Answer:
(a) k = 1.76× 10⁶ N/m
(b) E = 3.52 × 10⁶ J
Explanation:
(a)
The period (T) of a spring = 2π√(m/k)
where m = mass of the spring in kg, k = spring constant.
T = 2π√(m/k)..................... equation 1
making k the subject of the equation,
k = 4π²(m)/T².............................. equation 2
Where m = 4.00 × 10⁵ kg, T = 3.00 s, π = 3.143
Substituting these values into equation 2
k = 4(3.143)²(4.0×10⁵)/3²
k = (1.58 × 10⁷)/9
k = 1.76× 10⁶ N/m
(b)
The energy stored(E) in a spring = 1/2ke²
Where k = spring constant, e = extension.
E = 1/2ke²
k = 1.76× 10⁶ N/m, e= 2.00 m
∴E = 1/2(1.76× 10⁶)(2)²
E = 2 × 1.76 × 10⁶
E = 3.52 × 10⁶ J
One might be tempted to say that exothermic processes are always spontaneous since the system is emitting energy (heat) in order to reach a (preferred) lower energy state. However, as we have just investigated, the spontaneous process for polymers is endothermic. This reveals that we must consider entropy changes when determining the nature of spontaneity. The most probable configuration of a system and its surroundings, naturally, is the one that will be observed. The condition for spontaneity can be recast using the concept of the free energy of the system, where a change in free energy results both from changes in the enthalpy (which includes internal potential and kinetic energies) and the entropy (the number of states accessible to the system). Δ G = Δ H − T Δ S.
An unknown chemical reaction undergoes an enthalpy change of Δ H =17 kJ/mol while the entropy increases by Δ S =50 J/(mol * K).
Above what temperature (in Kelvin) does this reaction occur spontaneously?
Answer : This reaction occur spontaneously at temperature above in kelvins is, 340 K
Explanation : Given,
[tex]\Delta H[/tex] = 17 KJ/mole = 17000 J/mole
[tex]\Delta S[/tex] = 50 J/mole.K
Gibbs–Helmholtz equation is :
[tex]\Delta G=\Delta H-T\Delta S[/tex]
As per question the reaction is spontaneous that means the value of [tex]\Delta G[/tex] is negative or we can say that the value of [tex]\Delta G[/tex] is less than zero.
[tex]\Delta G<0[/tex]
The above expression will be:
[tex]0>\Delta H-T\Delta S[/tex]
[tex]T\Delta S>\Delta H[/tex]
[tex]T>\frac{\Delta H}{\Delta S}[/tex]
Now put all the given values in this expression, we get :
[tex]T>\frac{17000J/mole}{50J/mole.K}[/tex]
[tex]T>340K[/tex]
Therefore, this reaction occur spontaneously at temperature above in kelvins is, 340 K
Final answer:
The reaction occurs spontaneously above 340 Kelvin, determined by applying the Gibbs free energy formula (ΔG = ΔH - TΔS) and ensuring the values for ΔH and ΔS are in compatible units.
Explanation:
The question requires us to determine above what temperature a reaction occurs spontaneously, given an enthalpy change (ΔH) of 17 kJ/mol and an entropy change (ΔS) of 50 J/(mol·K). To find this, we use the formula for Gibbs free energy change (ΔG = ΔH - TΔS), where a reaction is spontaneous when ΔG < 0.
First, we need to ensure that both values are in the same units, so we convert ΔH from kJ to J: 17 kJ/mol = 17000 J/mol. Then, we solve for T in the equation ΔG < 0, substituting ΔH and ΔS into the equation:
0 > 17000 J/mol - T(50 J/(mol·K)),
implying T > 17000 J/mol / 50 J/(mol·K) = 340 K.
Therefore, the reaction occurs spontaneously above 340 K.
A sinusoidally-varying voltage V(t)=V0sin(2pift) with amplitude V0 = 10 V and frequency f = 100 Hz is impressed across the plates of a circular-shaped parallel plate air-gap capacitor of radius a = 1.0 cm and plate separation d = 0.01 mm. The amplitude of Maxwell's displacement current ID flowing across the gap between the plates of this capacitor is?
The amplitude of Maxwell's displacement current flowing across the gap between the plates of the capacitor is 1.753 x 10^-7 A.
Explanation:The amplitude of Maxwell's displacement current flowing across the gap between the plates of this capacitor can be determined using the formula for the displacement current, which is given by ID = ε0AdV/dt, where ε0 is the permittivity of free space, A is the area of the capacitor plates, and dV/dt is the rate of change of voltage with respect to time.
In this case, the area A of the circular plates is equal to πa^2, where a is the radius of the plates. Therefore, A = π(0.01 m)^2 = 0.000314 m^2. The rate of change of voltage can be obtained from the equation V(t) = V0sin(2πft), where V0 is the amplitude of the voltage, f is the frequency, and t is the time.
Substituting the given values V0 = 10 V and f = 100 Hz into the equation, we have V(t) = 10sin(2π(100)t). Differentiating this equation with respect to time gives dV/dt = 2000πcos(2π(100)t). Plugging in the value of t = 0 (for maximum displacement current) into the equation, we find that cos(2π(100)(0)) = cos(0) = 1.
Therefore, the amplitude of the Maxwell's displacement current ID is given by ID = (8.85 x 10^-12 F/m)(0.000314 m^2)(2000π)(1) = 1.753 x 10^-7 A.
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A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.107 mm wide. At the point in the pattern which is an angular distance of 3.09° from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad.
a) What is the wavelenght of the radiation?
b) What is the intensity at this point, if the intensity at the center of the central maximum is I/O?
Answer:
a. λ = 647.2 nm
b. I₀ 9.36 x 10⁻⁵
Explanation:
Given:
β = 56.0 rad , θ = 3.09 ° , γ = 0.170 mm = 0.170 x 10⁻³ m
a.
The wavelength of the radiation can be find using
β = 2 π / γ * sin θ
λ = [ 2π * γ * sin θ ] / β
λ = [ 2π * 0.107 x 10⁻³m * sin (3.09°) ] / 56.0 rad
λ = 647.14 x 10⁻⁹ m ⇒ λ = 647.2 nm
b.
The intensity of the central maximum I₀
I = I₀ (4 / β² ) * sin ( β / 2)²
I = I₀ (4 / 56.0²) * [ sin (56.0 /2) ]²
I = I₀ 9.36 x 10⁻⁵
To find the wavelength of the radiation and the intensity at a specific point in a single-slit diffraction pattern, we can use formulas and calculations based on the given information.
Explanation:To find the wavelength of the radiation, we can use the formula for single-slit diffraction:
sin(θ) = mλ/w
Where θ is the angle from the center of the central maximum, m is the order of the bright fringe (m = 1 for the first bright fringe), λ is the wavelength, and w is the width of the slit. Rearranging the formula, we get:
λ = wsin(θ)/m
Plugging in the values, we have:
λ = (0.107 mm)(sin(3.09°))/(1)
Calculating this gives us the wavelength of the radiation.
To find the intensity at the given point, we can use the formula for intensity in single-slit diffraction:
I/I0 = (sin(θ)/θ)2
The given point is at an angular distance of 3.09°, so we can use this formula to calculate the intensity.
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Adam and Bobby are twins who have the same weight. Adam drops to the ground from a tree at the same time that Bobby begins his descent down a frictionless slide. If they both start at the same height above the ground, how do their kinetic energies compare when they hit the ground? a. Bobby has twice the kinetic energy as Adam More information is required to compare their kinetic energies. b. Adam has greater kinetic energy than Bobby. c. They have the same kinetic energy. d. Bobby has greater kinetic energy than Adam.
Answer:c
Explanation:
it is given that Adam and Boby starts from same height so their total Energy at top is [tex]T_t[/tex]
[tex](T_{top})_a=[/tex]Potential Energy of Adam
[tex](T_{top})_b=[/tex]Potential Energy of boby
when they fall a height h their speed at bottom will be
[tex]v=\sqrt{2gh}[/tex]
which will be same for both Adam and Boby
Energy at bottom
[tex](T_b)_a=[/tex]Kinetic Energy of Adam
[tex](T_b)_b=[/tex]Kinetic Energy of boby
Since velocity is same therefore kinetic Energy is same for Adam and boby
thus option c is correct
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: =E−Ryn2 In this equation Ry stands for the Rydberg energy, and n stands for the principal quantum number of the orbital that holds the electron. (You can find the value of the Rydberg energy using the Data button on the ALEKS toolbar.) Calculate the wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron from an orbital with =n10 to an orbital with =n8. Round your answer to 3 significant digits.
The energy change in an electron's transition in a hydrogen atom can be calculated using the Bohr formula. This energy corresponds to the energy of the emitted photon in the transition, and can be used to calculate the wavelength of the light emitted during this transition.
Explanation:The energy change associated with the transition of an electron between two energy levels in a hydrogen atom can be calculated using the equation E = 13.6 eV/n². In this equation, 'E' represents the energy of the electron, and 'n' is the quantum number of the orbit that the electron occupies. If we consider a transition of an electron from an orbital with n = 10 to an orbital with n = 8, we can calculate the energy change (∆E) using the difference in energies of these two states. This energy change corresponds to the energy of the emitted photon when the electron transition occurs. The energy of the photon can be connected to its wavelength through the equation E = hf, where 'h' is Planck's constant, and 'f' is the frequency, which is related to the wavelength (λ) by the speed of light (c) as f = c/λ. Therefore, you can first calculate ∆E using the given energy equation, then use the result to find 'f' using E = hf, and finally substitute 'f' in f = c/λ to find λ, the wavelength of the emitted light.
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Ray baked a cake. The total mass of the cake was equal to the total mass of the ingredients. This is an example of what?
Answer:
The Law of Conservation of Mass-Energy
Explanation:
It states that nothing can be created or destroyed, which is why the cakes mass is the same as the ingredients.
Answer : This is an example of law of conservation of mass.
Explanation :
Law of conservation of mass : It states that mass can neither be created nor destroyed but it can only be transformed from one form to another form.
The balanced chemical reaction always follow the law of conservation of mass.
Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.
For example :
[tex]2H_2+O_2\rightarrow 2H_2O[/tex]
This reaction is a balanced chemical reaction in which number of atoms of hydrogen and oxygen are equal on the both side of the reaction. So, this reaction obey the law of conservation of mass.
As per question, the total mass of the cake was equal to the total mass of the ingredients. That means, they obey the law of conservation of mass.
Hence, this is an example of law of conservation of mass.
Stars form from clouds of gas and dust. As a protostar gravitationally contracts within its parent cloud, "conservation of angular momentum" says thatA) the protostar rotates more slowly.b) the protostar\'s rotation does not change.c) the protostar\'s axis of rotation will change direction.d) the protostar rotates more quickly.
Answer:
D) True. the protostar rotates more quickly.
Explanation:
If the system is isolated, the angular momentum must be retained.
Initial
L₀ = I w₀
Final
[tex]L_{f}[/tex] = [tex]I_{f}[/tex] [tex]w_{f}[/tex]
L₀ = [tex]L_{f}[/tex]
I w₀ = [tex]I_{f}[/tex][tex]w_{f}[/tex]
[tex]w_{f}[/tex] = I /[tex]I_{f}[/tex] w₀
In general, the radius of the cloud decreases significantly to form the star, the moment of inertia must decrease, so the angular velocity must increase
Let's examine the answers
A) False. The opposite happens
B) False. Speed changes
C) False. For this there must be an external force, which does not exist
D) True. You agree with the above
Final answer:
The correct answer to the question of what occurs to a protostar as it contracts due to the 'conservation of angular momentum' is that the protostar rotates more quickly (d). This is analogous to a figure skater pulling their arms in to spin faster and is supported by observations of star-forming regions like the Orion Nebula.
Explanation:
Stars form from clouds of gas and dust. As these clouds collapse under their own gravity, they form protostars, which spin due to the conservation of angular momentum. Conservation of angular momentum dictates that as the radius of a spinning object decreases, its rotation speed must increase to conserve angular momentum. This concept is similar to a figure skater who spins faster when they pull their arms in. Therefore, when a protostar gravitationally contracts within its parent cloud, it rotates more quickly because as it shrinks, the protostar's rate of spin increases to conserve angular momentum.
This increase in rotation speed results in the formation of a spinning accretion disk around the equator, which is easier to observe in some regions, such as the Orion Nebula or the Taurus star-forming region. Observations from telescopes like the Hubble Space Telescope support this understanding of the role of angular momentum in star formation. In conclusion, the correct answer to the question is (d) the protostar rotates more quickly.
A device plugged into a wall outlet experiences an alternating current. The RMS current value is 2 A. The peak current is ___.
Answer:
The peak value of current will be 2.828 A.
Explanation:
Given that
Value of RMS current I(rms) = 2 A
Lets take peak current = I(p)
As we know that relationship between RMS and peak current given as
[tex]I(rms)=\dfrac{I(p)}{\sqrt2}[/tex]
Now by putting the values in the above equation
[tex]I(rms)=\dfrac{I(p)}{\sqrt2}[/tex]
[tex]2\ A=\dfrac{I(p)}{\sqrt2}[/tex]
[tex]I(p)=2\sqrt2[/tex]
As we know that
[tex]\sqrt{2}=1.414[/tex]
Therefore
I(p)=2.828 A
The peak value of current will be 2.828 A.
A flight attendant pulls her 70.0 N flight bag a distance of 253 m along a level airport floor at a constant velocity. The force she exerts is 40.0 N at an angle of 52 degrees above the horizontal. Find the following: a) the work she does on the flight bag b) the work done on the flight bag c) the coefficient of kinetic friction between the flight bag and the floor.
Answer:
6230.49413 J
0 J
0.63998
Explanation:
F = Force = 40 N
[tex]\theta[/tex] = Angle = 52°
Work done is given by
[tex]W=Fscos50\\\Rightarrow W=40\times 253\times cos52\\\Rightarrow W=6230.49413\ J[/tex]
The work she does on the flight bag is 6230.49413 J
The work done on the flight bag will be the opposite of the work done by the flight attendant
[tex]W=-6230.49413\ J[/tex]
So net work will be
[tex]W_n=6230.49413-6230.49413\\\Rightarrow W_n=0[/tex]
The net work done on the flight bag is 0 J
Coefficient of friction is given by
[tex]\mu=\dfrac{F_f}{F_N}\\\Rightarrow \mu=\dfrac{F_f}{F_g-F_{app}}\\\Rightarrow \mu=\dfrac{40cos52}{70-40sin52}\\\Rightarrow \mu=0.63998[/tex]
The coefficient of friction is 0.63998
(a) The work done by the attendant on the flight bag is 6230.49 J.
(b) The work on the flight bag is 0 J .
(c) The coefficient of kinetic friction between the flight bag and the floor is 0.35.
Given data:
The weight of flight bag is, W = 70.0 N.
The distance covered by the bag is, d = 253 m.
The magnitude of force exerted on bag is, F = 40.0 N.
The angle of inclination with horizontal is, [tex]\theta =52^{\circ}[/tex].
Work done defined as the product of force and distance covered due to applied force.
(a)
The work done on the flight bag is given as,
[tex]W'=F \times dcos\theta[/tex]
Solving as,
[tex]W'=40 \times 253cos52\\W'=6230.49 \;\rm J[/tex]
Thus, the work done by the attendant on the flight bag is 6230.49 J.
(b)
The work done on the flight bag will be the opposite of the work done by the flight attendant. So,
W'' = - W
W'' = - 6230.49 J
Then net work done on the flight bag is,
[tex]W_{net}=W'+W''\\W_{net}=6230.49 - 6230.49\\W_{net}=0[/tex]
Thus, the net work on the flight bag is 0 J .
(c)
The expression for the frictional force is given as,
[tex]f = \mu \times W[/tex]
[tex]\mu[/tex] is the coefficient of kinetic friction. And the frictional force is due to the horizontal component of applied force. Then,
[tex]f=Fcos\theta[/tex]
So,
[tex]Fcos\theta= \mu \times W\\\\40 \times cos52=\mu \times 70\\\\\mu=\dfrac{40 \times cos52}{70} \\\\\mu = 0.35[/tex]
Thus, the coefficient of kinetic friction between the flight bag and the floor is 0.35.
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A resistor with R=300 ? and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate at which electrical energy is dissipated in the resistor is 216 W. (a) What is the impedance Z of the circuit? (b) What is the amplitude of the voltage across the inductor? (c) What is the power factor?
Answer:
impedance Z = 416.66 ohm
voltage across inducance V = 346.99 V
power factor = 0.720
Explanation:
given data
resistor R = 300
voltage amplitude = 500 V
resistor = 216 W
to find out
impedance and amplitude of the voltage and power factor
solution
we apply here average power formula that is
average power = I²×R ............1
I = [tex]\frac{Vrms}{Z}[/tex]
so
average power = ([tex]\frac{Vrms}{Z}[/tex])²×R
Vrms = [tex]\frac{1}{\sqrt{2} }[/tex] × Vmax
Z = V × [tex]\sqrt{\frac{R}{2P} }[/tex]
Z = 500 × [tex]\sqrt{\frac{300}{2*216} }[/tex]
impedance Z = 416.66 ohm
and
we know voltage across inductor is here express as
V = I × X .............2
so here X will be by inductance
Z² = R² + (X)²
(X)² = 416.66² - 300²
X = 289.15 ohm
and I = [tex]\frac{V}{Z}[/tex]
I = [tex]\frac{500}{416.66}[/tex]
I = 1.20 A
so from equation 2
V = 1.20 × 289.15
voltage across inducance V = 346.99 V
and
average power = Vmax × Imax × cos∅
tan∅ = [tex]\frac{289.15}{300}[/tex]
tan∅ = 43.95°
so power factor is
power factor = cos43.95°
power factor = 0.720
The problem involves the use of Ohm's law, impedance, power, and a power factor in an AC circuit. The impedance of the circuit, the voltage across the inductor, and the power factor can be calculated using given values, the principle of impedance and the relationships among AC voltage, current, resistance and power.
Explanation:
The question involves an AC circuit composed of a resistor and inductor in series connected to an AC voltage source. We have a resistor with a resistance (R) of 300 Ω and a dissipated power (P) of 216W. The voltage amplitude (Vo) of the AC source is 500 V. It is important to remember that in this context, impedance (Z), which has a unit of ohms, represents the total resistance in an AC circuit and can be calculated using Ohm's law for AC circuits.
(a) To find the impedance Z of the circuit, we consider that the power P is given by the relation P = Vo^2/R, substituting for P, and R, we can solve for Vo, which will be sqrt(P*R). Then, the rms voltage (V) is given by Vo/sqrt(2). Our current I would be P/V. Finally, applying Ohm's law, Z=V/I would give us the impedance.
(b) The voltage across the inductor can be found by using Pythagoras' Theorem in the context of an AC circuit, VL = sqrt(Vo^2 - VR^2), where VR is the voltage across the resistor (equal to I* R).
(c) Lastly, the power factor can be found as the cosine of the phase angle θ, which can also be defined as R/Z. We'd first calculate θ = arccos(R/Z), and then find the power factor as cos(θ).
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Two 4.0-kg blocks are tied together with a compressed spring between them. They are thrown from the ground with an initial velocity of 35 m/s, 45° above the horizontal. At the highest point of the trajectory they become untied and spring apart. About how far below the highest point is the center of mass of the two-block system 2.0 s later, before either fragment has hit the ground?
Answer:
The distance is 20 m.
Explanation:
Given that,
Mass of two blocks = 4.0 kg
Initial velocity = 35 m/s
Angle = 45°
Time = 2 sec
After, they become untied, center of mass trajectory will not change:
We need to calculate the distance
Using equation of motion
[tex]s=ut-\dfrac{1}{2}gt^2[/tex]
Where, u = initial velocity
g = acceleration due to gravity
t = time
[tex]s=0\times2-\dfrac{1}{2}\times9.8\times2^2[/tex]
[tex]s=-19.6 = -20\ m[/tex]
[tex]|s|=20\ m[/tex]
Hence, The distance is 20 m.
The center of mass of the two-block system is approximately 4.4 meters below the highest point 2.0 seconds later.
Explanation:First, we need to determine the initial vertical velocity of the two-block system. We can use the initial velocity and the launch angle to find the vertical component of the velocity:
Vertical velocity = initial velocity * sin(angle) = 35 m/s * sin(45°)
Then, we can use the formula for vertical displacement to find how far below the highest point the center of mass is:
Vertical displacement = (initial vertical velocity * time) - (0.5 * acceleration * time^2)
Since the blocks are at their highest point, the vertical displacement is equal to zero. We can rearrange the formula to solve for time:
Time = (initial vertical velocity) / acceleration
Finally, we can substitute the values into the formula for vertical displacement:
Vertical displacement = (initial vertical velocity * time) - (0.5 * acceleration * time^2)
The center of mass of the two-block system is approximately 4.4 meters below the highest point 2.0 seconds later.
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Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. They are used to ignite nuclear fusion, for example. Such a laser may produce an electromagnetic wave with a maximum electric field strength of 1.52\times 10^{11}~\text{V/m}1.52×10 11 V/m for a time of 1.00 ns. What energy does it deliver on a 1.00~\mathrm{mm^2}1.00 mm 2 area?
Answer:
30643 J
Explanation:
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]
t = Time taken = 1 ns
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
[tex]E_0[/tex] = Maximum electric field strength = [tex]1.52\times 10^{11}\ V/m[/tex]
A = Area = [tex]1\ mm^2[/tex]
Magnitude of magnetic field is given by
[tex]B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{1.52\times 10^{11}}{3\times 10^8}\\\Rightarrow B_0=506.67\ T[/tex]
Intensity is given by
[tex]I=\dfrac{cB_0^2}{2\mu_0}\\\Rightarrow I=\dfrac{3\times 10^8\times 506.67^2}{2\times 4\pi \times 10^{-7}}\\\Rightarrow I=3.0643\times 10^{19}\ W/m^2[/tex]
Power, intensity and time have the relation
[tex]E=IAt\\\Rightarrow E=3.0643\times 10^{19}\times 1\times 10^{-6}\times 1\times 10^{-9}\\\Rightarrow E=30643\ J[/tex]
The energy it delivers is 30643 J
A 26-g steel-jacketed bullet is fired with a velocity of 630 m/s toward a steel plate and ricochets along path CD with a velocity 500 m/s. Knowing that the bullet leaves a 50-mm scratch on the surface of the plate and assuming that it has an average speed of 600 m/s while in contact with the plate, determine the magnitude and direction of the impulsive force exerted by the plate on the bullet.
Utilizing the principles of impulse and momentum, the magnitude of the impulsive force exerted by the steel plate on the bullet is approximately 40,576.47 N, and its direction is opposite to the bullet's initial motion.
Explanation:Impulsive Force on a Ricocheting Bullet
The problem involves finding the impulsive force exerted by the steel plate on a steel-jacketed bullet that ricochets off its surface. To determine the impulsive force, we utilize the concepts of impulse and momentum. The bullet has an initial velocity of 630 m/s and exits with a velocity of 500 m/s after ricocheting. While in contact with the plate, we are given an average speed of 600 m/s. The displacement during contact is 50 mm, equivalent to 0.050 meters.
To find the time of contact, we use the formula for distance, which is: distance = average speed * time, or time = distance / average speed. Hence, the time of contact is 0.050 meters / 600 m/s = 0.0000833 seconds (or 83.3 microseconds). Now, using the change in momentum (Δp = mass * change in velocity) and the impulse-momentum theorem (impulse = Δp), we can find the impulse. Considering the magnitude of the velocities and the mass of the bullet (26 g or 0.026 kg), the change in velocity is (500 m/s - 630 m/s) = -130 m/s. Therefore, Δp = 0.026 kg * -130 m/s = -3.38 kg*m/s. Given that impulse equals the average force times the time of the impact (Impulse = average force * time), we can find the average force as Impulse/time = -3.38 kg*m/s / 0.0000833 s = -40,576.47 N. The negative sign indicates that the force on the bullet is directed opposite to its initial motion, which is consistent with a ricochet.
The magnitude of the impulsive force exerted by the plate on the bullet is approximately 40,576.47 N, and its direction is opposite to the bullet's initial motion, demonstrating the principle of conservation of momentum during collisions.
An oil film with index of refraction 1.64 is trapped between two pieces of glass with index of refraction 1.43. No light is reflected by such a film when 578 nm light falls on it at normal incidence. What is the nonzero minimum thickness of the oil film that will satisfy these conditions? Answer in units of nm.
Final answer:
The minimum nonzero thickness of an oil film with index of refraction 1.64, between glass with index 1.43 to achieve no reflection for 578 nm light at normal incidence, is approximately 88 nm. This results from the condition for destructive interference.
Explanation:
When light of wavelength 578 nm hits the film at normal incidence and no light is reflected, it means that the reflected light waves from the top and bottom surfaces of the oil film are exactly out of phase, causing destructive interference.
For destructive interference to occur, the path difference between the two reflected waves must be an odd multiple of half the wavelength in the medium, which is given by the formula:
Path Difference (in the medium) = 2 × n × thickness of the film = [tex](m + \(\frac{1}{2}\)) \times \(\frac{\lambda}{n}\)[/tex], where m = 0, 1, 2, ...
Considering that there is a [tex]\(\frac{\lambda}{2}\)[/tex] phase shift when light reflects off a medium with a lower index of refraction to a higher one, and the film's index of refraction is greater than that of the surrounding glass, we take m = 0 for the smallest non-zero thickness. Thus, the minimum thickness of the oil film is:
Thickness = [tex]\(\frac{\lambda}{4 * n}\)[/tex] = [tex]\(\frac{578 nm}{4 \times 1.64}\)[/tex] = 88.16 nm
The nonzero minimum thickness of the oil film that will satisfy the conditions of no reflected light is therefore approximately 88 nm.
You place 100 grams of ice, with a temperature of −10∘C, in a styrofoam cup. Then you add an unknown mass of water, with a temperature of +10∘C, and allow the system to come to thermal equilibrium. For the calculations below, use the following approximations. The specific heat of solid water is 2 joules / gram, and the specific heat of liquid water is 4 joules / gram. The latent heat of fusion of water is 300 joules / gram. Assume no heat is exchanged with the surroundings. Express your answers in grams, using two significant digits.
If the final temperature of the system is -5∘C , how much water was added? ______________ grams
Answer:
Mass of water 2.9g
Explanation:
Ice
[tex]m_{ice}=100g[/tex]
[tex]c_{ice}=2J/g.K[/tex]
[tex]T_{ice,initial}=-10\°C[/tex]
[tex]T_{ice,final}=T_{equilibrium}=-5\°C[/tex]
Water
[tex]c_{water}=4J/g.K[/tex]
[tex]T_{water,initial}=10\°C[/tex]
[tex]T_{water,final}=0\°C[/tex]
[tex]T_{equilibrium}=-5\°C[/tex]
[tex]l_{water}=300J/g[/tex]
[tex]m_{water}=?g[/tex]
Step 1: Determine heat gained by ice
[tex]Q_{ice}=m_{ice}c_{ice}(T_{ice,final}-T_{ice,initial})[/tex]
[tex]Q_{ice}=100*2*(-5--10)[/tex]
[tex]Q_{ice}=1000J[/tex]
Step 2; Determine heat lost by water
[tex]Q_{water}=m_{water}c_{water}(T_{water,initial}-T_{water,final})+m_{water}l_{water}[/tex]
[tex]Q_{water}=m_{water}*4*(10-0)+m_{water}*300[/tex]
[tex]Q_{water}=40m_{water}+300m_{water}[/tex]
[tex]Q_{water}=340m_{water}[/tex]
Step 3: Heat gained by ice is equivalent to heat lost by water
[tex]Q_{ice}=Q_{water}[/tex]
[tex]1000=340m_{water}[/tex]
[tex]m_{water}=2.9g[/tex]
Two protons, with equal kinetic energy, collide head-on. What is the minimum kinetic energy Kp of one of these protons necessary to make a pion-antipion pair? The rest energy of a pion is 139.6MeV.
Answer:
[tex]K_p=139.6\ MeV[/tex]
Explanation:
It is given that,
The rest energy of a pion is 139.6 MeV. Here, two protons having equal kinetic energy collides elastically. We need to find the minimum kinetic energy of one of these protons necessary to make a pion-antipion pair.
It can be calculated using conservation of energy and momentum, the total energy of the particles gets converted into rest mass energy of new particles. So,
[tex]2K_p=2\times E_{\pi^+}[/tex]
[tex]K_p=\times E_{\pi^+}[/tex]
[tex]K_p=139.6\ MeV[/tex]
So, the minimum kinetic energy of one of these protons necessary to make a pion-antipion pair is 139.6 MeV. Hence, this is the required solution.
The size (radius) of an oxygen molecule is about 2.0 ×10−10 m. Make a rough estimate of the pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior at ordinary temperatures (T= 300K ). Assume that deviatons would be noticeable when volume of the gas per molecule equals the volume of the molecule itself.
Express your answer using one significant figure.
P = ? Pa
Answer:
P = 1 x 10⁸ Pa
Explanation:
given,
radius = 2.0 ×10⁻¹⁰ m
Temperature
T = 300 K
Volume of gas molecule =
[tex]V = \dfrac{4}{3}\pi r^3[/tex]
[tex]V = \dfrac{4}{3}\pi (2\times 10^{-10})^3[/tex]
V = 33.51 x 10⁻³⁰ m³
we know,
P V = 1 . k T
k = 1.38 x 10⁻²³ J/K
P(33.51 x 10⁻³⁰) = 1 . (1.38 x 10⁻²³) x 300
P = 1.235 x 10⁸ Pa
for 1 significant figure
P = 1 x 10⁸ Pa
It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hooke’s law and is depressed 0.75 cm by its maximum load of 120 kg. (a) What is the spring’s effective force constant? (b) A player stands on the scales and depresses it by 0.48 cm. Is he eligible to play on this under-85-kg team?
Answer:
(a)spring effective force constant =1568N/cm
(b) Yes
Explanation:
Hooke's law is represented mathematically as, F=ke
where
F is force applied to elastic material,k= spring constante= extension(a).After the maximum load is exceeded, Hooke's law doesn't apply anymore. from the problem, its stated that a maximum load of 120kg will cause an extension of 0.75cm, we will use this to determine the spring constant.
F=mg, g=9.8[tex]m/s^{2}[/tex]
F= 120*9.8 =1176N
From Hooke's law, k=[tex]\frac{F}{e}[/tex]
k=[tex]\frac{1176}{0.75}[/tex]
k=1568N/cm
(b). The players who stands on the scale causes a 0.48cm extension.
e= 0.48cm, k= 1568N/cm
F=Ke
F= 1568*0.48
F= 752.64N
To calculate the mass of the player we divide this force by g=9.8[tex]m/s^{2}[/tex]
F=mg
m=F/g
[tex]m= \frac{ 752.64}{9.8} \\\\m=76.8kg[/tex]
since the rugby team is an under 85kg team, this player with 76.8kg mass is eligible
A 3.0-kg cylinder falls vertically from rest in a very tall, evacuated tube near the surface of the earth. What is its speed after the cylinder has fallen 6.0 m?
Answer:v=10.84 m/s
Explanation:
Given
mass of Cylinder [tex]m=3 kg[/tex]
height of cylinder [tex]h=6 m[/tex]
It is given that tube is evacuated so we can neglect air resistance so friction provided by the air is zero.
Since Energy cannot be destroyed but can be transformed from one form to another therefore Potential Energy of Cylinder is converted to Kinetic Energy of Cylinder
Potential Energy[tex]=mgh=3\times 9.8\times 6=176.4 J[/tex]
Kinetic Energy [tex]=\frac{1}{2}mv^2=\frac{1}{2}\times 3\times (v)^2[/tex]
[tex]176.4=\frac{1}{2}\times 3\times (v)^2[/tex]
[tex]v=\sqrt{117.6}[/tex]
[tex]v=10.84 m/s[/tex]
Speed of cylinder after the cylinder has fallen 6 m is 11 meter/second.
Energy conservation based problemWhat information do we have?
Mass of cylinder = 3 kg
Height = 6 m
Using energy conservation theroy
mgh = (1/2)mv²
gh = (1/2)v²
(9.8)(6) = (1/2)v²
Velocity = 11 m/s (Approx.)
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A 1.0 kg piece of copper with a specific heat of cCu=390J/(kg⋅K) is placed in 1.0 kg of water with a specific heat of cw=4190J/(kg⋅K). The copper and water are initially at different temperatures. After a sufficiently long time, the copper and water come to a final equilibrium temperature. Part A Which of the following statements is correct concerning the temperature changes of both substances? (Ignore the signs of the temperature changes in your answer.) Which of the following statements is correct concerning the temperature changes of both substances? (Ignore the signs of the temperature changes in your answer.) The temperature change of the copper is equal to the temperature change of the water. The temperature change of the water is greater than the temperature change of the copper. The temperature change of the copper is greater than the temperature change of the water.
Answer:
The temperature change of the copper is greater than the temperature change of the water.
Explanation:
deltaQ = mc(deltaT)
Where,
delta T = change in the temperature
m =mass
c = heat capacity
[tex]\frac{(deltaT)_{Cu}}{(deltaT)_{w}}=\frac{4190J/kg.K}{390J/kg.K}\\ \\(deltaT)_{Cu}=10.74(deltaT)_{w}[/tex]
The temperature change in the copper is nearly 11 times the temperature change in the water.
So, the correct option is,
The temperature change of the copper is greater than the temperature change of the water.
Hope this helps!
The temperature change of the copper is equal to the temperature change of the water.
Explanation:The temperature change of the copper is equal to the temperature change of the water.
When two substances come into contact, heat is transferred between them until they reach thermal equilibrium, where their temperatures are equal. According to the Law of Conservation of Energy, the heat lost by one substance is equal to the heat gained by the other substance. In this case, since all heat transfer occurs between the copper and water, the temperature change of the copper is equal to the temperature change of the water.
An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef- ficient based on area bl is CD = 0.06, estimate the power required to tow the banner. (b) For comparison, determine the power required if the airplane was instead able to tow a rigid flat plate of the same size. (c) Explain why one had a larger power requirement (and larger drag) than the other. (d) Finally, determine the power required if the airplane was towing a smooth spherical balloon with a diameter of 2 m.
Final answer:
We can estimate the power required to tow a banner by using the drag force formula. If the airplane was towing a rigid flat plate, the power required would be different due to a different drag coefficient. The power requirement and drag are larger for the banner due to its higher drag coefficient compared to a flat plate. No calculation can be made for a smooth spherical balloon without the drag coefficient.
Explanation:
(a) To estimate the power required to tow the banner, we can use the formula: Power = Drag force * velocity. The drag force can be calculated using the drag coefficient and the area of the banner. The area is given by multiplying the height (0.8 m) by the length (25 m). The drag force can then be multiplied by the velocity of the plane (150 km/hr converted to m/s) to obtain the power required.
(b) If the plane was towing a rigid flat plate of the same size, the drag coefficient would be different. The power required can be calculated using the new drag coefficient and the same formula as in part (a).
(c) The power requirement and drag are larger for the banner because the drag coefficient for the banner is higher compared to that of a rigid flat plate. This means that the banner experiences more air resistance, requiring more power to tow.
(d) To determine the power required to tow a smooth spherical balloon with a diameter of 2 m, we would need the drag coefficient associated with the balloon. Since it is not provided in the question, we cannot calculate the power required.
One of the greatest terrorism-related nuclear threats is from Select one: a. nuclear power plants. b. dirty bombs. c. nuclear warheads. d. None of these are correct.
Answer:
Dirty bomb
Explanation:
Among the nuclear bomb One type is a "dirty bomb." It combines a conventional explosive such as the dynamite with radioactive material which can spread when the system explodes. The explosion is releasing "dirty" bits of radioactive particles which are extremely harmful and can cause loss equivalent to a nuclear attack.
Question:The mechanical properties of cobalt may be improved by incorporating fine particles of tungsten carbide (WC). Given that the moduli of elasticity of these materials are, respectively, 200 GPa and 700 GPa, plot modulus of elasticity vs. the volume percent of WC in Co from 0 to 100 vol% using both upper- and lower-bound expressions to form a performance envelope into which the material will fall. Remember that the isostrain case is the upper-bound case and the isostress is the lower-bound case.Mechanical Properties:Mechanical properties help to identify and classify the materials. The most common Mechanical properties are ductility, tenacity, strength, impact resistance, hardness. The mechanical property does not depend on the quantity of the material.
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Final answer:
To explore how the incorporation of tungsten carbide (WC) into cobalt (Co) affects the composite material's modulus of elasticity, the moduli values are calculated for a range of WC volume percentages under both isostrain and isostress conditions, illustrating the performance envelope and guiding material design in engineering applications.
Explanation:
To plot the modulus of elasticity vs. the volume percent of WC in Co from 0 to 100 vol%, we use both upper- and lower-bound expressions to form a performance envelope. For the isostrain (upper-bound) case, the mixture's modulus of elasticity (Ecomposite, isostrain) can be calculated using the rule of mixtures formula: Ecomposite, isostrain = ECoVCo + EWCVWC, where E represents modulus of elasticity, and V represents volume fraction. For the isostress (lower-bound) case, the inverse of the composite's modulus of elasticity can be expressed as 1/Ecomposite, isostress = VCo/ECo + VWC/EWC. Here, ECo = 200 GPa is the modulus of elasticity for cobalt, and EWC = 700 GPa is for tungsten carbide.
As the volume percent of WC increases from 0 to 100%, the calculated Ecomposite, isostrain values will show a trend of increasing modulus due to the higher modulus of WC compared to Co. Conversely, the Ecomposite, isostress values will also follow an upward trend but at a different rate, illustrating the material's variation within the performance envelope depending on the volume fraction of WC. This graphical representation will help to visualize how incorporating tungsten carbide into cobalt affects the composite's mechanical performance, underlining the importance of material design and engineering applications.
For many years Colonel John P. Stapp, USAF, held the world's land speed record. He participated in studying whether a jet pilot could survive emergency ejection. On March 19, 1954, he rode a rocket-propelled sled that moved down a track at a speed of 632 mi/h. He and the sled were safely brought to rest in 1.40s
(a) Determine the negative acceleration he experienced.
__ m/s2
(b) Determine the distance he traveled during this negative acceleration.
__ m.
Answer:
(a) -202 m/s²
(b) 198 m
Explanation:
Given data
Initial speed (v₀): 283 m/s[tex]\frac{632mi}{h} .\frac{1609.34m}{1mi} .\frac{1h}{3600s} =283m/s[/tex]
Final speed (vf): 0 (rest)Time (t): 1.40 s(a) The acceleration (a) is the change in the speed over the time elapsed.
a = (vf - v₀)/t = (0 - 283 m/s)/ 1.40s = -202 m/s²
(b) We can find the distance traveled (d) using the following kinematic expression.
y = v₀ × t + 1/2 × a × t²
y = 283 m/s × 1.40 s + 1/2 × (-202 m/s²) × (1.40 s)²
y = 198 m
An empty capacitor is capable of storing 1.0 × 10-4 J of energy when connected to a certain battery. If the distance between the plates is halved and then filled with a dielectric (κ = 4.3), how much energy could this modified capacitor store when connected to the same battery?
Answer:
Explanation:
Energy stored in a capacitor
= 1/2 C₁V²
capacity of a capacitor
c = εK A / d
k is dielectric and d is distance between plates .
When the distance between the plates is halved and then filled with a dielectric (κ = 4.3)
capacity becomes 4.3 x 2 times
New capacity
C₂ = 8.6 C₁
Energy of modified capacitor
1/2 C₂ V²= 1/2 x 8.6 c x V²
Energy becomes
8.6 times.
Energy stored = 8.6 x 10⁻⁴ J
A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of the instrument. The plates are separated by 3 mm and the magnitude of the magnetic field is 0.3 T. What voltage between the plates will allow particles of speed 5.0 x 10^5 m/sa. 1200 V b. 3800 V c. 7500 V d. 190 V e. 380 V
Answer:
Voltage, V = 450 volts
Explanation:
It is given that,
Separation between plates, d = 3 mm = 0.003 m
magnitude of magnetic field, B = 0.3 T
Speed of the particle, [tex]v=5\times 10^5\ m/s[/tex]
The relation between the magnetic field, electric field and the velocity of the particle is given by :
[tex]v=\dfrac{E}{B}[/tex]
Also, [tex]E=\dfrac{V}{d}[/tex]
[tex]v=\dfrac{V}{Bd}[/tex]
[tex]V=vBd[/tex]
[tex]V=5\times 10^5\times 0.3\times 0.003[/tex]
V = 450 volts
So, the voltage between the plates will be 450 V. Hence, this is the required solution.
The Reynolds number, rho VD/mu, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionless, using both the FLT system and the MLT system for basic dimensions, and determine its value for methane flowing at a velocity of 4 m/s through a 2-in-diameter pipe.
Answer:
Re = 1 10⁴
Explanation:
Reynolds number is
Re = ρ v D /μ
The units of each term are
ρ = [kg / m³]
v = [m / s]
D = [m]
μ = [Pa s]
The pressure
Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]
μ = [Pa s] = [kg / m s²] [s] = [kg / m s]
We substitute the units in the equation
Re = [kg / m³] [m / s] [m] / [kg / m s]
Re = [kg / m s] / [m s / kg]
RE = [ ]
Reynolds number is a scalar
Let's evaluate for the given point
Where the data for methane are:
viscosity μ = 11.2 10⁻⁶ Pa s
the density ρ = 0.656 kg / m³
D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m
Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶
Re = 1.19 10⁴
The Reynolds number is a key parameter in fluid mechanics that signals laminar or turbulent flow. It is a dimensionless quantity determined by fluid properties and flow characteristics.
The Reynolds number, a dimensionless parameter, indicates whether flow is laminar or turbulent in fluid mechanics. It is defined as Re = rho VD/mu, where rho is the fluid density, V is the velocity, D is the diameter, and mu is the fluid viscosity. The value of Reynolds number for methane flowing at 4 m/s through a 2-in-diameter pipe is calculated using this formula.
I place a 500-g ice cube (initially at 0°C) in a Styrofoam box with wall thickness 1.0 cm and total surface area 600 cm2 .
If the air surrounding the box is at 20°C and after 4 hours the ice is completely melted, what is the conductivity of the Styrofoam material? (Lf = 80 cal/g)
a. 9.6 × 10−5cal/s⋅cm⋅°C
b. 2.8 × 10−6cal/s⋅cm⋅°C
c. 1.15 × 10−2cal/s⋅cm⋅°C
d. 2.3 × 10−4cal/s⋅cm⋅°C
Final answer:
The conductivity of the Styrofoam material is found using the heat required to melt the ice and the conduction formula, resulting in 2.8 × 10^⁻⁶cal/s·cm·°C.
Explanation:
To find the conductivity of the Styrofoam material, we need to calculate the amount of heat transferred to the ice cube to melt it completely and then relate this to the conductivity formula. First, we calculate the total heat required to melt the 500-g ice cube using the latent heat of fusion (Lf). The total heat (Q) required is the mass (m) of the ice times the latent heat of fusion (Lf).
Q = m × Lf = 500 g × 80 cal/g = 40000 cal
Next, we use the formula for heat conduction, which is Q = (k × A × ΔT × t) / d. Here, Q is the heat transferred, k is the thermal conductivity, A is the area through which heat is transferred, ΔT is the temperature difference between the inside and outside of the box, t is the time, and d is the thickness of the walls.
Since all values except k are known, we rearrange the equation to solve for k:
k = (Q × d) / (A × ΔT × t)
k = (40000 cal × 1.0 cm) / (600 cm² × 20°C × (4 × 3600 s))
Upon calculation, we find the conductivity k equals 2.8 × 10-6cal/s·cm·°C, which corresponds to option (b).
A team of dogs drags a 53.9 kg sled 1.62 km over a horizontal surface at a constant speed. The coefficient of friction between the sled and the snow is 0.234. The acceleration of gravity is 9.8 m/s 2 . Find the work done by the dogs. Answer in units of kJ.
Answer:
200.24 kJ
Explanation:
[tex]m[/tex] = mass of sled = 53.9 kg
[tex]d[/tex] = distance traveled by the sled = 1.62 km = 1620 m
[tex]\mu[/tex] = Coefficient of friction between sled and snow = 0.234
frictional force acting on the sled is given as
[tex]f = \mu mg[/tex]
[tex]F[/tex] = Applied force by the dogs on the sled
Since the sled moves at constant speed, the force equation for the motion of the sled is given as
[tex]F = f \\F = \mu mg[/tex]
[tex]W[/tex] = Work done by the dogs on the sled
Work done by the dogs on the sled is given as
[tex]W = F d\\W = \mu mg d\\W = (0.234) (53.9) (9.8) (1620)\\W = 200237.64 J\\W = 200.24 kJ[/tex]
The work done by the dogs is 200.238 kJ.
The work done by the dog is equal to the work done to move the sled through the distance and the work done against friction.
Formula:
W = dma+μmgd............. Equation 1Where:
W = work done by the dogsm = mass of the sleda = acceleration of the sledg = acceleration due to gravityμ = coefficient of frictionFrom the question,
Given:
m = 53.9 kga = 0 m/s² (move with constant speed)d = 1.62 km = 1620 mg = 9.8 m/s²μ = 0.234Substitute these values into equation 1
W = (53.9×0×1620)+(53.9×9.8×0.234×1620)W = 2002378 JW = 200.238 kJHence, The work done by the dogs is 200.238 kJ.
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Part A If the accuracy in measuring the position of a particle increases, what happens to the accuracy in measuring its velocity? If the accuracy in measuring the position of a particle increases, what happens to the accuracy in measuring its velocity? The accuracy in measuring its velocity also increases. The accuracy in measuring its velocity decreases. The accuracy in measuring its velocity remains the same. The accuracy in measuring its velocity becomes uncertain.
Answer:
The correct answer is the speed accuracy decreases
Explanation:
Before examining the final statements, let's review the uncertainty principle
Δx Δp> = h ’/ 2
h ’= h / 2π
This is because the process of measuring one quantity affects the measurement of the other.
Let's review the claims
The speed of the particle is proportional to the moment
p = mv
Therefore, if the position is measured more accurately (x) the accuracy of p must decrease
Δp = h ’/ 2 Δx
The correct answer is the speed accuracy decreases
According to the Heisenberg uncertainty principle in physics, if the accuracy in measuring a particle's position increases, the accuracy in measuring its velocity decreases. This is a fundamental characteristic of the quantum world due to the wave-particle duality of matter.
Explanation:The question refers to the Heisenberg uncertainty principle in physics, particularly quantum mechanics. This principle states that the more precisely the position of a particle is known, the less precisely its velocity (or momentum, to be more specific) can be known, and vice versa. Therefore, if the accuracy of measuring a particle's position increases, then the accuracy in measuring its velocity decreases.
This phenomenon isn't due to measurement techniques or technology. It's a fundamental limit defined by the nature of the quantum world. It derives from the wave-particle duality of matter, which means small particles like electrons behave both as particles and waves. If we measure the position very precisely, the particle acts more like a wave with an uncertain speed. Conversely, if we measure the speed very precisely, the particle acts more like a particle with an uncertain position.
For example, if we use an extremely short-wavelength electron probe to measure an electron's position, we'd be very accurate but massively disturb the electron's velocity in the process. Hence, increasing the precision in position measurement increases the uncertainty in velocity measurement.
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