to equal the magnitude of its momentum when it is moving on a highway at 34 m/s? Express your answer with the appropriate units. μA ? Ay = Value Units Submit Request Answer
From the height of 58.91 m would a car have to fall in order for the magnitude of its momentum to equal the magnitude of its momentum when it is moving on a highway at 34 m/s.
What is momentum?A motion of an item is said to have momentum when its mass multiplied by its speed. If you describe anything as having momentum, you are describing its mass and direction of motion. A truck loaded with products has a big mass and must slow down before a stop signal since stopping is exceedingly difficult due to the truck's enormous momentum and constant speed. Even though a bullet has a relatively little mass and a very high velocity, it nevertheless possesses a significant amount of momentum.
Magnitude of momentum of car on highway = m × 34 kg.m/sec
Here, mass remains the constant, speed of the car should be = 34
Thus, v² = v₀² = 2gd
v₀ = 0
v = 34 m/sec
so, d = v² /2g
d = (34)² / 2 × 9.81
d = 58.91 m
From the height of 58.91 m would a car have to fall in order for the magnitude of its momentum to equal the magnitude of its momentum when it is moving on a highway at 34 m/s.
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