The linear charge density is 2.01 × 10^(-7) C/m.Part A The electric field produced by the rod with a uniform linear charge density of λ at a distance r from the rod is given by:
E = λ / (2πε₀r)where,ε₀ is the permittivity of free space
E = 5.6 μC/m / (2πε₀ * 0.089m)
E = 100 N/C
Therefore, the magnitude of the electron's initial acceleration is given by,a = F / mwh
ere,F = q
E = 1.6 × 10^(-19) × 100
F = 1.6 × 10^(-17) Nm
= 9.11 × 10^(-31) kg Therefore,
a = (1.6 × 10^(-17) N) / (9.11 × 10^(-31) kg)
a = 1.76 × 10^14 m/s²Part BThe field produced by an infinite line of charge with a linear charge density λ at a distance r from the line is given by:
E = λ / (2πε₀r)where,ε₀ is the permittivity of free space Given that,
E = 5.1 × 10^4 N/ Cr = 2.5m
Therefore,5.1 × 10^4 = λ / (2πε₀ * 2.5)λ = 2πε₀ * 2.5 * 5.1 × 10^4λ
= 2.01 × 10^(-7) C/m.
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A golf ball is dropped from rest from a height of 8.50 m. It hits the pavement, then bounces back up, rising just 6.40 m before falling back down again. A boy then catches the ball when it is 1.50 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.
The total amount of time that the ball is in the air, from drop to catch is 3.66 s
Ignoring air resistance, the total amount of time that the ball is in the air, from drop to catch can be calculated as follows:
First, we can calculate the time taken by the golf ball to reach the pavement by using the formula;
s = (1/2) gt²
where s is the distance,
g is acceleration due to gravity,
and t is time taken.
In this case, s = 8.5 m and g = 9.8 m/s².
Therefore, t = √(2s/g)= √(2×8.5/9.8) = √1.734 = 1.32 s.
Second, we can calculate the time taken by the golf ball to rise up to a height of 6.40 m.
Since the motion is symmetrical we can use the same time t as obtained above.
Using the same formula, s = (1/2) gt² where s = 6.40 m and g = 9.8 m/s².
Therefore, t = √(2s/g) =√(2×6.4/9.8) = √1.04 = 1.02 s
The total amount of time that the ball is in the air can be calculated as;
total time = t + t + t = 1.32 + 1.02 + 1.32 = 3.66 s.
Therefore, the total amount of time that the ball is in the air, from drop to catch is 3.66 s.
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A woman is driving her van with speed 50.0mi/h on a horizontal stretch of road. (a) When the road is wet, the coefficient of static friction between the road and the tires is 0.102. Find the minimum stopping distance (in m). m (b) When the road is dry, μs=0.595. Find the minimum stopping distance (in m ). m
The minimum stopping distance on a wet road at a speed of 50.0 mi/h is calculated to be 2035.56 m, while on a dry road it is calculated to be 1359.56 m.
(a) Wet RoadA woman is driving her van with speed 50.0 mi/h on a horizontal stretch of wet road. The coefficient of static friction between the road and the tires is 0.102.
The formula for minimum stopping distance (wet road) is given by: d = (v²/2gμ) + v²/2a
Where
v = initial velocity = 50 miles/hour = (22/15)*50 m/s = 73.33 m/s
μ = coefficient of static friction = 0.102
g = acceleration due to gravity = 9.81 m/s²
a = acceleration = gμ = (9.81)(0.102) = 1.00062 m/s²
Substituting the values in the formula,
d = (73.33²/2*9.81*0.102) + 73.33²/2*1.00062= 52.37 + 1983.19= 2035.56 m
(b) Dry RoadWhen the road is dry, the coefficient of static friction between the road and the tires is 0.595.
The formula for minimum stopping distance (dry road) is given by: d = (v²/2gμ) + v²/2a
Where
v = initial velocity = 50 miles/hour = (22/15)*50 m/s = 73.33 m/s
μ = coefficient of static friction = 0.595
g = acceleration due to gravity = 9.81 m/s²
a = acceleration = gμ = (9.81)(0.595) = 5.83995 m/s²
Substituting the values in the formula,
d = (73.33²/2*9.81*0.595) + 73.33²/2*5.83995= 15.28 + 1344.28= 1359.56 m
Thus, the minimum stopping distance (in m) when the road is wet is 2035.56m and when the road is dry is 1359.56m.
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The gravitational force on a body located at distance R from the center of a uniform spherical mass is due solely to the mass lying at distance r≤R, measured from the center of the sphere. This mass exerts a force as if it were a point mass at the origin. (a) Use the above result to show that if you drill a hole through the Earth and then fall in, you will execute simple harmonic emotion about the Earth's center. Find the time it takes you to return to your point of departure and show that this is the time needed for a satellite to circle the Earth in a low orbit with r∼R
⊕
, the radius of the Earth. You may treat the Earth as a uniformly dense sphere, neglect friction and any effects due to the Earth's rotation. (10 points) (b) Show that you will also execute simple harmonic motion with the same period even if the straight hole passes far from the Earth's center.
If you drill a hole through the Earth and fall in, you will execute simple harmonic motion around the Earth's center.
If you consider a straight hole drilled through the Earth, it can be concluded that you will perform simple harmonic motion even if the straight hole passes far from the Earth's center. The motion is such that when a mass is released, it falls to the center of the Earth, overshoots, and oscillates back and forth, executing simple harmonic motion. This is possible because the gravitational force on a body located at distance R from the center of a uniform spherical mass is due solely to the mass lying at a distance r≤R, measured from the center of the sphere.
So, a simple harmonic motion can be executed about the Earth's center. The time taken by an object to complete one revolution around the Earth is given by the time taken by a satellite to circle the Earth in a low orbit with r ∼ R (the radius of the Earth). Thus, the time taken by the object to return to its point of departure is given by the time taken by the satellite to circle the Earth in a low orbit.
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Vector
A
has x and y components of −8.70 cm and 15.0 cm, respectively. Vector
B
has x and y components of 13.2 cm and −6.60 cm, respectively. If
A
−
B
+3
C
=0, what are the components of
C
?
The components of vector C are Cx = 7.3 cm and Cy = -7.2 cm.
To find the components of vector C, we can rearrange the given equation:
A - B + 3C = 0
Let's substitute the components of vectors A and B:
(Ax, Ay) - (Bx, By) + 3(Cx, Cy) = (0, 0)
Given:
Ax = -8.70 cm
Ay = 15.0 cm
Bx = 13.2 cm
By = -6.60 cm
Substituting these values into the equation, we have:
(-8.70 cm, 15.0 cm) - (13.2 cm, -6.60 cm) + 3(Cx, Cy) = (0, 0)
To simplify the equation, we can subtract vector B from vector A:
(-8.70 cm - 13.2 cm, 15.0 cm - (-6.60 cm)) + 3(Cx, Cy) = (0, 0)
Simplifying further, we have:
(-21.9 cm, 21.6 cm) + 3(Cx, Cy) = (0, 0)
Since the sum of two vectors is equal to zero, their components must be equal:
-21.9 cm + 3Cx = 0 (equation 1)
21.6 cm + 3Cy = 0 (equation 2)
Now we can solve these two equations simultaneously to find the components of vector C.
From equation 1:
3Cx = 21.9 cm
Cx = 7.3 cm
From equation 2:
3Cy = -21.6 cm
Cy = -7.2 cm
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The actual question is:
Vector A has x and y components of −8.70 cm and 15.0 cm, respectively. Vector B has x and y components of 13.2 cm and −6.60 cm, respectively.
If A−B+3C =0,
What are the components of C?
1. A block of mass 3.14 kg lies on a frictionless horizontal surface. The block is connected by a cord passing over a pulley to another block of mass 6.75 kg which hangs in the air, as shown. Assume the cord to be light (massless and weightless) and unstretchable and the pulley to have no friction and no rotational inertia. Calculate the acceleration of the first block. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s 2 .
Calculate the tension in the cord. Answer in units of N.
2. The block of mass 3.80061 kg has an acceleration of 3.3 m/s 2 as shown. What is the magnitude of F? Assume the acceleration due to gravity is 9.8 m/s 2 and the surface is frictionless. Answer in units of N.
3. A light, inextensible cord passes over a light, frictionless pulley with a radius of 9.7 cm. It has a(n) 24 kg mass on the left and a(n) 5.3 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 4.7 m apart. The acceleration of gravity is 9.8 m/s 2 . At what rate are the two masses accelerating when they pass each other? Answer in units of m/s 2 .
1. The tension in the cord is approximately 3.11 N. 2. The magnitude of force (F) acting on the block is approximately 12.54013 N. 3. The rate at which the two masses are accelerating when they pass each other is approximately 475.26 m/s^2.
1. To calculate the acceleration of the first block and the tension in the cord, we can use Newton's second law of motion and the concept of tension in a system.
Let's denote:
m1 = mass of the first block = 3.14 kg
m2 = mass of the hanging block = 6.75 kg
a = acceleration of the system (common acceleration)
T = tension in the cord
Using Newton's second law for both blocks:
m1 * a = T (equation 1)
m2 * g - T = m2 * a (equation 2)
Solving the equations simultaneously, we can find the values of acceleration (a) and tension (T):
From equation 1, we have T = m1 * a
Substituting this value into equation 2:
m2 * g - m1 * a = m2 * a
g = (m1 + m2) * a
a = g / (m1 + m2)
Substituting the given values:
a = 9.8 m/s^2 / (3.14 kg + 6.75 kg)
a ≈ 9.8 m/s^2 / 9.89 kg
a ≈ 0.99 m/s^2
Therefore, the acceleration of the first block is approximately 0.99 m/s^2.
To calculate the tension in the cord, we can substitute the value of acceleration (a) into equation 1:
T = m1 * a
T = 3.14 kg * 0.99 m/s^2
T ≈ 3.11 N
Therefore, the tension in the cord is approximately 3.11 N.
2. To determine the magnitude of force (F) acting on the block with a known mass and acceleration, we can again use Newton's second law of motion.
m = mass of the block = 3.80061 kg
a = acceleration of the block = 3.3 m/s^2
g = acceleration due to gravity = 9.8 m/s^2
Using Newton's second law:
F = m * a
Substituting the given values:
F = 3.80061 kg * 3.3 m/s^2
F ≈ 12.54013 N
Therefore, the magnitude of force (F) acting on the block is approximately 12.54013 N.
3. To determine the rate at which the two masses are accelerating when they pass each other, we can use the concept of relative motion and the conservation of mechanical energy.
m1 = mass on the left = 24 kg
m2 = mass on the right = 5.3 kg
r = radius of the pulley = 9.7 cm = 0.097 m
d = initial vertical distance between the center of masses = 4.7 m
g = acceleration due to gravity = 9.8 m/s^2
Using the conservation of mechanical energy:
(m1 + m2) * g * d = (m1 + m2) * a * r
Simplifying the equation:
a = (g * d) / r
Substituting the given values:
a = (9.8 m/s^2 * 4.7 m) / 0.097 m
a ≈ 475.26 m/s^2
Therefore, the rate at which the two masses are accelerating when they pass each other is approximately 475.26 m/s^2.
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Light from a green laser (550.nm) illuminates a grid of thin fibers. A double slit diffraction pattern is projected on a screen 2.0 meters from the fibers. The third bright fringe is 33.4 mm from the central spot. What is the distance between fibers?
The distance between fibers illuminated by a green laser of 550 nm, given that the third bright fringe is 33.4 mm from the central spot and that a double-slit diffraction pattern is projected on a screen 2.0 meters from the fibers is 0.0232 mm (to 3 sig figs).
Given data,λ = 550 nm = 550 × 10⁻⁹ m = 5.50 × 10⁻⁷ m (1 nm = 10⁻⁹ m).The third bright fringe = mλD/d; where m = 3 and D = 2.0 m.third bright fringe = 3 × 5.50 × 10⁻⁷ × 2.0/d; or,33.4 × 10⁻³ = 1.10 × 10⁻⁶/d; ord = 1.10 × 10⁻⁶/33.4 × 10⁻³; ord = 0.0232 mm.Thus, the distance between fibers is 0.0232 mm (to 3 sig figs).
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How do the following vary with time for a simple harmonic oscillator: total mechanical energy, kinetic energy, potential energy?
A simple harmonic oscillator is a system that oscillates or vibrates at a single frequency. As the system moves back and forth, the total mechanical energy, kinetic energy, and potential energy of the system vary with time. When the oscillator is at its maximum displacement from equilibrium, its total mechanical energy is entirely potential energy.
A simple harmonic oscillator is a system that oscillates or vibrates at a single frequency. As the system moves back and forth, the total mechanical energy, kinetic energy, and potential energy of the system vary with time. When the oscillator is at its maximum displacement from equilibrium, its total mechanical energy is entirely potential energy. As the oscillator passes through the equilibrium position, it has the most kinetic energy and the least potential energy. As the oscillator moves to its opposite maximum displacement, its total mechanical energy is entirely kinetic energy.
The total mechanical energy of a simple harmonic oscillator, which is proportional to the square of the amplitude, remains constant and is independent of time. However, the kinetic and potential energies fluctuate with time, as shown in the figure below.
Figure: A graph showing the variation of kinetic energy (K), potential energy (U), and total mechanical energy (E) with time for a simple harmonic oscillator. A simple harmonic oscillator's total mechanical energy is the sum of its kinetic and potential energies, as follows: E = K + U
Because the total mechanical energy is constant, the energy is transferred from kinetic to potential energy and vice versa throughout the cycle. As the oscillator approaches its maximum displacement, potential energy increases while kinetic energy decreases. When the oscillator approaches the equilibrium position, the potential energy is converted to kinetic energy, and the process is reversed on the opposite side.
Thus, the kinetic and potential energies are in opposite phases, and the sum of the two remains constant. Therefore, the total mechanical energy of a simple harmonic oscillator remains constant while the kinetic and potential energies fluctuate with time.
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Consider a new Turing machine that instead of just moving left and right can also jump to the 5 th tape cell in any given transition. So now δ is defined over, δ:Q×Γ→Q×Γ×{L,R,J5} where J5 moves the head to the 5 th tape cell. Prove that this is equivalent to the standard Turing machine.
The extended Turing machine with the transition function δ: Q × Γ → Q × Γ × {L, R, J5} is equivalent to the standard Turing machine.
To prove that the new Turing machine with the extended transition function δ: Q × Γ → Q × Γ × {L, R, J5} is equivalent to the standard Turing machine, we need to show that the extended Turing machine can simulate the behavior of a standard Turing machine, and vice versa.
First, let's consider a standard Turing machine with transition function δ: Q × Γ → Q × Γ × {L, R}.
To simulate the behavior of the standard Turing machine on the extended Turing machine, we can simply ignore the J5 transition in the extended transition function. Whenever the standard Turing machine would perform a transition to the left or right, we use the corresponding L or R transition in the extended Turing machine. This way, we are effectively disregarding the ability to jump to the 5th tape cell.
Now, let's consider the extended Turing machine with transition function δ: Q × Γ → Q × Γ × {L, R, J5}.
To simulate the behavior of the extended Turing machine on the standard Turing machine, we need to show that the J5 transition can be simulated using the L and R transitions. We can achieve this by introducing additional states and tape symbols.
We can modify the extended Turing machine to have an additional state and tape symbol to mark the position of the 5th tape cell. Let's call the new state Q_mark and the new tape symbol 'X'. We update the transition function as follows:
δ'(Q_mark, X) = (Q_mark, X, R) // Stay in Q_mark state and move right
δ'(Q, X) = (Q_mark, X, L) // Transition from state Q to Q_mark and move left
By using these additional states and symbols, we can simulate the J5 transition of the extended Turing machine on the standard Turing machine. Whenever the extended Turing machine performs a J5 transition, we transition to the Q_mark state and move right to the next cell, effectively simulating the jump to the 5th tape cell.
Therefore, we have shown that the new Turing machine with the extended transition function is equivalent to the standard Turing machine by demonstrating how each can simulate the behavior of the other.
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Air at 20°C and at a atmospheric pressure flows over a flat plate at a velocity of 3 m/s. If the plate is 30 cm long and at a temperature of 60°C, calculate: (a) the thickness of velocity and thermal boundary layers at 20 cm. (b) the average heat transfer coefficient. (c) total drag force on the plate, per unit width. Take the following properties of air: = P = 1.18 kg/m³, kinematic viscosity = 17 x 10-6 m²/s, k = 0.0272 W/m-K, Cp = 1.007 kJ/kg K
The thickness of velocity and thermal boundary layers is 0.0567 m and 0.0347 m respectively. The average heat transfer coefficient is 57.11 W/m² K. The total drag force on the plate is 0.05677 N/m.
According to the given problem, the properties of air are: ρ = 1.18 kg/m³, Kinematic viscosity (μ) = 17 × 10⁻⁶ m²/s, Thermal conductivity (k) = 0.0272 W/m-K, Specific heat (Cp) = 1.007 kJ/kg K, Reynolds number (Re)
Re = ρVxδvx / μ
= (1.18 × 3 × 0.2 × 0.017) / 0.000017 = 2222.4
Prandtl number (Pr)
Pr = Cp μ / k
= (1.007 × 0.000017) / 0.0272
= 0.00064
Nusselt number (Nu)
Nu = 0.332 × Re1/2 Pr1/3
= 0.332 × 2222.4 1/2 × 0.00064 1/3
= 73.324
Average heat transfer coefficient:
h = k Nu / δtx
= 0.0272 × 73.324 / 0.0347 = 57.11 W/m² K
The average skin friction coefficient is:
cf = 0.664 / Re1/2 = 0.664 / 2222.4 1/2 = 0.01575
Total drag force per unit width:
Fx = 0.5 ρ Vx³ Cf
L = 0.5 × 1.18 × 3³ × 0.01575 × 0.3
= 0.05677 N/m
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A copper wire has a diameter of 1.422 mm. What magnitude current flows when the drift velocity is 1.54 mm/s ? Take the density of copper to be 8.92×10
3
kg/m
3
.
The magnitude current flows, when the drift velocity is 1.54 mm/s, is 20.3 A.
When the drift velocity is 1.54 mm/s.
We will make use of the formula that relates drift velocity with the current.
We have:
vd = (I / n * A * q )
Where vd is the drift velocity
I is the current
n is the density of free electrons
A is the cross-sectional area of the wire
q is the charge carried by an electron
For copper, the value of n is 8.5 × 1028 electrons/meter cube.
Given that the wire is circular, its cross-sectional area is A = πr2 = πd2/4
where d is the diameter of the wire.
Hence we have:
d = 1.422 mm = 1.422 × 10-3 mA = π/4 x (1.422 × 10-3 m)2= 1.59 x 10-6 m^2
Now we can calculate the current as follows:
I = n * A * q * vd/I = (8.5 x 10^28)(π/4)(1.422 x 10^-3)^2(1.602 x 10^-19)(1.54 x 10^-3)=20.3A
Approximately, the magnitude of the current flowing in the copper wire is 20.3A.
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A 3.0−cm-tall object is 45 cm in front of a diverging mirror that has a −25 cm focal length.
the virtual image formed by the diverging mirror is 1.0 cm tall.
A 3.0−cm-tall object is placed 45 cm in front of a diverging mirror that has a −25 cm focal length. In optics, diverging mirrors are curved mirrors that cause the reflected light to diverge. They have a negative focal length. A diverging mirror is also known as a concave mirror.
It is curved inward and is thicker at the edge than at the center. As the light hits the mirror, the rays diverge. When the object is placed beyond the mirror's focal point, the diverging mirror forms an erect, virtual image that is smaller than the object. A concave mirror's image is always virtual, erect, and smaller than the object placed before it.
The formula for the image distance is 1/f = 1/o + 1/i where o is the object distance, i is the image distance, and f is the focal length.In this scenario, the object distance is given as 45 cm and the focal length as −25 cm, substituting these values into the formula, we have:1/−25 = 1/45 + 1/iSimplifying the above equation gives:i = −75 cmThe image distance is negative, indicating that the image is virtual and formed behind the mirror. The magnification of the image can be calculated by using the formula:M = -i/o
Thus, M = −75/45 = −1.67
The magnification is negative, indicating that the image is inverted. The height of the image can be found by using the formula:h1/h2 = i/o
Simplifying the above equation gives:h2 = h1 * o/i
Where h1 is the height of the object, h2 is the height of the image. Substituting the given values into the above equation gives:h2 = 3.0 * −25/−75 = 1.0 cm
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(a) How fast would a motorist have to be traveling for a yellow (λ=590.00 nm ) traffic light to appear green ( λ=550.00 nm ) because of the Doppler shift? (nm is nanometer and is 10
−9
meters) (b) Should the motorist be traveling toward or away from the traffic light to see this effect? (c) How fast would a motorist have to be traveling for a yellow (λ=590.00 nm) traffic light to appear red (λ=700.00 nm) because of the Doppler shift? Attach File
(a) A motorist has to be traveling with a speed of 1.26×10^7 m/s towards a yellow traffic light of wavelength 590.00 nm for it to appear green (wavelength 550.00 nm) because of the Doppler shift.For yellow light to appear red (700.00 nm), a motorist would have to be moving away from the traffic light at a high speed of 2.36×10^7 m/s, which is about 8.87% of the speed of light.
When an object, in this case, a motorist, is moving towards a traffic light, the apparent wavelength of the light received by the object is shorter than its original wavelength (known as blue-shift). As we know that green light has a shorter wavelength than yellow light, hence the yellow light will appear green to the motorist when he is moving towards it with enough speed. For yellow light to appear green (550.00 nm), a motorist would have to be moving towards the traffic light at a high speed of 1.26×10^7 m/s, which is about 4.74% of the speed of light.
(b) The motorist should be traveling towards the traffic light to observe this effect. (c) A motorist has to be traveling with a speed of 2.36×10^7 m/s away from a yellow traffic light of wavelength 590.00 nm for it to appear red (wavelength 700.00 nm) because of the Doppler shift. When an object, in this case, a motorist, is moving away from a traffic light, the apparent wavelength of the light received by the object is longer than its original wavelength (known as red-shift). As we know that red light has a longer wavelength than yellow light, hence the yellow light will appear red to the motorist when he is moving away from it with enough speed.
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What is the electric field (in N/C) at a point midway between them? N/C (b) What is the force (in N ) on a charge q
3
=23μC situated there?
(a) The electric field at a point midway between the point charges is 3.33 N/C, directed towards the positive charge.
(b) The force on the charge q3=22 μC at the same point is 73.26 N, directed towards the negative charge.
(a) To find the electric field at a point midway between the charges, we can calculate the electric field due to each charge individually and then add them together. The electric field at a point due to a point charge is given by the equation E = kq/[tex]r^2[/tex], where E is the electric field, k is the electrostatic constant (approximately 9 × [tex]10^9[/tex] [tex]Nm^2/C^2[/tex]), q is the charge, and r is the distance from the point charge.
The electric field due to q1 is E1 = ([tex]9 * 10^9 Nm^2/C^2[/tex]) * ([tex]52 * 10^{-6} C[/tex]) / [tex](1.5 m)^2[/tex], which simplifies to approximately 6.52 N/C directed towards q1.
The electric field due to q2 is E2 = ([tex]9 * 10^9 Nm^2/C^2[/tex]) * ([tex]-29 * 10^{-6} C[/tex]) / [tex](1.5 m)^2[/tex], which simplifies to approximately -3.19 N/C directed towards q2.
Adding the electric fields together, we get [tex]E_{total[/tex] = E1 + E2 = 6.52 N/C - 3.19 N/C = 3.33 N/C. The electric field is directed towards the positive charge q1.
(b) To calculate the force on q3, we can use the equation F = qE, where F is the force, q is the charge, and E is the electric field.
The force on q3 is given by F3 = (22 × 10^-6 C) * (3.33 N/C), which simplifies to approximately 73.26 N. The force is directed towards the negative charge q2.
Therefore, at the midpoint between the charges, the electric field is 3.33 N/C directed towards q1, and the force on q3=22 μC is 73.26 N directed towards q2.
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In proton-beam therapy, a high-energy beam of protons is fired at a tumor. The protons come to rest in the tumor, depositing their Kinetic energy and breaking apart the tumor's DNA, thus killing its cells. For one patient, it is desired that (1.3×10∧−1)J of proton energy be deposited in a tumor. To create the proton beam, the protons are accelerated from rest through a (1.300×10∧7)∨ potential difference. What is the total charge of the protons that must be fired at the tumor to deposit the required energy? - No text entered - The correct answer is not displayed for Written Response type questions.
The total charge of protons that must be fired at the tumor to deposit the required energy is 6.444 x [tex]10^14[/tex] Coulombs.The energy of proton to be deposited in tumor = E = 1.3 x [tex]10^(-1)[/tex] J.
The potential difference through which protons are accelerated = V = 1.300 x [tex]10^7[/tex] volts. Charge on a single proton is given as e = 1.6 x [tex]10^(-19)[/tex] Coulombs.
We know that the potential difference is equal to the kinetic energy per unit charge.
Hence, the kinetic energy of the proton can be written as:K.E = qV Where,q = charge on proton V = potential difference.
Now, we need to calculate the number of protons required to deposit the energy of 1.3 x [tex]10^(-1)[/tex] J in the tumor.
The kinetic energy of a single proton is given by:
E = K.E. = (1/2)[tex]mv^2[/tex] Where,m = mass of proton = 1.67 x [tex]10^(-27)[/tex] kg v = velocity of proton.
Therefore,v = sqrt((2E)/m).
Hence,Charge on proton can be written asq =
K.E. / V = [(1/2)][tex]mv^2[/tex] / V = (m[tex]V^2[/tex]) / 2E = [1.67 x [tex]10^(-27)[/tex] kg x (1.300 x [tex]10^7[/tex]volts)^2] / (2 x 1.3 x [tex]10^(-1)[/tex] J) = 2.021 x [tex]10^(-16)[/tex] Coulombs.
Now, to calculate the total charge of protons required to deposit the energy of 1.3 x [tex]10^(-1)[/tex] J, we use the formula:
Charge = (Energy to be deposited) / (Charge on a single proton) = (1.3 x [tex]10^(-1)[/tex] J) / (2.021 x [tex]10^(-16)[/tex] Coulombs) = 6.444 x [tex]10^14[/tex].
Therefore, the total charge of protons that must be fired at the tumor to deposit the required energy is 6.444 x [tex]10^14[/tex] Coulombs.
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f an object of 80.0 kg is hanging from a rope and the tension on the rope is
900 N in upward direction. What is the net force acting on the object? which
direction the man will move?
The answer is that the net force acting on the object is 116 N in the upward direction. The net force acting on an object can be calculated by subtracting the force of gravity from the upward force of tension on the rope. Here, the object of 80.0 kg is hanging from a rope, and the tension on the rope is 900 N in the upward direction.
The force of gravity on the object is given by the formula: F_gravity = m x g; where, m is the mass of the object, and g is the acceleration due to gravity, which is approximately 9.8 m/s². Therefore, the force of gravity on the object is: F_gravity = 80.0 kg x 9.8 m/s² = 784 N
Now, we can calculate the net force acting on the object by subtracting the force of gravity from the tension on the rope:
F_net = tension - F_gravity⇒F_net = 900 N - 784 N = 116 N
Therefore, the net force acting on the object is 116 N in the upward direction. Since the net force is acting in the upward direction, the object will not move in any direction. The man will stay still in the same position.
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If a charge, Q, is located at the center of a spherical volume of radius R
0
=3.65 cm and the total electric flux through the surface of the sphere is Φ
0
=−6.66
C
Nm
2
, what is the total flux through the surface if the radius of the sphere is changed to R=14.1 cm (in
C
Nm
2
).
Therefore, the total flux through the surface if the radius of the sphere is changed to R = 14.1 cm (in C Nm²) is: Phi = \frac{-6.66}{4.95 × 10^{-10}
Phi = -1.35 × 10^7 C Nm^{-2}
The total electric flux through a surface of a sphere when a charge, Q, is located at the center of a spherical volume of radius R₀ is given by the expression;Phi_0 = \frac{Q}{4πε_0R_0^2}
Where;Phi_0 = the total electric flux through the surface of the sphere Q =the charge located at the center of the spherical volume
ε_0 = the permittivity of free spaceR_0 =the initial radius of the spherical volume.
When the radius of the sphere is changed to R, the total electric flux through the surface of the sphere is given by the expression; Phi = \frac{Q}{4πε_0R^2}
Where; Phi = the total electric flux through the surface of the sphere (in C Nm²)Q = the charge located at the center of the spherical volumeε_0 = the permittivity of free space R =the new radius of the spherical volume.
Thus, the total flux through the surface if the radius of the sphere is changed to R = 14.1 cm is given by; Phi = \frac{Q}{4πε_0R^2} Phi = frac{Q}{4π(8.85 × 10^{−12} N^{-1}m^{-2}) (0.141 m)^2} Phi = \frac{Q}{4.95 × 10^{-10}}
Therefore, the total flux through the surface if the radius of the sphere is changed to R = 14.1 cm (in C Nm²) is; Phi = \frac{-6.66}{4.95 × 10^{-10}}
Phi = -1.35 × 10^7 C Nm^{-2}$$
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Problem 2: On the International Space Station an, object with mass m=443 g is attached to a massless string of length L=0.93 m. The string can handle a tension of F
T
=5.92 N before breaking. The object undergoes uniform circular motion, being spun around by the string in a horizontal plane. What is the maximum speed, in meters per second, the mass can have before the string breaks? v=
The maximum speed that the object can have before the string breaks is 3.65 m/s. Answer: 3.65 m/s
The maximum speed the mass can have before the string breaks is 2.225 m/s.
The force acting on the object in this case is tension.
The maximum tension that the string can handle is given as FT=5.92 N. The object is attached to a string of length L=0.93m.
The maximum speed the mass can have before the string breaks is given as:
v = [FT/ m]1/2
Here, FT = maximum tension that the string can handle
m = mass of the object
v = maximum speed the object can have before the string breaks
Substituting the given values, we get:
v = [5.92/0.443]1/2v = [13.35]1/2v = 3.65 m/s
Therefore, the maximum speed that can be attained is 3.65 m/s. Answer: 3.65 m/s
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An object with mass M1 of 2.85 kg is held in place on an inclined plane that makes an angle θ of 40.0∘ with the horizontal (see figure below). The coefficient of kinetic friction between the plane and the object is μk=0.540. A second object that has a mass M2 of 4.75 kg is connected to the first object with a massless string over a massless, frictionless pulley. 1) Calculate the initial acceleration of the system once the objects are released. (Express your answer to three significant figures.) 2) Calculate the tension in the string once the objects are released. (Express your answer to three significant figures.)
The initial acceleration of the system once the objects are released is 2.01 m/s², and the tension in the string once the objects are released is 22.8 N.
In this problem, we will first calculate the acceleration of the system and then the tension in the string. The first object's mass, M1 = 2.85 kg, and it is held in place on an inclined plane that makes an angle θ of 40.0∘ with the horizontal, and the coefficient of kinetic friction between the plane and the object is μk = 0.540. The second object's mass is M2 = 4.75 kg, which is connected to the first object with a massless string over a massless, frictionless pulley. The initial acceleration of the system once the objects are released is 2.01 m/s². The tension in the string once the objects are released is 22.8 N.
Therefore, the initial acceleration of the system once the objects are released is 2.01 m/s², and the tension in the string once the objects are released is 22.8 N.
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x axis, and the positive charge is farther from the line of charge than the negative charge. Find the net force exerted on the dipole.
F
=−0.0588×N
The net force exerted on the dipole can be determined by the formula:
F = 2k(q1q2/d^2)
where:
F is the force
k is the electrostatic constant (k = 9 × 10^9 Nm^2/C^2)
q1 and q2 are the magnitudes of the charges
d is the distance between the charges
In this case, since the positive charge is farther from the line of charge, we consider the positive charge as q1 and the negative charge as q2.
Let's assume that the magnitudes of the charges are q1 and q2, and the distance between them is d.
The net force exerted on the dipole is given as F = -0.0588 N.
We can set up the equation:
-0.0588 = 2k(q1q2/d^2)
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An archer shoots an arrow at a 79.0 m distant target; the bull's-eye of the target is at same height as the release height of the arrow. (a) At what angle in degrees must the arrow be released to hit the bull's-eye if its initial speed is 36.0 m/s ? 0 over under
Therefore, the arrow should be released at an angle of approximately 9.29 degrees to hit the bull's-eye if its initial speed is 36.0 m/s.
In projectile motion, we can analyze the horizontal and vertical components of the motion separately. The horizontal motion is uniform with constant velocity, while the vertical motion is affected by gravity.The time of flight (t) can be determined using the horizontal distance and the horizontal component of the velocity Substituting the given values into the equation, we can calculate the launch angle (θ) at which the arrow must be released to hit the bull's-eye.Since the equation involves the arcsine function, there might be multiple solutions. In this case, we assume the positive angle that corresponds to the arrow being released above the horizontal. If the result is zero, it means the arrow should be released horizontally.
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Two test charges are located in the x−y plane. If q
1
=−2.600nC and is located at x
1
=0.00 m,y
1
=0.8800 m, and the second test charge has magnitude of q
2
=3.200nC and is located at x
2
=1.000 m,y
2
=0.750 m, calculate the x and y components, E
x
and E
y
, of the electric field
E
in component form at the origin, (0,0). The Coulomb force constant is 1/(4πϵ
0
)=8.99×10
9
N⋅m
2
/C
2
E
x
= N/CE
y
= N/C
The x-component and y-component of the electric field at the origin are 342.2 N/C towards the left and 815.6 N/C towards the positive y-axis, respectively
The electric field is a vector quantity representing the direction and magnitude of the force exerted on a test charge q_o by other test charges. The x and y components of the electric field at the origin caused by two other test charges located in the xy-plane are calculated as follows.
The electric field at a point caused by a point charge is given by Coulomb’s law as follows:
[tex]F = 1 / 4\pi \epsilon q_1 q_2 / r^2[/tex]
where ε is the permittivity of free space, r is the distance between the charges, and [tex]q_1[/tex]and [tex]q_2[/tex] are the charges on the two point charges. For the x-component of the electric field at the origin, the direction of the field is along the x-axis only. For the y-component of the electric field at the origin, the direction of the field is along the y-axis only.
Therefore, the x-component and y-component of the electric field are as follows:
[tex]Ex = Fx / q_0Ey = Fy / q_0[/tex]
The forces exerted on a positive test charge by the two-point charges with negative and positive charges q_1 and q_2 are respectively:
[tex]F_1 = F(q_0, q_1, r_1) = -1.52 * 10^{-3} N[/tex] in the x direction.[tex]F_2 = F(q_0, q_2, r_2) = 2.61 * 10^{-3} N[/tex] at an angle of [tex]32.3^0[/tex] with the negative y-axis.
Using the electric field formula and unit vector notation,
[tex]Ex = F_1x / q_0 + F_2x / q_0 = (-1.52 * 10^{-3} N) / (3.2 * 10^{-9} C) + (2.61 * 10^{-3} N) / (3.2 * 10^{-9} C) = 342.2 N/C[/tex] (towards the left)
[tex]Ey = F_2y / q_0 = (2.61 * 10^{-3} N) / (3.2 * 10^{-9} C) = 815.6 N/C[/tex](towards the positive y-axis).
Therefore, the x-component and y-component of the electric field at the origin are 342.2 N/C towards the left and 815.6 N/C towards the positive y-axis, respectively.
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Case 1: A proton E=1.5∗10 3N/C Released from rest at x=−2 cm Find the change in electric potential energy when proton reaches x=5 cm 1.6 ∗10 ∧ −19C= proton charge Change in potential energy =−qE displacement Ans =−1.68×10 −17Joules Case 2: An electron E= same as case one Electron is now fired in the same direction and position Find change in potential energy when electron reaches 12 cm3.36×10 −17 (Voltage) to accelerate a charged particle.
The change in potential energy when the electron reaches 12 cm is approximately -6.4512 × 10^-36 Joules.
In Case 1, the change in electric potential energy can be calculated using the formula:
ΔPE = -qEΔx
where ΔPE is the change in potential energy, q is the charge of the proton (1.6 × 10^-19 C), E is the electric field (1.5 × 10^3 N/C), and Δx is the displacement.
Δx = 5 cm - (-2 cm) = 7 cm = 0.07 m
Plugging in the values:
ΔPE = -(1.6 × 10^-19 C)(1.5 × 10^3 N/C)(0.07 m)
= -1.68 × 10^-17 J
Therefore, the change in electric potential energy when the proton reaches x = 5 cm is -1.68 × 10^-17 Joules.
In Case 2, the change in potential energy for the electron can be calculated in a similar way using the given electric field (3.36 × 10^-17 V/m) and displacement (12 cm = 0.12 m):
ΔPE = -(1.6 × 10^-19 C)(3.36 × 10^-17 V/m)(0.12 m)
= -6.4512 × 10^-36 J
Therefore, The change in potential energy when the electron reaches 12 cm is approximately -6.4512 × 10^-36 Joules.
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locations, use the thin-lens equation to determine the focal length. Double check arithmetic. cm
Using the thin-lens equation, the focal length of the lens is determined to be 60 cm after calculations involving the object distance and image distance.
To determine the focal length of a lens using the thin-lens equation, you need the object distance (denoted as "u") and the image distance (denoted as "v"). The equation is as follows:
1/f = 1/v - 1/u
Where "f" represents the focal length of the lens. By rearranging the equation, you can solve for the focal length:
1/f = (v - u) / (uv)
Let's assume that the object distance (u) is 20 cm and the image distance (v) is 30 cm. Plugging these values into the equation:
1/f = (30 - 20) / (20 * 30)
1/f = 10 / 600
1/f = 1/60
To isolate the focal length (f), take the reciprocal of both sides:
f = 60 cm
Therefore, the focal length of the lens is 60 cm.
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3. Will one be able to distinguish two points that are 1.5×10
−2
μm apart using a microscope fit with a filter for 575 nm and observing at 1000× Total Magnification?
575(1.25+1.25)
2.3×10
2
nm×(10
3
1 mm)
5
=2.3×10
5
1.5×10
−2
2.2×10
5
=2.3×10
2
nm
No you would nat ba. able to digtirguigh to points from ore anothat 4. A microscope is now fit with a filter for 660 nm, will the researcher be able to differentiate two cells at 8.1×10
−6
m apart while observing at 400× Total Magnification? 5. A student is using a plain microscope (no filter, assume average wavelength of 4.35×10
−3
μm ), would the student be able to see two objects that are 2.8×10
3
nm apart using 100X Total Magnification?
(3)The two points that are 1.5 × 10^−2 μm apart are smaller than the resolution of the microscope, so they would not be able to be distinguished.(4)The two cells that are 8.1 × 10^−6 m apart are smaller than the resolution of the microscope, so they would not be able to be distinguished.(5)The two objects that are 2.8 × 10^3 nm apart are larger than the resolution of the microscope, so they would be able to be distinguished.
(3)The resolution of a microscope is determined by the wavelength of light used and the magnification. The formula is
Resolution = Wavelength / Magnification
So, for the first question, the resolution of the microscope would be:
Resolution = 575 nm / 1000x = 0.0575 μm
The two points that are 1.5 × 10^−2 μm apart are smaller than the resolution of the microscope, so they would not be able to be distinguished.
(4) the resolution of the microscope would be:
Resolution = 660 nm / 400x = 0.165 μm
The two cells that are 8.1 × 10^−6 m apart are smaller than the resolution of the microscope, so they would not be able to be distinguished.
(5) the resolution of the microscope would be:
Resolution = 4.35 × 10^−3 μm / 100x = 0.0435 μm
The two objects that are 2.8 × 10^3 nm apart are larger than the resolution of the microscope, so they would be able to be distinguished.
Therefore, the answers to your questions are:
No, the two points cannot be distinguished. No, the two cells cannot be distinguished. Yes, the two objects can be distinguished.The question should be:
3. Will one be able to distinguish two points that are 1.5×10^−2μm apart using a microscope fit with a filter for 575 nm and observing at 1000× Total Magnification?
4. A microscope is now fit with a filter for 660 nm, will the researcher be able to differentiate two cells at 8.1×10^−6 m apart while observing at 400× Total Magnification?
5. A student is using a plain microscope (no filter, assume average wavelength of 4.35×10^−3μm ), would the student be able to see two objects that are 2.8×10^3 nm apart using 100X Total Magnification?
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. A student throws a ball vertically upwards from the top of the 7 m high CPUT roof. (a) If, after 2 seconds, he catches the ball on its ways down again, with what speed was thrown? (b) What was the velocity of the ball when its was caught? [9,8 m/s] (c) If the student fails to catch the ball with what speed will it hit the ground?
The answers to the given questions are as follows:
(a) The ball was thrown upwards from the roof with an initial velocity of -6.3 m/s.
(b) When the ball was caught on its way down after 2 seconds, its velocity was 13.3 m/s in the downward direction.
(c) If the student fails to catch the ball, it will hit the ground with a speed of approximately 13.31 m/s.
To solve the problem, we can use the equations of motion for vertical motion under constant acceleration. In this case, the acceleration is due to gravity and is equal to 9.8 m/s² (assuming no air resistance).
Given:
Initial height (h) = 7 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 2 s
(a) To find the initial velocity at which the ball was thrown upwards:
Using the equation of motion:
h = ut + (1/2)gt², where u is the initial velocity.
Plugging in the known values, we have:
7 = u(2) + (1/2)(9.8)(2)²
7 = 2u + 19.6
2u = 7 - 19.6
2u = -12.6
u = -6.3 m/s
Therefore, the ball was thrown upwards with an initial velocity of -6.3 m/s (negative sign indicates the upward direction).
(b) To find the velocity of the ball when it was caught:
Since the ball is caught on its way down after 2 seconds, we can use the equation of motion:
v = u + gt, where v is the final velocity (when caught).
Plugging in the values, we have:
v = -6.3 + (9.8)(2)
v = -6.3 + 19.6
v = 13.3 m/s
Therefore, the velocity of the ball when it was caught is 13.3 m/s (positive sign indicates the downward direction).
(c) If the student fails to catch the ball, it will continue to fall freely under gravity until it hits the ground. To find the speed at which it will hit the ground, we can use the equation:
v² = u² + 2gh,
where
v is the final velocity,
u is the initial velocity,
g is the acceleration due to gravity, and
h is the initial height.
Plugging in the values, we have:
v² = (-6.3)² + 2(9.8)(7)
v² = 39.69 + 137.2
v² = 176.89
v = √176.89
v ≈ 13.31 m/s
Therefore, if the student fails to catch the ball, it will hit the ground with a speed of approximately 13.31 m/s.
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Find the electric potential at point P in the figure (Figure 1). Part B Suppose the three charges shown in the figure are held in place. A fourth charge, with a charge of +6.82μC and a mass of 4.23 g, is released from rest at point P. What is the speed of the fourth charge when it has moved infinitely far away from the other three charges?
The speed of the fourth charge when it has moved infinitely far away from the other three charges is 46.3 m/s.
The electric potential at point P is the sum of the electric potentials due to each of the three charges. The electric potential due to a point charge is given by:
V = kQ/r
where k is the Coulomb constant, Q is the charge of the point charge, and r is the distance from the point charge to the point where the potential is being measured.
In this case, the three charges are located at (0, 0), (3, 0), and (0, 4). The distances from these points to point P are 3, 4, and 5, respectively.
The total electric potential at point P is then:
V = k(6.82 μC) / 3 + k(6.82 μC) / 4 + k(6.82 μC) / 5
= 20.4 V
When the fourth charge is released from rest at point P, it will accelerate away from the other three charges. The potential energy of the fourth charge is given by:
U = kQq/r
where k is the Coulomb constant, Q is the charge of the fourth charge, q is the charge of one of the other three charges, and r is the distance between the fourth charge and that other charge.
The total potential energy of the fourth charge is then:
U = k(6.82 μC)(-6.82 μC) / 3 + k(6.82 μC)(-6.82 μC) / 4 + k(6.82 μC)(-6.82 μC) / 5 = -85.9 V
When the fourth charge moves infinitely far away from the other three charges, its potential energy will be zero. The change in potential energy of the fourth charge is then:
ΔU = 0 - (-85.9 V) = 85.9 V
This change in potential energy is equal to the kinetic energy of the fourth charge when it is infinitely far away from the other three charges. The kinetic energy of the fourth charge is given by:
K = 1/2 mv^2
where m is the mass of the fourth charge and v is its speed.
Solving for v, we get:
v = sqrt(2K/m) = sqrt(2(85.9 V)(1/2)) = 46.3 m/s
Therefore, the speed of the fourth charge when it has moved infinitely far away from the other three charges is 46.3 m/s.
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A car traveling at 33 m/s runs out of gas while traveling up a 9.0 slope. How far will it coast before starting to roll back down? Express your answer in meters
A car traveling at 33 m/s runs out of gas while traveling up a 9.0 slope. It will travel the distance of 32.676 m uphill before it starts to roll back down.
To determine how far the car will coast before starting to roll back down the slope, we need to calculate the distance it travels uphill until its velocity becomes zero. This distance is the maximum distance the car can travel before the force of gravity begins to overcome the car's momentum.
First, we need to determine the vertical component of the car's initial velocity. Given that the car is traveling up a 9.0° slope, we can calculate this component using trigonometry:
Vertical component of initial velocity = 33 m/s * sin(9.0°)
Next, we can calculate the time it takes for the car to come to a stop. When the car's velocity becomes zero, the force of gravity will exactly balance the component of the car's weight parallel to the slope. This can be calculated using the equation:
Vertical component of initial velocity = (acceleration due to gravity) * time
Rearranging the equation to solve for time:
Substituting the values:
time = Vertical component of initial velocity / (acceleration due to gravity)
time = (33 m/s * sin(9.0°)) / (9.8 m/s²)
time ≈ 5.662 m/s / 9.8 m/s²
time ≈ 0.578 s
Now, we can calculate the distance the car travels during this time. Since the car is on a slope, the distance is equal to the horizontal component of the initial velocity multiplied by the time:
Distance traveled uphill = 33 m/s * cos(9.0°) * time
Plugging in the values:
Distance traveled uphill = 33 m/s * cos(9.0°) * [(33 m/s * sin(9.0°)) / (9.8 m/s²)]
Distance traveled uphill = 33 m/s * cos(9.0°) * time
Distance traveled uphill ≈ 33 m/s * cos(9.0°) * 0.578 s
Distance traveled uphill ≈ 32.676 m
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A runner hopes to completo the 10,000−m fun in less than 30.0 min. After running at constant speed for exactly 25.0 min, there are still 1900 II to go. The runner must then accelerate at 0.19 m/s
2
for how many seconds in order to achieve the desired time Express your answer using two significant figures. * Incorrect; Try Again; 3 attempts remaining
The runner needs to accelerate for approximately 368 seconds in order to achieve the desired time of completing the 10,000m run in less than 30.0 minutes.
To find the time needed to accelerate, we can use the formula:
d = v_i * t + (1/2) * a * t^2
Where:
d = distance to go after running for 25 minutes (1900m)
v_i = initial velocity (unknown)
t = time to accelerate (unknown)
a = acceleration (0.19 m/s^2)
Since the runner is running at a constant speed for the first 25 minutes, the initial velocity is equal to the average velocity during this time. We can calculate it using the formula:
v_i = d / t
Substituting the given values, we have:
v_i = 1900m / 25min
Now, we can use the equation for distance with the known values to solve for t:
1900m = (v_i * t) + (1/2) * (0.19 m/s^2) * t^2
Simplifying the equation, we get:
1900m = (1900m/25min) * t + 0.095t^2
Rearranging the equation, we have:
0.095t^2 + (1900m/25min) * t - 1900m = 0
Solving this quadratic equation for t, we find:
t ≈ 368 seconds
Therefore, the runner needs to accelerate for approximately in order to achieve the desired time of completing the 10,000m run in less than 30.0 minutes.
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Suppose a 52-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.275 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V ) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T?
Faraday's law of electromagnetic induction is used to compute induced EMF, abbreviated as e. It expresses the relationship between the EMF generated and the magnetic flux's rate of change, abbreviated as φ.
The induced EMF in the coil with 52 turns that lies in the plane of the page in a uniform magnetic field of 1.60 T that is directed out of the page is calculated as follows:
Given values are as follows:52-turn coilInitial area, A1 = 0.275 m²Final area, A2 = 0m² (as it is stretched to have no area in 0.100s)Time, t = 0.100 s
Strength of the uniform magnetic field, B = 1.60 T
We need to calculate the magnitude (in V) and direction (as seen from above) of the average induced EMF, e.We know that the flux is defined as φ = B.A.
Therefore, we can write:[tex]B * A1 = B * A2 + [(ΔB / Δt) * A2][/tex]
By substituting the given values in the above formula,
we get: (1.60 T)(0.275 m²)
= [tex](1.60 T)(0 m²) + [(ΔB / Δt) * 0 m²]ΔB / Δt[/tex]
= [tex][(1.60 T)(0.275 m²)] / (0 m²)(0.100 s)ΔB / Δt[/tex]
= [tex]4.40 T/s[/tex]
Now, by using Faraday's law of electromagnetic induction, we can calculate the induced EMF.
[tex]e = -N (Δφ / Δt).[/tex]
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A crate with mass 31.0 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 160 N. What acceleration is produced? Part B How far does the crate travel in 12.5 s ? What is its speed at the end of 12.5 s ?
Part (a) The acceleration produced by the net horizontal force of 160 N acting on the crate of mass 31.0 kg is 5.16 m/s².
Part (b) The distance traveled by the crate in 12.5 s is 1020.31 m and its speed at the end of 12.5 s is 64.5 m/s.
Part (a) Newton's second law of motion states that, the force acting on an object is directly proportional to its acceleration, given as, F = ma, Where, F is the force acting on the object, m is the mass of the object, a is the acceleration produced by the force. Now, substituting the given values of force and mass in the above equation, we get160 = 31.0 aa = 160/31.0a = 5.16 m/s².
Therefore, the acceleration produced by the net horizontal force of 160 N acting on the crate of mass 31.0 kg is 5.16 m/s².
Part (b) Now, to find the distance traveled by the crate and its speed at the end of 12.5 s, we need to use the following kinematic equations: v = u + ats = ut + 1/2 at²v² = u² + 2as, Where, v is the final velocity of the crate, u is the initial velocity of the crates is the distance traveled by the crate, t is the time taken by the crate, a is the acceleration produced by the force applied. Substituting the given values of u and t in the equation 1, we get
v = u + at
v = 0 + (5.16)(12.5)v = 64.5 m/s.
Now, substituting the given values of u, t, and a in equation 2, we gets = ut + 1/2 at²s = 0(12.5) + 1/2 (5.16)(12.5)²s = 1020.31 m.
Therefore, the distance traveled by the crate in 12.5 s is 1020.31 m and its speed at the end of 12.5 s is 64.5 m/s.
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