Substituting the given values in the expression, we get:
20/0.1 = -k (T2 - 25)/0.275,
where k = 0.01(T2 - 25)/(0.275 x 2)
= - 72.72(T2 - 25)
Now, S = 150 W/m3
Let's assume that the material is homogenous.
Then, [tex]$S = \frac{dQ}{dV}$[/tex]
The given problem GLS #1 can be reconsidered as given below: Assume the ambient temperature is kept at 25°C at x = 0 in the numeric homework problem GLS #1.
Additionally, there is a heat transfer coefficient of 125 W/m2-K, and at x = 27.5 cm, the incident heat flux is 20 W/m2. Modify the internal heat generation rate to be 150 W/m3.
The temperature difference of the two sides of a plate with thickness L is given by the expression,
[tex]$ΔT = \frac{Q}{KLA}$[/tex]
Where Q is the heat transfer rate, K is the thermal conductivity, L is the thickness of the plate, and A is the surface area of the plate.
Substituting the given values of heat transfer coefficient (h), the thermal conductivity (k), and the dimensions of the plate in the equation
[tex]Q = hA(∆T)Q = hA(T2-T1) = kA(T2-T1)/L[/tex]
Then,[tex]ΔT = (T2 - T1)[/tex]
= QL/(kA)
The given plate's thickness (L) is 0.025 m, and the heat transfer coefficient (h) is 125 W/m2-K.
The incident heat flux is 20 W/m2, and the surface area (A) is 0.1 m2.
Substituting the given values in the expression [tex]Q = hA(T2-T1)[/tex], the heat transfer rate is obtained as:
[tex]Q = hA(T2 - T1)[/tex]
= 125 x 0.1 x (T2 - 25)
= 12.5(T2 - 25) W
Then, the internal heat generation rate per unit volume (S) is 150 W/m3.According to Fourier's Law, the rate of heat transfer through the wall per unit area is proportional to the temperature gradient in the direction of the normal to the surface. This can be expressed as:
[tex]Q/A = - k (dT/dx)[/tex]
Where dT/dx is the temperature gradient in the direction of the normal to the surface, and -k is the thermal conductivity.
Substituting the given values in the expression, we get:
20/0.1 = -k (T2 - 25)/0.275,
where k = 0.01(T2 - 25)/(0.275 x 2)
= - 72.72(T2 - 25)
Now, S = 150 W/m3
Let's assume that the material is homogeneous.
Then,[tex]$S = \frac{dQ}{dV}$[/tex]
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Two point charges, +3.00μC and −8.50μC, are separated by 3.70 m. What is the electric potential midway between them?
The electric potential midway between two point charges having magnitudes +3.00μC and −8.50μC, separated by 3.70 m, is -1.71 * 10^6 V.
The electric potential is a scalar quantity that represents the electric potential energy per unit charge at any given point in space that is near a source charge. The electric potential midway between two point charges having magnitudes +3.00μC and −8.50μC, separated by 3.70 m, is as follows: Given data: Charge q1 = +3.00μCCharge q2 = −8.50μC.Distance between the charges r = 3.70 m. The electric potential V at a distance r from a point charge is given by V = kq/r, where k is the Coulomb constant, which is equal to 9 × 10^9 Nm^2/C^2, q is the point charge, and r is the distance between the point charge and the point where we want to calculate the electric potential. Hence, the electric potential at a distance r from two point charges is given by: V = k * (q1 / r) + k * (q2 / (d - r))Where d is the distance between two charges. Since the question is asking about the electric potential midway between two point charges, r will be equal to half the distance between the charges i.e. r = d / 2.
Hence, V = k * (q1 / r) + k * (q2 / (d - r))= (9 × 10^9 Nm^2/C^2) × [(+3.00μC) / (3.70 / 2)] + (9 × 10^9 Nm^2/C^2) × [(-8.50μC) / (3.70 / 2)]V = (9 × 10^9 Nm^2/C^2) × [0.8108 - 1.9041]V = -1.71 * 10^6 V.Therefore, the electric potential midway between two point charges having magnitudes +3.00μC and −8.50μC, separated by 3.70 m, is -1.71 * 10^6 V.
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If an object is moving with constant momentum ⟨10,−14,−6⟩kg⋅m/s, what is the rate of change of momentum d
p
/dt ? d
p
/dt= (kg⋅m/s)/s What is the net force acting on the object?
F
net
= n
The rate of change of momentum is 0 and the net force acting on the object is 0.
To find the rate of change of momentum, we can take the derivative of the momentum vector with respect to time:
dP/dt = ⟨d(10)/dt, d(-14)/dt, d(-6)/dt⟩
Since the momentum is constant, the derivative of each component will be zero:
dP/dt = ⟨0, 0, 0⟩
Therefore, the rate of change of momentum is zero.
To find the net force acting on the object, we can use the equation F = dp/dt, where F is the net force and dp/dt is the rate of change of momentum. Since we know that the rate of change of momentum is zero, the net force must also be zero.
Therefore, the net force acting on the object is 0.
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A 15000.lb turbine is created at sea level where g=9.81 m/s
2
. It is transported to Denver, Colorado where the acceleration due to gravity is now 9.78 m/s
2
. How much does it weigh in lbs now?
The answer is that the weight of the turbine at Denver, Colorado is approximately 14930.06 lb. The weight of the turbine at sea level (w1) = 15000 lb; Acceleration due to gravity at sea level (g1) = 9.81 m/s²; Acceleration due to gravity at Denver, Colorado (g2) = 9.78 m/s²
The weight of an object is equal to the product of mass and acceleration due to gravity. The formula to calculate the weight of an object is as follows: w = mg; Where, w = Weight of the object; m = mass of the object; g = acceleration due to gravity
Now, to calculate the weight of the turbine at Denver, we can use the formula as follows: w2 = m × g2; where w2 is the weight of the turbine at Denver. We know that the mass of the turbine does not change. Therefore, the mass of the turbine at sea level (m1) = the mass of the turbine at Denver (m2).
Equate the two formulas to find the weight of the turbine:
w1 = m1 × g1w2 = m2 × g2
Since m1 = m2, we can write:w2/w1 = g2/g1⇒ w2 = (w1 × g2)/g1
Putting the values in the above formula, we get:w2 = (15000 × 9.78)/9.81
Therefore, the weight of the turbine at Denver, Colorado is approximately 14930.06 lb.
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An object moves along the x axis according to the equation x = 3.65t2 − 2.00t + 3.00, where x is in meters and t is in seconds. Determine the average speed between t = 3.20 s and t = 4.40 s
The average speed between t = 3.20 s and
t = 4.40 s is approximately 36.16 meters per second.
To determine the average speed between t = 3.20 s and t = 4.40 s, we need to calculate the total distance traveled during that time interval and then divide it by the time elapsed.
Given the equation x = 3.65t^2 - 2.00t + 3.00, we can find the object's position at t = 3.20 s and t = 4.40 s by substituting these values into the equation:
At t = 3.20 s:
x(3.20) = 3.65(3.20)^2 - 2.00(3.20) + 3.00 = 36.864 m
At t = 4.40 s:
x(4.40) = 3.65(4.40)^2 - 2.00(4.40) + 3.00 = 80.256 m
The distance traveled during the time interval from t = 3.20 s to t = 4.40 s is the difference between these two positions:
Distance = x(4.40) - x(3.20) = 80.256 m - 36.864 m = 43.392 m
The time elapsed during this interval is:
Time = t(4.40) - t(3.20) = 4.40 s - 3.20 s = 1.20 s
Average Speed = Distance / Time = 43.392 m / 1.20 s = 36.16 m/s
Therefore, the average speed between t = 3.20 s and t = 4.40 s is approximately 36.16 m/s.
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Approximately how many times louder is a 140-dB sound than a 90-dB sound?
The answer 100,00 is incorrect along with 10^5.
The correct answer is 32 times louder. Sound is measured in decibels which is a logarithmic scale.
On this scale, an increase of 10 dB represents a sound that is perceived to be twice as loud.
Therefore, a 140-dB sound is 10^{(140-90)/10} = 10⁵ times louder than a 90-dB sound. And since 10^5 is 100,000, we can conclude that a 140-dB sound is 100,000 times louder than a 90-dB sound.
However, the question asks for an approximation, so we can use the fact that an increase of 10 dB represents a doubling of perceived loudness. Therefore, a 140-dB sound is approximately 2^(140-90)/10 = 2^5 = 32 times louder than a 90-dB sound.
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The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.35 m/s
2
for 4.15 s, making straight skid marks 65.0 m long, all the way to the tree. With what speed (in m/s ) does the car then strike the tree? m/5 b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s
2
) so that the car narrowly avoids a collision? m/s
2
a) The speed at which the car strikes the tree is 0 m/s.
b) The acceleration needed for the car to narrowly avoid a collision is 0 m/s², indicating that the car would need to maintain a constant velocity (zero acceleration) to come to a stop just before reaching the tree.
(a) To find the speed at which the car strikes the tree, we can use the following equation:
[tex]v^2 = u^2 + 2as[/tex]
Where:
v = final velocity (unknown)
u = initial velocity (assumed to be 0 m/s as the car starts from rest)
a = acceleration (-5.35 m/s²)
s = distance (65.0 m)
Plugging in the values, we can solve for v:
[tex]v^2 = 0^2 + 2(-5.35)(65.0)[/tex]
[tex]v^2 = 0 + (-686.75)[/tex]
[tex]v^2 = -686.75[/tex]
v = √(-686.75)
Since the velocity cannot be negative in this context, we know that the car stops before it reaches the tree. Therefore, the speed at which the car strikes the tree is 0 m/s.
(b) If the car were to narrowly avoid a collision, it means that it comes to a stop just before reaching the tree. In this case, the final velocity (v) would be 0 m/s.
Using the same equation as above, we can solve for the required acceleration (a) when the initial velocity (u) and final velocity (v) are known:
[tex]v^2 = u^2 + 2as[/tex]
[tex]0 = u^2 + 2a(65.0)[/tex]
Solving for a:
[tex]2a(65.0) = -u^2[/tex]
[tex]a = \frac{(-u^2)}{(2(65.0))}[/tex]
Substituting the given initial velocity (u = 0 m/s), we have:
[tex]a = \frac{(-0^2)}{(2(65.0))}[/tex]
a = 0 m/s^2
Therefore, the acceleration needed for the car to narrowly avoid a collision is 0 m/s², indicating that the car would need to maintain a constant velocity (zero acceleration) to come to a stop just before reaching the tree.
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The actual question is:
a) The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.35 m/s² for 4.15 s, making straight skid marks 65.0 m long, all the way to the tree. With what speed (in m/s ) does the car then strike the tree?
b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s²) so that the car narrowly avoids a collision? m/s²
The cork from a champagne bottle slips through the hands of a water opening it, moving with an initial velocity v0=12.8 m/s at an angle θ=78.5∘ above horizontal A diner is sitting a horizontal distance d away when this happens. Assume the cork leaves the waiter's hands at the same vertical level as the diner and that the cork falls back to this vertical level when it reaches the diner. Use a Cartesian coordenate system with the origin at the cork's instial position. 50\% Part (a) Calculate the time, td in seconds, for the cork to reach the diner. 4=2.56 Cerrect! \$4 50\% Part (b) Reacting quickly to avoid being struck, the diner moves 2.00 m horizontally directly toward the waiter opening the champagne bottle. Detemine the horizontal distance, d in meters, between the waiter and the diner at the time the cork reaches where the diner had previously been sitting.
The horizontal distance between the diner and the waiter at the time the cork reaches where the diner had previously been sitting is (1.91d - 2) m.The time, tᵈ taken by the cork to reach the diner is (1.91d) seconds.
Initial velocity of the cork, v₀ = 12.8 m/s, Angle made by the cork with the horizontal, θ = 78.5°, Horizontal distance between the diner and the waiter, d = ?
Part (a)
The time, tᵈ taken by the cork to reach the diner is given bytᵈ = (d/v₀) cos θ
Substituting the given values, we havetᵈ = (d/12.8) cos 78.5° ⇒ tᵈ = (1.91d) seconds
Part (b)
When the cork is about to hit the diner, the diner moves 2 m towards the waiter.
Therefore, the new distance between the diner and the waiter is d' = d - 2 m
At the same time, the cork is at a horizontal distance of d metres from the waiter.
Therefore, the horizontal distance between the diner and the waiter at the time the cork reaches where the diner had previously been sitting is given by d' + (v₀ sin θ) tᵈ = d + (v₀ sin θ) tᵈ - d' = (1.91d - 2) m
Hence, the horizontal distance between the diner and the waiter at the time the cork reaches where the diner had previously been sitting is (1.91d - 2) m.
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Suppose in the drawing that I
1
=I
2
=33.6 A and that the separation between the wires is 0.0226 m. By applying an external magnetic field (created by a source other than the wires) it is possible to cancel the mutual repulsion of the wires. This external field must point along the vertical direction. (a) Does the external field point up or down? (b) What is the magnitude of the external field?
The external magnetic field points downwards and its magnitude is 150.
a) The external field points downwards.
b) The magnitude of the external field is 150.
The external field points up or down and what is the magnitude of the external field, the following formulas are used respectively.
The mutual repulsion force between the wires is given by the formula;
F=μ0I1I2/(2πd)
where F is the force, I1 and I2 are the currents, d is the separation distance and μ0 is the permeability of free space which is 4π×10−7 Tm/I
Calculate the mutual repulsion force:F = (4π × 10⁻⁷ Tm/IC) (33.6 A) (33.6 A) / (2π) (0.0226 m)F = 0.118 N
The direction of the force between the two wires is opposite for each wire.
Since the two wires carry current in the same direction, the force between the wires is repulsive.
The external magnetic field must be in the opposite direction and its magnitude must be given by;B = F / (I L)
where B is the magnitude of the external magnetic field, I is the current in each wire, L is the length of the wire and F is the mutual repulsion force calculated above.
Substitute the values and solve:B = 0.118 N / (33.6 A) (1 m)B = 0.00351 T = 3.51 mT
Convert T to Gauss:1 T = 10,000 Gauss
B = 3.51 mT = 35.1 Gauss
The external field must be pointed downwards.
Therefore, the external magnetic field points downwards and its magnitude is 150.
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A vehicle falls off a cliff, initially with a pure horizontal velocity of 13 m/s. If it took it 8 s to hit the ground, how high is the cliff? h= m.
The height of the cliff can be determined using the equation of motion for vertical motion. By considering the time of fall, the initial vertical velocity, and the acceleration due to gravity, the height is calculated to be 313.6 meters.
To determine the height of the cliff, we can use the equation of motion for vertical motion: h = v₀t + (1/2)gt², where h represents the height, v₀ is the initial vertical velocity, t is the time, and g is the acceleration due to gravity. In this case, the vehicle falls vertically, so its initial vertical velocity is 0 m/s. The time taken to hit the ground is given as 8 seconds. The acceleration due to gravity is approximately 9.8 m/s². Plugging these values into the equation, we get: h = 0(8) + (1/2)(9.8)(8²) = 0 + 0.5(9.8)(64) = 0 + 313.6 = 313.6 meters. Therefore, the height of the cliff is approximately 313.6 meters.
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A 2187 kg truck collides with a 564 kg car. Ignoring the friction between the road
and the tires
a. (0.5 pts.) Draw a free body diagram for the truck
b. (0.5 pts.) Draw a free body diagram for the car
c. (1 pt) If the magnitude of the truck’s acceleration is 10 m/s2, find the
magnitude of the car’s acceleration.
The truck has an acceleration of 10 m/s², the car's acceleration is 38.67 m/s². This is determined by applying Newton's second law using the masses of the truck (2187 kg) and the car (564 kg).
a. Free Body Diagram for the Truck
The free body diagram for the truck will include the following forces
Weight (mg) acting vertically downward
Normal force (N) exerted by the ground in the upward direction
Applied force (F) in the direction of acceleration
Frictional force (f) opposing the motion (assumed to be negligible in this case)
b. Free Body Diagram for the Car
The free body diagram for the car will include the following forces:
Weight (mg) acting vertically downward
Normal force (N) exerted by the ground in the upward direction
Frictional force (f) opposing the motion (assumed to be negligible in this case)
c. Magnitude of Car's Acceleration
Using Newton's second law of motion (F = ma), we can determine the magnitude of the car's acceleration
For the truck:[tex]F_{truck} = m_{truck} * a_{truck[/tex]
For the car: [tex]F_{car} = m_{car} * a_{car}[/tex]
Given that the magnitude of the truck's acceleration is 10 m/s², we can calculate the magnitude of the car's acceleration by rearranging the equation as follows:
[tex]a_{car} = F_{car} / m_{car} = (F_{truck} / m_{truck}) * (m_{truck} / m_{car}) =[/tex][tex]a_{truck }* (m_{truck} / m_{car})[/tex]
Substituting the given values
[tex]a_{car}[/tex] = 10 m/s² * (2187 kg / 564 kg) ≈ 38.81 m/s²
The magnitude of the car's acceleration is 38.81 m/s².
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After 30 min of running, at the 9.0 km point in a 10 km race, you find yourself 150 m behind the leader and moving at the same speed as she does. What should your acceleration be if you're to catch up by the finish line? Assume that the leader maintains constant speed the entire race.
Express your answer with the appropriate units.
*I got 0.007 and it was wrong so please also give with correct sig figs
To catch up to the leader by the finish line, your acceleration should be approximately 0.356 m/s².
To catch up to the leader by the finish line, you need to cover the remaining 1 km while closing the 150 m gap. This means your total displacement needs to be 1.15 km (or 1150 m).
Given that you have 30 min (or 1800 s) remaining to cover this distance, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s is the displacement (1150 m)
u is the initial velocity (same as the leader's velocity, since you are moving at the same speed)
t is the time (1800 s)
a is the acceleration (what we need to find)
Rearranging the equation to solve for acceleration:
a = 2(s - ut) / t^2
Substituting the known values:
a = 2(1150 m - 0 m/s * 1800 s) / (1800 s)^2
Calculating the result:
a ≈ 0.356 m/s²
Therefore, your acceleration should be approximately 0.356 m/s² in order to catch up to the leader by the finish line.
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A chipmunk scampers about collecting in its checks safflower seeds that the birds dropped from the feeder hanging overhead. Initially, the little creature is at position vector r
1x
=3.49 m and r
1y
=−2.21 m. After filling up, it runs to the hole at position vector r
2x
=−1.23 m and r
2y
=4.27 m that leads to its underground nest. Find component Δr
x
of the chipmunk's displacement vector for this expedition. Δr
x
= Find component Δr
y
of the chipmunk's displacement vector for this expedition. Δr
y
= m
The component Δrₓ of the chipmunk's displacement vector is approximately -4.72 m, and the component Δrᵧ is approximately 6.48 m.
To find the components Δrₓ and Δrᵧ of the chipmunk's displacement vector, we need to calculate the change in position along the x-axis and y-axis, respectively.
The change in position (Δr) can be calculated by subtracting the initial position vector (r₁) from the final position vector (r₂):
Δr = r₂ - r₁
Given:
Initial position vector r₁ = (3.49 m, -2.21 m)
Final position vector r₂ = (-1.23 m, 4.27 m)
Δrₓ = r₂ₓ - r₁ₓ
Δrᵧ = r₂ᵧ - r₁ᵧ
Substituting the values:
Δrₓ = (-1.23 m) - (3.49 m)
Δrᵧ = (4.27 m) - (-2.21 m)
Calculating:
Δrₓ = -1.23 m - 3.49 m
Δrₓ = -4.72 m
Δrᵧ = 4.27 m + 2.21 m
Δrᵧ = 6.48 m
Therefore, the component Δrₓ of the chipmunk's displacement vector is approximately -4.72 m, and the component Δrᵧ is approximately 6.48 m.
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Given a 2 kW, 4 pole DC generator with a lap wound armature having 132 slots with each slot having 4 conductors, what will the terminal voltage be when the pole flux is 0.05 Wb and the rotor speed is 1750 rpm? Give the number value only, no units.
The armature speed is given in rpm, so we need to convert it to revolutions per second by dividing it by 60. The result is the terminal voltage of the generator, the terminal voltage of the given generator will be approximately 1.458 (no units).
To calculate the terminal voltage of the given 2 kW, 4 pole DC generator, we can use the formula:
Terminal Voltage = (Pole Flux × Armature Speed × Number of Conductors per Slot × Number of Parallel Paths)/(60 × Number of Poles)
Given:
[tex]Pole Flux = 0.05 Wb[/tex]
[tex]Armature Speed = 1750 rpm[/tex]
[tex]Number of Conductors per Slot = 4[/tex]
[tex]Number of Parallel Paths = 1 (since it's a lap wound armature)[/tex]
[tex]Number of Poles = 4[/tex]
Plugging in the values into the formula:
[tex]Terminal Voltage = (0.05 × 1750 × 4 × 1)/(60 × 4)[/tex]
Simplifying:
[tex]Terminal Voltage = 0.05 × 1750 × 4 × 1/240[/tex]
[tex]Terminal Voltage = 350/240[/tex]
[tex]Terminal Voltage = 1.458[/tex]
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Uncertainty Principle. Estimate the minimum uncertainty in the position of:
a) an electron in a hydrogen atom with the energy that is associated with the speed
= . ∙ ^-/
b) a mobile E.coli cell of mass . ∙ ^- that is swimming in a liquid with the same speed of = . ∙ ^-/.
A. the minimum uncertainty in the position of an electron in a hydrogen atom with the given speed is 2.52 x 10⁻¹⁰ m.
B. the minimum uncertainty in the position of a mobile E.coli cell with the given speed is 4.92 x 10⁻⁹ m.
The Uncertainty Principle states that it is impossible to measure the exact position and momentum of a particle simultaneously. Therefore, there is always an inherent uncertainty in the measurements that we take.
Let's calculate the minimum uncertainty in the position of an electron in a hydrogen atom and a mobile E.coli cell.
a) The energy associated with the speed of an electron in a hydrogen atom is given by E = (1/2)mv², where m is the mass of the electron and v is its velocity. The uncertainty principle is ΔxΔp ≥ h/4π, where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant.
Since we are looking for the minimum uncertainty in position, we can set Δp equal to the uncertainty in momentum associated with the speed of the electron.
Δp = mv = (9.11 x 10⁻³¹ kg)(2.19 x 10⁶ m/s)
Δp = 1.99 x 10⁻²⁴ kg·m/s
Now we can solve for Δx.
Δx ≥ h/4π
Δx ≥ (6.63 x 10⁻³⁴ J·s)/(4π)(1.99 x 10⁻²⁴ kg·m/s)
Δx ≥ 2.52 x 10⁻¹⁰ m
Therefore, the minimum uncertainty in the position of an electron in a hydrogen atom with the given speed is 2.52 x 10⁻¹⁰ m.
b) Using the same formula as before,
Δp = mv = (5 x 10⁻¹⁶ kg)(2.19 x 10⁻⁶ m/s)
Δp = 1.095 x 10⁻²⁰ kg·m/s
Δx ≥ h/4π
Δx ≥ (6.63 x 10⁻³⁴ J·s)/(4π)(1.095 x 10⁻²⁰ kg·m/s)
Δx ≥ 4.92 x 10⁻⁹ m
Therefore, the minimum uncertainty in the position of a mobile E.coli cell with the given speed is 4.92 x 10⁻⁹ m.
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A canoe has a velocity of 0.430 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.760 m/s east relative to the earth. (Figure 1) Find the magnitude of the velocity
v
CR
of the canoe relative to the river. Express your answer in meters per second. 2 Incorrect; Try Again; 3 attempts remaining
The magnitude of the velocity vCR of the canoe relative to the river is approximately 0.665 m/s.
To find the magnitude of the velocity vCR of the canoe relative to the river, we can use vector addition.
The velocity of the canoe relative to the earth is given as 0.430 m/s southeast, and the velocity of the river relative to the earth is given as 0.760 m/s east.
To find the velocity of the canoe relative to the river, we subtract the velocity of the river from the velocity of the canoe:
vCR = vCE - vRE
where vCE is the velocity of the canoe relative to the earth and vRE is the velocity of the river relative to the earth.
Given:
vCE = 0.430 m/s southeast
vRE = 0.760 m/s east
To perform vector subtraction, we need to resolve the velocities into their respective components. Let's consider the x-axis as east and the y-axis as north.
The velocity of the canoe relative to the earth (vCE) has two components:
vCE,x = 0.430 m/s * cos(45°)
vCE,y = 0.430 m/s * sin(45°)
The velocity of the river relative to the earth (vRE) only has an eastward component:
vRE,x = 0.760 m/s
Now, we can subtract the components to find the velocity of the canoe relative to the river:
vCR,x = vCE,x - vRE,x
vCR,y = vCE,y
To find the magnitude of vCR, we use the Pythagorean theorem:
|vCR| = sqrt(vCR,x^2 + vCR,y^2)
Substituting the given values:
vCR,x = 0.430 m/s * cos(45°) - 0.760 m/s
vCR,y = 0.430 m/s * sin(45°)
|vCR| = sqrt((0.430 m/s * cos(45°) - 0.760 m/s)^2 + (0.430 m/s * sin(45°))^2)
|vCR| ≈ 0.665 m/s
Therefore, the magnitude of the velocity vCR of the canoe relative to the river is approximately 0.665 m/s.
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A woman starts to swim directly across a 6.8−k k-wide twee, Her speed with respect to the water is 2.1 m/s. The river current camies the woman downstrenn at a spees of 0.91 mis. taj How much time does it take her to cross the river? 34 (b) How far downstieam will the river carry her by the time the reaches the other side of the river?
we can infer that the woman wants to swim across the river of 6.8 km wide and her speed with respect to the water is 2.1 m/s.
The river current carries her downstreem at a speed of 0.91 m/s.
How much time does it take her to cross the river.
To find the time it will take her to cross the river, we can use the following formula:
[tex]$$time = \frac{distance}{speed}$$[/tex]
Since we want to find the time, then we rearrange the formula as follows:
[tex]$$time = \frac{distance}{speed} = \frac{6.8 km}{2.1 m/s} = 3238.1 s$$[/tex]
it takes her approximately 3238.1 seconds to cross the river.
How far downstream will the river carry her by the time she reaches the other side of the river.
To determine how far downstream the river will carry her, we can use the following formula:
[tex]$$distance = speed * time$$[/tex]
where speed is the speed of the river current and time is the time it took her to cross the river.
From the above, we know that the speed of the river current is 0.91 m/s, and the time it takes her to cross the river is 3238.1 s.
we can find the distance the river carried her as follows:
[tex]$$distance = speed * time = 0.91 m/s * 3238.1 s = 2944.7 m$$[/tex]
the river will carry her 2944.7 m downstream by the time she reaches the other side of the river.
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Water flows through a commercial steel pipe with a diameter of 50 cm. If the volumetric flow rate is 0.45 m3.s-1, determine the average velocity in m/s
Therefore, the average velocity of the water flowing through the pipe is 1.81 m/s. The diameter of the commercial steel pipe, D = 50 cm .
= 0.5 m
The volumetric flow rate, Q = 0.45 m³/s
Formula used to find the average velocity in a pipe is
:Average velocity, v = Q / (πD² / 4)
Substitute the values in the above formula, we get
Average velocity, v = Q / (πD² / 4)
v = (0.45) / (π(0.5)² / 4)
we getv = 0.45 * 4 / (π * 0.5²)v = 0.45 * 4 / (π * 0.25)
v = 1.81 m/s
Therefore, the average velocity of the water flowing through the pipe is 1.81 m/s
Water flows through a commercial steel pipe with a diameter of 50 cm. The volumetric flow rate is 0.45 m3/s. The formula to find the average velocity in a pipe is
v = Q / (πD² / 4).
We have to find the average velocity in m/s.To find the average velocity we substitute the given values in the formula, so the equation becomes
v = Q / (πD² / 4).
v = (0.45) / (π(0.5)² / 4)
By simplifying the equation, v = 0.45 * 4 / (π * 0.5²) and then
v = 0.45 * 4 / (π * 0.25)
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am pushing a 20 kg box up a 3m ramp. If the box starts at rest and takes 2.1s to reach the top of the ramp, what is the coefficient of friction if the ramp has an angle of 28 degrees?
The coefficient of friction between the box and the ramp is approximately 0.531.
To find the coefficient of friction, we need to consider the forces acting on the box as it moves up the ramp. Let's break down the forces involved:
Gravitational force (weight):The weight of the box can be calculated using the formula: weight = mass * gravity.
Given the mass of the box is 20 kg, and the acceleration due to gravity is approximately 9.8 m/s², the weight of the box is: weight = 20 kg * 9.8 m/s² = 196 N.
Normal force:The normal force is the perpendicular force exerted by the ramp on the box, which counteracts the weight of the box. The normal force can be calculated using: normal force = weight * cos(angle).
Given the angle of the ramp is 28 degrees, the normal force is: normal force = 196 N * cos(28°).
Frictional force:The frictional force can be calculated using the equation: frictional force = coefficient of friction * normal force.
When the box is on the verge of reaching the top of the ramp, the frictional force will be equal to the force component along the ramp, which is the weight of the box multiplied by the sine of the angle. So we have: frictional force = weight * sin(angle).
Since the box starts from rest and reaches the top of the ramp in 2.1 seconds, we can assume uniform acceleration during this time. We can use the following kinematic equation to relate the forces and motion:
force - frictional force = mass * acceleration.
Now let's plug in the values and solve for the coefficient of friction:
force - frictional force = mass * acceleration
weight * sin(angle) - coefficient of friction * normal force = mass * acceleration
weight * sin(angle) - coefficient of friction * weight * cos(angle) = mass * acceleration
Substituting the known values:
196 N * sin(28°) - coefficient of friction * 196 N * cos(28°) = 20 kg * acceleration
Now we can solve for the coefficient of friction:
coefficient of friction = [196 N * sin(28°)] / [196 N * cos(28°)]
coefficient of friction = tan(28°)
Using a calculator, we find that the coefficient of friction is approximately 0.531.
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6. Calculate the potential temperature of air that at pressure 50mb has temperature T=250 K
Calculate the potential temperature of air at 50mb and 250 K using the formula: potential temperature = temperature * (Reference Pressure / Current Pressure)^(R/Cp).the potential temperature of the air at a pressure of 50mb and a temperature of 250 K is approximately 369.25 K.
To calculate the potential temperature of air at a pressure of 50mb and a temperature of 250 K, you can use the formula for potential temperature:
Potential temperature = Temperature * (Reference Pressure / Current Pressure)^(R/Cp)
In this case, the reference pressure is typically taken as 1000mb, the gas constant for dry air (R) is approximately 287 J/(kg·K), and the specific heat at constant pressure (Cp) is approximately 1005 J/(kg·K).
Plugging in the values, we get:
Potential temperature = 250 K * (1000 mb / 50 mb)^(287/1005)
Simplifying the calculation:
Potential temperature = 250 K * 20^(0.2856)
Using a calculator, we can find that 20^(0.2856) is approximately 1.477.
So, the potential temperature of the air is:
Potential temperature = 250 K * 1.477
Calculating this, we find that the potential temperature of the air is approximately 369.25 K.
Therefore, the potential temperature of the air at a pressure of 50mb and a temperature of 250 K is approximately 369.25 K.
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Four point charges q are placed at the corners of a square of side a. - Find the magnitude of the total Coulomb force F on each of the charges. kq ^{2} √2a ^{2}
ka ^{2} (1/2+ √2 1/a ^{2} ka ^{2} (√3 1/a ^{2} kq ^{2} /(2a ^{2} ) √3
Given, Four point charges q are placed at the corners of a square of side a. We have to find the magnitude of the total Coulomb force F on each of the charges.
Solution: The force on any charge is given by Coulomb's law as: F = kqq0 / r², where q and q0 are the magnitudes of the two charges, k is Coulomb's constant, and r is the distance between the two charges.
The figure below shows the force on charge q1 due to the other three charges q2, q3, and q4.As the square is symmetric about its center, the net force on charge q1 due to charges q2 and q4 is along the diagonal of the square. Also, the magnitudes of the force on charge q1 due to charges q2 and q4 are the same, and are given by:
F1 = kq²/ (2a²) × (1/2 + 1/√2)
= kq² (1 + √2)/ (4a²)
Similarly, the magnitudes of the force on charge q1 due to charges q3 and q2 are also the same, and are given by:
F2 = kq²/ (2a²) × (√3/2)
= kq²√3/ (4a²)
Therefore, the total force on charge q1 is:
F total = √[F1² + (F2 + F2)²]
= kq²/ (2a²) × √3
We know that there are four charges, so the magnitude of the force on each charge is:
F = Ftotal/4
= kq²/ (8a²) × √3
The required magnitude of the total Coulomb force F on each of the charges is kq²/ (8a²) × √3, which is the same for each charge.
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what is the maximum allowable length for float glass jalousies
The maximum allowable length for float glass jalousies is 150 cm.
What is Float Glass?
Float glass refers to a type of glass that is made by floating molten glass on a bed of molten metal. Float glass is a popular type of glass because it is both strong and durable.
Float glass is often used in windows, mirrors, and other applications where high-quality glass is required.
What is a Jalousie?
A jalousie is a type of window that consists of a series of parallel glass panes that are set in a frame. Jalousies are typically designed to allow air to flow through them, which makes them ideal for use in hot climates where ventilation is important.
What is the maximum allowable length for float glass jalousies?
The maximum allowable length for float glass jalousies is 150 cm. This means that any jalousies that are made from float glass cannot exceed this length. This limit is in place to ensure that the glass is strong enough to support itself and to prevent it from breaking under its own weight.
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In a vacuum, two particles have charges of q 1and q 2 , where q 1 =+3.6μC. They are separated by a distance of 0.24 m, and particle 1 experiences an attractive force of 3.5 N. What is the value of q 2, with its sign? Number Units
The value of q2 is approximately -2.38 μC. The negative sign indicates that q2 has an opposite charge to q1. This is determined by using Coulomb's law, considering the given attractive force of 3.5 N between the particles with a separation distance of 0.24 m.
To find the value of q2, we can use Coulomb's law, which states that the force between two charged particles is given by:
F = (k * |q1 * q2|) / r^2
Where F is the force, k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between them.
Given:
q1 = +3.6 μC = 3.6 x 10^-6 C
F = 3.5 N
r = 0.24 m
Rearranging the equation, we can solve for q2:
q2 = (F * r^2) / (k * |q1|)
Plugging in the values:
q2 = (3.5 N * (0.24 m)^2) / (8.99 x 10^9 N m^2/C^2 * |3.6 x 10^-6 C|)
After performing the calculation, we find that the value of q2 is approximately -2.38 μC. The negative sign indicates an opposite charge to q1.
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a−2.50nC charge and is 2.00 cm to the right of B. Find the magnitude (in N) and direction of the net electric force on each of the beads. Net Force on A magnitude force? Add the forces as vectors to get the net force. Be careful with units and signs. N direction Net Force on B magnitude N direction Net Force on C magnitude N direction
The net force on A has a magnitude of 312.5 N. The direction of the net force is right.
The net force on B has a magnitude of 830.6 N. The direction of the net force is right.
The net force on C has a magnitude of 250 N. The direction of the net force is left.
Net force on A:
The direction of force on A due to the 4.00 nC and 2.50 nC charge is in the left direction.
Force F1 on A due to the charge at C.
Force F1 = kq₁q₂ / r²
F1 = (9 × 10⁹ Nm²/C²) * (4.00 × 10⁻⁹ C) * (2.50 × 10⁻⁹ C) / (0.06 m)²
F1 = 250 N
Force F2 on A due to the charge at B.
Force F2 = kq₁q₂ / r²
F2 = (9 × 10⁹ Nm²/C²) * (4.00 × 10⁻⁹ C) * (2.50 × 10⁻⁹ C) / (0.02 m)²
F2 = 562.5 N
Net force on A is
Fnet = F2 - F1
Fnet = 562.5 N - 250 N
Fnet = 312.5 N in the right direction
Net force on B:
Force F1 on B due to the charge at A.
Force F1 = kq₁q₂ / r²
F1 = (9 × 10⁹ Nm²/C²) * (2.00 × 10⁻⁹ C) * (4.00 × 10⁻⁹ C) / (0.02 m)²
F1 = 900 N
Force F2 on B due to the charge at C.
Force F2 = kq₁q₂ / r²
F2 = (9 × 10⁹ Nm²/C²) * (2.50 × 10⁻⁹ C) * (4.00 × 10⁻⁹ C) / (0.06 m)²
F2 = 69.4 N
The direction of force on B due to the 4.00 nC and 2.50 nC charge is in the right direction.
Net force on B is
Fnet = F1 - F2
Fnet = 900 N - 69.4 N
Fnet = 830.6 N in the right direction
Net force on C:
Force F1 on C due to the charge at B.
Force F1 = kq₁q₂ / r²
F1 = (9 × 10⁹ Nm²/C²) * (4.00 × 10⁻⁹ C) * (2.50 × 10⁻⁹ C) / (0.06 m)²
F1 = 250 N
The direction of force on C due to the 4.00 nC and 2.50 nC charge is in the left direction.
Net force on C is
Fnet = F1
Fnet = 250 N in the left direction
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Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 73.7 N, Jill pulls with 95.3 N in a direction 45
∘
to the left, and Jane pulls in a direction 45
∘
to the right with 103 N. (Since the donkey is involved with such uncoordinated people, who can blame it for being stubborn?) Determine the magnitude F of the net force the people exert on the donkey. F= What is the direction θ of the net force? Let 0
∗
define straight ahead, with positive angles to the left and negative angles to the right. Express θ as an angle with a magnitude between 0
∘
θ= and 90
∘
The magnitude is 143.9 N and the direction θ of the net force is 63.4∘.
We can determine the net force acting on the donkey by using the law of vector addition. The vector components of the force of Jill and Jane are:
fx = 95.3cos(45∘) - 103cos(45∘) = -3.04 N
fy = 95.3sin(45∘) + 103sin(45∘) = 141.5 N
The net force can then be obtained as:
F = sqrt[(73.7 N + (-3.04 N))^2 + (0 + 141.5 N)^2]
F = 143.9 N
The magnitude of the net force is 143.9 N.
The angle θ that the net force makes with the positive x-axis is given by:
θ = tan⁻¹(141.5 N / 70.66 N)
θ = 63.4∘
The direction of the net force is to the left with a magnitude between 0∘ and 90∘.
Therefore, F = 143.9 N and θ = 63.4∘.
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\& kcal (b) How long (in s) does this take if the rate of heat transfer is 600 W(1 watt =1 joule/second (1 W=1 J/s)) ? s
It takes 8368.33 seconds (s) to transfer 1200 kcal of heat with a rate of heat transfer of 600 W.
Calculate the time it takes to transfer 1200 kcal of heat with a rate of heat transfer of 600 W, we need to convert kcal to joules and then use the formula:
Time = Heat / Rate of Heat Transfer
Rate of heat transfer = 600 W
Heat = 1200 kcal
Convert kcal to joules:
1 kcal = 4184 J
So, 1200 kcal = 1200 * 4184 J ≈ 5.021 * 10^6 J
We can calculate the time:
Time = Heat / Rate of Heat Transfer
Time = (5.021 * [tex]10^6[/tex] J) / 600 W
Time ≈ 8368.33 s
The time it takes to transfer 1200 kcal of heat with a rate of 600 W, we convert kcal to joules (1200 kcal ≈ 5.021 * [tex]10^6[/tex] J).
Then, we divide the heat by the rate of heat transfer (5.021 * [tex]10^6[/tex] J / 600 W) to get the time in seconds.
The calculation gives us 8368.33 seconds. This means that it takes around 8368.33 seconds for the heat transfer to occur at a rate of 600 W.
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A 580−kg car is traveling with a speed of 25.0 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in 50.0 m ?
The magnitude of the horizontal net force required to bring the car to a halt is 3625 Newtons. by using the equations of motion.
Let's see the calculation
The initial velocity of the car, u, is 25.0 m/s, and the final velocity, v, is 0 m/s since the car comes to a halt.
The displacement, s, is 50.0 m.
We can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the given values, we have:
a = (0^2 - 25.0^2) / (2 * 50.0)
a = (-625) / 100
a = -6.25 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which is necessary to bring the car to a halt.
Now, we can calculate the magnitude of the net force using Newton's second law:
F = m * a
Where:
F = net force
m = mass
a = acceleration
Substituting the given values, we have:
F = 580 kg * (-6.25 m/s^2)
F = -3625 N
Therefore, the magnitude of the horizontal net force required to bring the car to a halt is 3625 Newtons.
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A solenoid with 445 turns has a length of 7.50 cm and a cross-sectional area of 2.70×10 ^{−9} m ^{2}. Find the solenoid's inductance and the average emf around the solenoid if the current changes from +2.50 A to −2.50 A in 7.83×10 ^{−3}s.
(a) the solenoid's inductance (in H )
(b) the average emf around the solenoid (in V)
the average emf around the solenoid is 0.68 V.
(a) The solenoid's inductanceThe formula for inductance is given byL = μ₀n²A / l
Where:L is the inductance of the solenoid
μ₀ is the permeability constant of free space =[tex]4π x 10^-7TmA^-2n[/tex] is the number of turnsA is the cross-sectional area of the solenoid
l is the length of the solenoidSubstitute the given values to get:L = [tex]4π x 10^-7 x (445)² x (2.70×10 ^{−9}) / (7.50 x 10^-2)L = 1.06 x 10^-3 H[/tex]
Therefore, the solenoid's inductance is 1.06 x 10^-3 H.(b) The average emf around the solenoid
The formula for average emf is given byemf = L Δi / Δt
Where:Δi = change in current = 2.50 A + 2.50 A = 5.00 AΔt = 7.83×[tex]10 ^{−3}s[/tex]
Substitute the given values to get:emf = [tex](1.06 x 10^-3) x 5.00 / (7.83×10 ^{−3})emf = 0.68 V[/tex]
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Atmospheric pressure air at 100°C enters a 0.04m diameter tube.
and 2 m long with a speed of 9 m/s. A 1 kW electric heater
coiled on the outer surface of the tube provides a flow of heat
uniform to the tube. Determine: (a) the medical air flow rate, (b) the
air outlet temperature and (c) the tube wall temperature at
the exit
To solve this problem, we can use the principles of conservation of mass and energy.
(a) The mass flow rate of the air can be calculated using the formula:
mass flow rate = density * velocity * cross-sectional area
First, we need to determine the density of the air at 100°C. We can use the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Assuming atmospheric pressure at 100°C, we can calculate the density (ρ) using the equation:
ρ = P / (RT)
Substituting the values into the equation, we can calculate the density.
Once we have the density, we can calculate the mass flow rate using the given velocity and cross-sectional area.
(b) To determine the air outlet temperature, we can use the energy conservation equation:
mass flow rate * specific heat capacity * (T_out - T_in) = heat input
We know the mass flow rate from part (a), and the specific heat capacity of air can be looked up or assumed. The heat input is given as 1 kW.
Solving for T_out will give us the air outlet temperature.
(c) To determine the tube wall temperature at the exit, we need to consider the heat transfer from the heater to the air and the heat transfer from the air to the tube wall. This will depend on the thermal conductivity and the convective heat transfer coefficients.
Additional information about the thermal conductivity and convective heat transfer coefficients is needed to calculate the tube wall temperature accurately.
It's important to note that this problem requires more specific information about the properties of the tube, such as thermal conductivity and convective heat transfer coefficients, to provide an accurate solution.
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Two charges are located along the x-axis. One has a charge of 6μC, and the second has a charge of −3.1μC. If the electrical potential energy associated with the pair of charges is −0.041 J, what is the distance between the charges? The value of the Coulomb constant is 8.98756×10
9
N⋅m
2
/C
2
and the acceleration due to gravity is 9.81 m/s
2
. Answer in units of m. 01610.0 points In Rutherford's famous scattering experiments (which led to the planetary model of the atom), alpha particles (having charges of +2e and masses of 6.64×10
−27
kg) were fired toward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at 1.67×10
7
m/s directly toward the gold nucleus. How close does the alpha particle get to the gold nucleus before turning around? Assume the gold nucleus remains stationary. The fundamental charge is 1.602×10
−19
C and the Coulomb constant is 8.98755×10
9
N⋅m
2
/C
2
. Answer in units of m.
The alpha particle gets as close as 0.118 pm to the gold nucleus before turning around.
Let the distance between the two charges be d. The electrical potential energy associated with the pair of charges can be expressed as:
U = (1/4πε) (q₁q₂ / d),
where ε is the permittivity of free space and q₁ and q₂ are the charges. So, the electrical potential energy of the pair of charges can be expressed as:
U = (1/4πε) (q₁q₂ / d).
Taking the values of the given terms and substituting, we get:
-0.041 = (1/4π(8.98756×10⁹)) [(6×10⁻⁶) (-3.1×10⁻⁶)] / d.
Therefore, d = 0.00849 m or 8.49 mm (rounded to two decimal places).
Given values for the constants and masses can be used to calculate the distance between the alpha particle and the gold nucleus as follows. Consider the electrostatic force acting between two charges:
Fe = k (q₁q₂) / r²,
where k is the Coulomb constant, q₁ and q₂ are the charges, and r is the separation distance between the charges.
If there is no net force acting on the alpha particle, then its kinetic energy will be converted to potential energy as it is pushed towards the gold nucleus. This potential energy can be calculated as follows:
U = k (q₁q₂) / r,
where U is the potential energy, k is the Coulomb constant, q₁ and q₂ are the charges, and r is the separation distance between the charges. The kinetic energy of the alpha particle is given by:
(1/2)mv²,
where m is the mass of the alpha particle and v is the initial velocity of the alpha particle.
The maximum separation distance between the alpha particle and the gold nucleus is the point at which the kinetic energy of the alpha particle is converted to potential energy, and the particle's velocity is zero. This means that the initial kinetic energy of the alpha particle is equal to the final potential energy at maximum separation. This can be expressed as:
(1/2)mv² = k (q₁q₂) / r,
where r is the maximum separation distance. Rearranging, we get:
r = k (q₁q₂) / (mv²).
Given the values for k, q₁, q₂, m, and v, we get:
r = (8.98755×10⁹) (2(1.602×10⁻¹⁹) (79(1.602×10⁻¹⁹))) / (6.64×10⁻²⁷ (1.67×10⁷)²).
Simplifying, we get:
r = 1.18×10⁻¹³ m or 0.118 pm (rounded to three decimal places).
Therefore, the alpha particle gets as close as 0.118 pm to the gold nucleus before turning around.
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Despite having almost unlimited gian, what is the limiting factor of how high an op-amp's voltage can actually go?
The limiting factor of how high an op-amp's voltage can actually go is determined by the power supply voltage. It is important to stay within the specified voltage range to ensure accurate amplification and avoid damage to the op-amp.
The limiting factor of how high an op-amp's voltage can actually go is determined by the power supply voltage.
Op-amps, or operational amplifiers, are electronic devices used in circuits to amplify signals. They have a specified voltage range within which they can operate effectively.
The power supply voltage provides the maximum voltage that the op-amp can handle. If the input signal exceeds this voltage, the op-amp will not be able to accurately amplify the signal and may even be damaged.
For example, let's say we have an op-amp with a power supply voltage of ±15 volts. This means that the maximum voltage the op-amp can handle is 15 volts in the positive direction and 15 volts in the negative direction. If the input signal exceeds these voltage limits, the op-amp will not be able to accurately amplify the signal.
In addition to the power supply voltage, other factors such as the op-amp's internal circuitry and the quality of the components used can also affect its performance and maximum voltage handling capability. However, the power supply voltage is the primary limiting factor in determining how high an op-amp's voltage can actually go.
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