a) The velocity of the object when it is 26.0 m above the ground is -12.74 m/s. b) the total distance the object travels during the fall is 8.38 m.
(a) For finding the velocity of the object when it is 26.0 m above the ground, use the equation of motion:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the object is freely falling, the acceleration is equal to the acceleration due to gravity, which is approximately [tex]9.8 m/s^2[/tex]. The initial velocity is zero because the object starts from rest.
Plugging in the values:
[tex]v = 0 + (9.8 m/s^2)(1.30 s) = 12.74 m/s[/tex]
Since the object is moving downward, assign a negative sign to the velocity. Therefore, the velocity of the object when it is 26.0 m above the ground is -12.74 m/s.
(b) For calculating the total distance the object travels during the fall, use the equation of motion:
[tex]s = ut + (1/2)at^2[/tex],
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Again, the initial velocity is zero.
Plugging in the values:
[tex]s = 0 + (1/2)(9.8 m/s^2)(1.30 s)^2 = 8.38 m[/tex].
Therefore, the total distance the object travels during the fall is 8.38 m.
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In an old-style television picture tube (not in a modern flat-panel TV) electrons are boiled out of a very hot metal filament placed near a negative metal plate (see the figure). These electrons start out nearly at rest and are accelerated toward a positive metal plate. They pass through a hole in the positive plate on their way toward the picture screen. If the high-voltage supply in the television set maintains a potential difference of 14900 volts between the two plates, what speed do the electrons reach? (You can use the nonrelativistic approximation here)
The speed that the electrons reach in the given scenario is 2.28 × 10⁷ m/s.
In an old-style television picture tube (not in a modern flat-panel TV), electrons are boiled out of a very hot metal filament placed near a negative metal plate. These electrons start out nearly at rest and are accelerated toward a positive metal plate. They pass through a hole in the positive plate on their way toward the picture screen. If the high-voltage supply in the television set maintains a potential difference of 14900 volts between the two plates, the speed that the electrons reach can be calculated as follows:
Step 1: Determine the electric potential energy. The electric potential energy of a point charge at a point with a voltage V is:U = qV Where, U is the electric potential energy q is the magnitude of the electric charge V is the voltage. The magnitude of the charge of an electron is e = 1.60 × 10⁻¹⁹ C.U = eVU = (1.60 × 10⁻¹⁹ C)(14900 V)U = 2.38 × 10⁻¹⁵ J.
Step 2: Find the kinetic energy, Kinetic energy is defined as: K.E. = 1/2mv²Where,K.E. is the kinetic energy of the electron, m is the mass of the electron v is the velocity of the electron. The mass of an electron is m = 9.11 × 10⁻³¹ kg. K.E. = 1/2mv²K.E. = 1/2(9.11 × 10⁻³¹ kg)(v²)K.E. = 4.57 × 10⁻³² v²
Step 3: Equate the potential energy to the kinetic energyThe electric potential energy gained by the electron is equal to the kinetic energy gained:U = K.E.2.38 × 10⁻¹⁵ J = 4.57 × 10⁻³² v²v² = 5.20 × 10¹⁶ m²/s²v = 2.28 × 10⁷ m/s. Therefore, the speed that the electrons reach is 2.28 × 10⁷ m/s.
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8. Blue light (λ=460 nm) is incident on a piece of potassium (ϕ=2.20eV). What is the maximum kinetic energy of the ejected photoelectrons? (1eV=1.60×10 ^−19J, or h=4.14×10 ^−15eV^∗sec ) ) a. 1.0eV b. 0.50eV c. 0.25eV d. 4.9eV
,The correct option is (c) 0.25 eV.
The maximum kinetic energy of the ejected photoelectrons is 0.25 eV. According to the Einstein's photoelectric effect, the photoelectron's kinetic energy is equal to the difference between the energy of the incident photon and the energy required to remove the electron (also known as the work function, denoted by φ).
The maximum kinetic energy (KEmax) of the ejected photoelectron can be determined using the equation:
[tex]\[KEmax = E - \phi\][/tex]
Where E is the energy of the incident photon and ϕ is the work function of the metal (in electron volts).
Now, we will calculate the maximum kinetic energy (KEmax) of the ejected photoelectrons using the given values:
Given:
Energy of the incident photon, E = [tex]\(\frac{hc}{\lambda} = \frac{(6.626 × 10^{-34} \, \text{J s})(3.0 × 10^8 \, \text{m/s})}{(460 × 10^{-9} \, \text{m})} = 4.31 × 10^{-19} \, \text{J}\)Work function, φ = 2.20 eV[/tex]
Maximum kinetic energy of the ejected photoelectron, \(KEmax = E - φ\)
[tex]\(= 4.31 × 10^{-19} \, \text{J} - (2.20 \, \text{eV} × 1.60 × 10^{-19} \, \text{J/eV})\)[/tex]
[tex]\(= 4.31 × 10^{-19} \, \text{J} - 3.52 × 10^{-19} \, \text{J}\)[/tex]
[tex]\(= 0.79 × 10^{-19} \, \text{J}\)[/tex]
[tex]\(= 0.79 \, \text{eV}\)[/tex]
Therefore, the maximum kinetic energy of the ejected photoelectrons is 0.79 eV or \[tex](0.79 × 1.60 × 10^{-19} \, \text{J} = 1.26 × 10^{-19} \, \text{J} \approx 0.25 \, \text{eV}\) (rounded off to two decimal places).[/tex]
Therefore, the correct option is (c) 0.25 eV.
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Two point charges are separated by 25.0 cm (see (Figure 1)). Assume that q1 = -6.25 nC and q2 = -10.5 nC.
1. Find the net electric field these charges produce at point A.
2. Find the net electric field these charges produce at point B.
3. What would be the magnitude of the electric force this combination of charges would produce on a proton at A?
4. What would be the direction of the electric force this combination of charges would produce on a proton at A?
The net electric field produced at point A is -1.155 × 10^6 N/C towards left. The net electric field produced at point B is -4.225 × 10^5 N/C towards right.
The magnitude of the electric force this combination of charges would produce on a proton at A is 1.848 N. The direction of electric force experienced by the proton at point A is the towards right.
The given two point charges are separated by a distance of 25 cm. Therefore, let us first calculate the distance between the two point charges. The point A is at a distance of x from the point charge q1. Therefore, it is at a distance of 25.0 - x from the point charge q2. The expressions for the electric fields produced by the charges q1 and q2 can be given as follows,
[tex]E1 = [k \times q1]/r1^2[/tex] and
[tex]E2 = [k \times q2]/r2^2[/tex]
where k is the Coulomb's constant, q1 is the charge on the first point charge, q2 is the charge on the second point charge, r1 is the distance between the point charge q1 and point A, r2 is the distance between the point charge q2 and point A.1. The net electric field at point A:
For the point A, x = 15 cm
Therefore, the distance between the point charge q1 and point A,
r1 = 15 cm
= 0.15 m
The distance between the point charge q2 and point A,
r2 = 25 - 15
= 10 cm
= 0.1 m
Substituting the given values in the expressions for the electric fields, we get,
[tex]E1 = [9 \times 10^9 \times (-6.25 \times 10^{-9})]/(0.15^2)[/tex]
[tex]= -2.1 \times 10^5\ N/C[/tex] (towards right)
and [tex]E2 = [9 \times 10^9 \times (-10.5 \times 10^{-9})]/(0.1^2)[/tex]
[tex]= -9.45 \times 10^5\ N/C[/tex] (towards left)
The net electric field is given by the vector sum of the electric fields produced by the individual charges.
E = E1 + E2
= -2.1 × 10^5 N/C - 9.45 × 10^5 N/C
= -1.155 × 10^6 N/C (towards left)
Therefore, the net electric field produced at point A is -1.155 × 10^6 N/C towards left.
2. The net electric field at point B:
For the point B, x = 10 cm
Therefore, the distance between the point charge q1 and point B,
r1 = 10 cm
= 0.1 m
and the distance between the point charge q2 and point B,
r2 = 25 - 10
= 15 cm
= 0.15 m
Substituting the given values in the expressions for the electric fields, we get,
E1 = [9 × 10^9 × (-6.25 × 10^-9)]/(0.1^2)
= -5.625 × 10^5 N/C (towards right)
and
E2 = [9 × 10^9 × (-10.5 × 10^-9)]/(0.15^2)
= -1.4 × 10^5 N/C (towards left)
The net electric field is given by the vector sum of the electric fields produced by the individual charges.
E = E1 + E2
[tex]= -5.625 \times 10^5\ N/C - (-1.4 \times 10^5\ N/C)[/tex]
[tex]= -4.225 \times 10^5\ N/C[/tex] (towards right)
Therefore, the net electric field produced at point B is -4.225 × 10^5 N/C towards right.
3. Magnitude of the electric force at point A:
For this, we need to calculate the electric field at point A first, which we have already calculated in part 1.
E = -1.155 × 10^6 N/C
The electric force experienced by a proton of charge q at this point is given by the expression
[tex]F = q.E[/tex]
[tex]= (1.6 \times 10^{-19}) \times (-1.155 \times 10^{6})[/tex]
= -1.848 N
Therefore, the magnitude of the electric force this combination of charges would produce on a proton at A is 1.848 N.
4. Direction of the electric force at point A:
From the above calculations, it is clear that the electric field is directed towards left and the charge on the proton is positive. Therefore, the direction of electric force experienced by the proton at point A is the towards right.
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To find the net electric field and electric force at points A and B, the electric field due to each charge needs to be calculated. The direction of the electric force on a proton can be determined by the sign of the charges.
Explanation:1. To find the net electric field at point A, we need to calculate the electric field due to each charge and then add them together. The electric field due to a point charge can be calculated using the formula E = k * (q / r^2), where E is the electric field, k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge to the point.
2. To find the net electric field at point B, you can follow the same steps as in the previous solution.
3. To find the electric force on a proton at point A, you can use the formula F = q * E, where F is the electric force, q is the charge of the proton (1.6 x 10^-19 C), and E is the electric field at point A.
4. The direction of the electric force on a proton at point A can be determined by the sign of the charge. Since both charges q1 and q2 are negative, the electric force on the proton will be in the opposite direction of the electric field.
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(a) The elevator starts from rest and accelerates downward with \( a=1.35 \mathrm{~m} / \mathrm{s}^{2} \). What are the tensions in the two strings in newtons? \[ \begin{array}{ll} T_{1}= & \mathrm{N}
Given,Acceleration of the elevator, a = 1.35 m/s²
The tension in the string is given as T1 and T2.Let the mass of the elevator be m.
Therefore, force acting on the elevator = mg.
The net force acting on the elevator is given by F=mg-ma.
Therefore, the force F= m(g-a)
Since the elevator is moving downward, the tension in the string T1 will be greater than T2 as T1 will be supporting the entire weight of the elevator and T2 will only support a part of the weight.
The tension in the string is given by,T1 - T2 = m(g-a).......(1)
and T1 + T2 = mg.......(2)
Solving equation (1) and (2), we get,T1 = 5886.2 N and T2 = 3924.1 N
Therefore, the tension in the first string T1 = 5886.2 N and the tension in the second string T2 = 3924.1 N.
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The engineer of a passenger train traveling at 24 m/s sights a freight train whose caboose is 170 m ahead on the same track. The freight train is traveling at 12 m/s in the same direction a the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of 0.100 m/s∧2 in a direction opposite to the passenger train's velocity, while the freight train continues with constant speed. Take x=0 at the location of the front of the passenger train at the moment the engineer applies the brakes. Unfortunately, the two trains will collide. (a) When will they collide? s (b) Where will thev collide? m (c) If the engineer of the passenger train brakes hard enough, there will NOT be a collision. What's the minimum braking acceleration (magnitude) so that the trains don't collide? m/s/s
The passenger and freight trains collide in about 2.41 seconds, approximately 56.5 meters from the passenger train's starting point. To prevent a collision, the passenger train needs a minimum braking acceleration of around 0.14 m/s².
To solve this problem, we need to determine the time of collision, the position of collision, and the minimum braking acceleration required to avoid a collision.
(a) Time of collision:
We can use the equation of motion for the passenger train to find the time it takes for the trains to collide. The equation is given by:
x = x₀ + v₀t + (1/2)at²,
where x is the distance traveled, x₀ is the initial position, v₀ is the initial velocity, t is time, a is acceleration, and t² represents t squared.
The initial position of the passenger train, x₀, is 0, the initial velocity, v₀, is 24 m/s, and the acceleration, a, is -0.100 m/s² (opposite direction to the velocity). The distance travelled, x, is 170 m. Plugging these values into the equation, we can solve for t:
170 = 0 + (24)t + (1/2)(-0.100)t².
Simplifying and solving the quadratic equation, we find t ≈ 2.41 seconds. Therefore, the passenger and freight trains will collide approximately 2.41 seconds after the passenger train's engineer applies the brakes.
(b) Position of collision:
To find the position of collision, we can use the equation:
x = x₀ + v₀t + (1/2)at².
Using the same values as before, except substituting t with 2.41 seconds, we can find the position, x. Plugging in the values, we get:
x = 0 + (24)(2.41) + (1/2)(-0.100)(2.41)².
Calculating this expression gives us x ≈ 56.5 meters. Therefore, the trains will collide approximately 56.5 meters from the initial position of the passenger train.
(c) Minimum braking acceleration to avoid collision:
To avoid a collision, the passenger train must decelerate with a magnitude equal to or greater than the acceleration of the freight train. The acceleration of the freight train is 0 m/s² since it continues with a constant speed.
Using the equation of motion, we can find the minimum braking acceleration required for the passenger train to avoid a collision. We set x equal to the initial separation between the trains (170 m) and solve for a:
170 = 0 + (24)(t) + (1/2)(a)(t)².
Simplifying this equation, we find:
0.5at² + 24t - 170 = 0.
Solving this quadratic equation, we find two possible values for a: approximately 0.14 m/s² and -0.34 m/s². Since we are interested in the minimum braking acceleration, we consider the positive value of 0.14 m/s² as the minimum braking acceleration required for the passenger train to avoid a collision.
In summary, the passenger and freight trains will collide approximately 2.41 seconds after the passenger train's engineer applies the brakes, at a position approximately 56.5 meters from the initial position of the passenger train. To avoid a collision, the minimum braking acceleration required for the passenger train is approximately 0.14 m/s².
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The collision occurs in 14.3 seconds, 85.7 meters away. The passenger train requires a minimum braking acceleration of 0.529 m/s².
The collision time is determined by solving the equation of motion for the passenger train's displacement. By plugging in the values and solving the quadratic equation, we find the time of collision. The collision point is calculated by finding the distance traveled by the passenger train during that time. To prevent collision, the passenger train's displacement must be less than or equal to the initial distance. Setting up the equation and solving for acceleration yields the minimum required required value.
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A mobile with SHM has an amplitude of 1.20 m and an angular speed of 5 rad/s, a) at what position x from the equilibrium position will the mobile have a speed of 180 m/s? b) Calculate the total energy if the mass of the mobile is 2.0kg
The total energy of the system is 8136 J.
Amplitude, A = 1.2 m
Angular Speed, ω = 5 rad/s
Maximum velocity, vmax = 180 m/s
Mass, m = 2.0 kg(a)
We need to find the position from the equilibrium position where the mobile has a speed of 180 m/s.
According to the question, maximum velocity, vmax = 180 m/s
Maximum velocity is given by:vmax = AωSo,180 = 1.2 × 5vmax = 6 m/s
The velocity of the mobile is maximum at mean position.
Now, we will find the displacement from the mean position.
x = Acos(ωt)
Where,x = displacement at any time t
A = amplitude
ω = angular frequency
t = time
The velocity of the mobile is maximum at mean position. So, we need to find the time when the velocity of the mobile is maximum. That is, when the mobile is at the mean position.
At mean position, x = 0m = A cos(ωt)0 = 1.2 cos (5t)cos(5t) = 0t = nπ/2ω = 5 rad/st = (π/2)/5 = 0.314 sThe time when the mobile is at the mean position is 0.314 s
At time t = 0.314 s, the displacement is given by:
x = Acos(ωt)x = 1.2 cos (5 × 0.314)x = 1.2 cos 1.57x = 0m(b)
We need to find the total energy if the mass of the mobile is 2.0kg.
The potential energy of the system is given by:
U = (1/2) kA²where,k = spring constantk = mω²
The mass of the mobile, m = 2.0 kgω = 5 rad/sk = mω²k = 2.0 × 5²k = 50 N/mU = (1/2) × 50 × (1.2)²U = 36 J
The kinetic energy of the system is given by:K = (1/2)mv²
Where,m = 2.0 kgv = vmax/2
v = 90 m/sK = (1/2) × 2.0 × (90)²K = 8100 J
Total energy of the system = Potential energy + Kinetic energy
E = U + KE = 36 + 8100E = 8136 J
Therefore, the total energy of the system is 8136 J.
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A truck covers 40.0 m in 9.20 s while uniformly slowing down to a final velocity of 3.00 m/s. (a) Find the truck's original speed. (b) Find its acceleration. Y
(a) The truck's original speed is approximately 0.732 m/s.
(b) The truck's acceleration is approximately 0.293 m/s².
To find the truck's original speed [tex](\(v_i\))[/tex] and acceleration (a), we'll use the given values and the equations mentioned earlier.
(a) Calculation for the truck's original speed:
We have the equation:
[tex]\[v_i = v_f - a \cdot t\][/tex]
Substituting the known values:
[tex]\(v_f = 3.00 \, \text{m/s}\),\(t = 9.20 \, \text{s}\),[/tex]
and rearranging the equation, we get:
[tex]\[v_i = 3.00 \, \text{m/s} - a \cdot 9.20 \, \text{s}\][/tex]
(b) Calculation for the truck's acceleration:
We have the equation:
[tex]\[d = v_i \cdot t + \frac{1}{2} a \cdot t^2\][/tex]
Substituting the known values:
[tex]\(d = 40.0 \, \text{m}\),\(t = 9.20 \, \text{s}\),\\and rearranging the equation, we get:\\\[40.0 \, \text{m} = v_i \cdot 9.20 \, \text{s} + \frac{1}{2} a \cdot (9.20 \, \text{s})^2\][/tex]
Now, we have two equations:
[tex]Equation 1: \(v_i = 3.00 \, \text{m/s} - a \cdot 9.20 \, \text{s}\)\\Equation 2: \(40.0 \, \text{m} = v_i \cdot 9.20 \, \text{s} + \frac{1}{2} a \cdot (9.20 \, \text{s})^2\)[/tex]
We can solve these equations simultaneously to find the values of [tex]\(v_i\)[/tex]and a. Let's proceed with the calculations.
From Equation 1:
[tex]\(v_i = 3.00 - 9.20a\)[/tex]
Substituting this expression for [tex]\(v_i\)[/tex] into Equation 2:
[tex]\(40.0 = (3.00 - 9.20a) \cdot 9.20 + \frac{1}{2} a \cdot (9.20)^2\)[/tex]
Expanding and rearranging the equation:
[tex]\(40.0 = 27.60 - 84.64a + 0.5a \cdot 84.64\)[/tex]
Combining like terms:
[tex]\(0.5a \cdot 84.64 - 84.64a = 27.60 - 40.0\)\\\(-42.32a = -12.40\)[/tex]
Dividing both sides by -42.32:
[tex]\(a = \frac{-12.40}{-42.32}\)[/tex]
Calculating \(a\):
[tex]\(a \approx 0.293 \, \text{m/s}^2\)[/tex]
Substituting the value of a back into Equation 1 to find [tex]\(v_i\)[/tex]:
[tex]\(v_i = 3.00 - 9.20 \cdot 0.293\)[/tex]
Calculating \(v_i\):
[tex]\(v_i \approx 0.732 \, \text{m/s}\)[/tex]
Therefore, the truck's original speed is approximately [tex]\(0.732 \, \text{m/s}\)[/tex] and its acceleration is approximately [tex]\(0.293 \, \text{m/s}^2\).[/tex]
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The lowest note on a five-string bass guitar has a frequency of 31 Hz. The vibrating length of string is 89 cm long. What is the wave speed on this string? Express your answer with the appropriate units.
The wave speed on the string of a five-string bass guitar is approximately 2.78 meters per second (m/s).
To find the wave speed on the string, we can use the formula v = λf, where v is the wave speed, λ (lambda) is the wavelength, and f is the frequency.
First, let's find the wavelength (λ). The vibrating length of the string is given as 89 cm, which is equivalent to 0.89 meters. For the lowest note on the string, the wavelength corresponds to twice the length of the string, as it represents a full oscillation from the highest to the lowest point and back. Therefore, the wavelength is λ = 2 * 0.89 meters.
Next, we are given the frequency (f) as 31 Hz. Now we can substitute the values into the wave speed formula: v = (2 * 0.89 meters) * (31 Hz).
Calculating the expression, we find that the wave speed on the string of the bass guitar is approximately 2.78 m/s. This represents the speed at which the wave propagates along the string, resulting in the production of sound.
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A gust of wind begins to rotate the blades of a windmill to an angular speed of 6.3rad/s with a constant gular acceleration of 2.2rad/s2. How many revolutions does it take for the blade's angular speed to ach the speed of 6.3rad/s ? A wooden block with a mass of 5 kg oscillates horizontally in the positive direction on a spring with a iod T and an amplitude of 0.48 cm. The block is at the equilibrium position when time equals 0 . Where he block when the time equals 3 T/4 ?
Since cosine of π/2 is zero, the block will be at its maximum displacement from the equilibrium position when t = 3T/4. The exact value of this displacement depends on the specific value of T, which is not provided in the question.
To find the number of revolutions it takes for the blade's angular speed to reach 6.3 rad/s, we need to calculate the time it takes for the angular speed to reach that value. We can use the formula:
ω = ω₀ + αt
Where:
ω = final angular speed = 6.3 rad/s
ω₀ = initial angular speed = 0 rad/s (since the blades start from rest)
α = angular acceleration = 2.2 rad/s²
t = time
Solving for time, we have:
t = (ω - ω₀) / α
t = (6.3 rad/s - 0 rad/s) / 2.2 rad/s²
t ≈ 2.86 s
Now, to find the number of revolutions, we need to divide the time by the period of one revolution:
Number of revolutions = t / T
Since the problem does not provide the period (T), we cannot calculate the exact number of revolutions without that information.
For the second part of the question, to determine where the block is located when the time equals 3T/4, we need to consider the equation of motion for simple harmonic motion:
x = A * cos(ωt)
Where:
x = displacement from equilibrium position
A = amplitude of the oscillation = 0.48 cm
ω = angular frequency = 2π / T
t = time
At t = 0, the block is at the equilibrium position, so x = 0.
To find the position at t = 3T/4, we substitute t = 3T/4 into the equation:
x = A * cos(ω * (3T/4))
Since cosine of π/2 is zero, the block will be at its maximum displacement from the equilibrium position when t = 3T/4. The exact value of this displacement depends on the specific value of T, which is not provided in the question.
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please show work Extra 73: A small block of mass \( m \) slides along a frictionless loop-the-loop. At what height \( h \) above the bottom of the loop should the block be released so that the normal force exerted on
To determine the height above the bottom of the loop at which the block should be released, we need to consider the forces acting on the block at different points of the loop. At the top of the loop, the normal force is directed downward, while at the bottom of the loop, the normal force is directed upward.
Let's analyze the forces acting on the block at the top and bottom of the loop.
1. Top of the loop:
At the top of the loop, the normal force (N) acts downward, and the gravitational force (mg) acts downward as well. The net force (F_net) must provide the centripetal force (F_c) required to keep the block moving in a circular path.
The centripetal force is given by:
F_c = m * a_c
Where m is the mass of the block and a_c is the centripetal acceleration.
Since the block is moving in a circular path at the top of the loop, the net force can be expressed as:
F_net = N - mg
Setting the net force equal to the centripetal force, we have:
N - mg = m * a_c
At the top of the loop, the centripetal acceleration is directed downward (opposite to the direction of the normal force), so we can write it as:
a_c = -g
Substituting this into the equation and solving for N, we get:
N - mg = m * (-g)
N = mg - mg
N = 0
Therefore, at the top of the loop, the normal force is zero. This implies that the block will lose contact with the track and will not stay on the loop.
2. Bottom of the loop:
At the bottom of the loop, the normal force (N) acts upward, and the gravitational force (mg) acts downward. The net force (F_net) must again provide the centripetal force (F_c) required to keep the block moving in a circular path.
The centripetal force is given by:
F_c = m * a_c
Where m is the mass of the block and a_c is the centripetal acceleration.
The net force can be expressed as:
F_net = N - mg
Setting the net force equal to the centripetal force, we have:
N - mg = m * a_c
At the bottom of the loop, the centripetal acceleration is directed upward (opposite to the direction of the gravitational force), so we can write it as:
a_c = g
Substituting this into the equation and solving for N, we get:
N - mg = m * g
N = mg + mg
N = 2mg
Therefore, at the bottom of the loop, the normal force is equal to twice the gravitational force acting on the block.
To determine the height (h) above the bottom of the loop at which the block should be released, we need to find the location where the normal force becomes zero. This occurs at the top of the loop. So, the block should be released at the top of the loop for the normal force to be zero.
In summary, the block should be released at the top of the loop, where the normal force is zero, to maintain contact with the loop throughout the motion.
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Required information A copper bar of thermal conductivity 401 W/(m⋅K) has one end at 124∘ C and the other end at 24.0∘ C. The length of the bar is 0.190 m, and the cross-sectional area is 1.00×10^−6m^2 . if two such bars were placed in parallel (side by side) with the ends in the same temperature baths, what would the rate of heat aduction P be? 0.211W Required information A copper bar of thermal conductivity 401 W/(m⋅K) has one end at 124∘ C and the other end at 24.0∘ C. The length of the bar is 0.190 m, and the cross-sectional area is 1.00×10^−6m^2 . f two such bars were placed in series (end to end) between the same constant-temperature baths, what would the rate of heat conduction P be? 0.211 W Required information A copper bar of thermal conductivity 401 W/(m⋅K) has one end at 124∘ C and the other end at 24.0∘ C. The length of the bar is 0.190 m, and the cross-sectional area is 1.00×10^−6m^2 . What is the rate of heat conduction P along the bar?
The rate of heat conduction, P, along the copper bar can be calculated using Fourier's law of heat conduction, which states that the rate of heat conduction is proportional to the temperature difference and the thermal conductivity of the material, and inversely proportional to the length and cross-sectional area of the bar.
For a single copper bar, the rate of heat conduction, P, can be calculated as follows:
P = (k * A * (T₁ - T₂)) / L,
where k is the thermal conductivity of copper (401 W/(m⋅K)), A is the cross-sectional area of the bar (1.00×10^−6 m²), T₁ is the temperature at one end (124°C), T₂ is the temperature at the other end (24.0°C), and L is the length of the bar (0.190 m).
Substituting the given values into the formula, we have:
P = (401 * 1.00×10^−6 * (124 - 24.0)) / 0.190 = 0.211 W.
Therefore, the rate of heat conduction along the copper bar is 0.211 W.
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A 200-N wagon is to be pulled up a 30
∘
incline at constant speed. A) How large a force parallel to the incline is needed if friction eff 1 point [Question above] B) Determine the normal force on the wagon.
To pull the 200-N wagon up a 30° incline at constant speed, a force parallel to the incline is needed to overcome the force of gravity and the friction force. The force required can be determined using the equation:
Force parallel to incline = Force of gravity + Friction force
The force of gravity is given by the weight of the wagon, which is equal to its mass multiplied by the acceleration due to gravity. The friction force can be calculated by multiplying the coefficient of friction between the wagon and the incline by the normal force.
The normal force on the wagon can be determined by considering the forces acting perpendicular to the incline. The normal force is equal in magnitude and opposite in direction to the component of the weight of the wagon that acts perpendicular to the incline.
The weight of the wagon can be decomposed into two components: one parallel to the incline and one perpendicular to the incline. The component perpendicular to the incline is given by the weight multiplied by the cosine of the angle of inclination. Thus, the normal force on the wagon is equal to the weight of the wagon multiplied by the cosine of the angle of inclination.
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(a) How much work (in J) does the electric field do on the electron? 23 (b) What is the change in potential energy (in J) of the entire system (vacuum tube plus electron)? ] (c) What is the velocity (in m/s ) of the electron after it moves the 2.80 cm distance? magnitude direction
(a) The work done by the electric field on the electron is 11.536 J. (b) The change in potential energy of the system is also 11.536 J. (c) After moving the distance of 2.80 cm, the velocity of the electron is approximately [tex]4.592 * 10^7 m/s[/tex].
(a) For calculating the work done by the electric field on the electron, use the formula for work done by an electric field:
[tex]work = electric field strength * distance * cosine(\theta)[/tex]
In this case, the electric field strength is given as 412 N/C, the distance is 2.80 cm (convert to meters: 0.028 m), and the angle [tex](\theta)[/tex] between the electric field and the direction of motion is 0 degrees (cos(0) = 1). Therefore, the work done is:
(412 N/C) * (0.028 m) * (1) = 11.536 J.
(b) The change in potential energy of the entire system can be calculated by using the formula:
change in potential energy = work done by the electric field.
In this case, the change in potential energy is also 11.536 J.
(c) For finding the velocity of the electron after moving the given distance, use the equation for kinetic energy:
kinetic energy = [tex](1/2) * mass * velocity^2[/tex]
Since the electron is initially at rest, the initial kinetic energy is zero. Therefore, the change in kinetic energy is equal to the work done by the electric field, which is 11.536 J. Can equate this to [tex](1/2) * mass * velocity^2[/tex] and solve for the velocity. Given that the mass of an electron is approximately [tex]9.11 * 10^{-31} kg[/tex], Rearrange the equation:
velocity = [tex]\sqrt((2 * change in kinetic energy) / mass)[/tex]
Plugging in the values:
velocity = [tex]\sqrt((2 * 11.536 J) / (9.11 * 10^{-31} kg)) = 4.592 * 10^7 m/s[/tex]
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The complete question is:
Inside a particular vacuum tube, there is a uniform magnetic field with a magnitude 412 N/C pointing in the positive x-direction. An electron, initially at rest, moves a distance of 2.80 cm in this field
(a) How much work (in J) does the electric field do on the electron?
(b) What is the change in potential energy (in J) of the entire system (vacuum tube plus electron)?
(c) What is the velocity (in m/s ) of the electron after it moves the 2.80 cm distance?
A transverse wave on a taut string is modeled with the wave function y(x,t)=0.5sin(14x−7.0t). What is the velocity and acceleration function of this wave? [use calculus] Find the amplitude, wavelength, period and speed of the wave
The given wave function for a transverse wave on a taut string is y(x,t) = 0.5sin(14x − 7.0t).We need to find the velocity and acceleration function of this wave. We can find the velocity function of the wave by differentiating the wave function with respect to time t and acceleration function of the wave by differentiating the velocity function with respect to time t.
Velocity function of the wavey(x, t) = 0.5sin(14x − 7.0t)Differentiating the above equation with respect to time t, we get;v(x, t) = dy(x, t)/dt = -0.5*7cos(14x - 7.0t)From the above equation, we can observe that the velocity of the wave is v(x, t) = -3.5cos(14x - 7.0t).Acceleration function of the wavev(x, t) = -3.5cos(14x - 7.0t)Differentiating the above equation with respect to time t,
we get;a(x, t) = dv(x, t)/dt = 24.5sin(14x - 7.0t)Given wave function, y(x,t) = 0.5sin(14x − 7.0t).Comparing the wave function with the standard wave function y(x,t) = Asin(kx − ωt), we get;Amplitude (A) = 0.5Wave number (k) = 14Angular frequency (ω) = 7.0The wavelength (λ) of the wave is given by λ = 2π/k = 2π/14 = π/7The period (T) of the wave is given by T = 2π/ω = 2π/7The speed of the wave is given by v = λf = ω/k. Here, f is the frequency of the wave.v = π/7*7 = π.
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An electron moves along the z-axis with v
z
=5.8×10
7
m/s. As it passes the origin, what are the strength and direction of the magnetic (1 cm , 0 cm,0 cm ) field at the following (x,y,z) positions? Express your answers using two significant figures. Enter your answers numerically separated by commas. Part B (0 cm,0 cm,2 cm) Express your answers using two significant figures. Enter your answers numerically separated by commas.
Therefore, the strength and direction of the magnetic field at (0 cm, 1 cm, 0 cm) is (0, 2.9 × 10⁻³ T, 0) and the strength and direction of the magnetic field at (0 cm, 0 cm, 2 cm) is (0, 0, 7.2 × 10⁻⁴ T).
The magnetic field that the electron experiences can be computed using the formula below:
F=Bqvsinθ
F is the force on the electron.
B is the strength of the magnetic field.
q is the electric charge of the electron.
v is the velocity of the electron.θ is the angle between the magnetic field and the velocity of the electron.
Because the electron is moving in the positive z-direction, the angle between the magnetic field and the velocity of the electron will be 90 degrees.
θ=90 degrees
=π/2 radians
The strength of the magnetic field can be determined by solving for B:
F=qvBsinθB
=F/qvsinθ
The charge on an electron is 1.6 × 10⁻¹⁹ C.
The velocity of the electron is 5.8 × 10⁷ m/s.
At the position (0 cm, 1 cm, 0 cm), the distance from the electron to the origin is
r = 1 cm
= 0.01 m.
Because the magnetic field is at position (1 cm, 0 cm, 0 cm), the position vector will be:
r = (0.01) i
At point B (0 cm, 0 cm, 2 cm), the distance between the electron and the origin is:
r = 2 cm = 0.02 m.
The position vector is:
r = (0) i + (0) j + (0.02) k
Now we can use the formula to calculate the magnetic field at each of the points.
(a) For position (0 cm, 1 cm, 0 cm)
F=qvBsinθB
=F/qvsinθB
=(1.6 × 10⁻¹⁹ C)(5.8 × 10⁷ m/s)/(0.01 m × 1.6 × 10⁻¹⁹ C × sin π/2)
= 2.9 × 10⁻³ T=0, 2.9 × 10⁻³ T, 0
(b) At position (0 cm, 0 cm, 2 cm)
F=qvBsinθB
=F/qvsinθB
=(1.6 × 10⁻¹⁹ C)(5.8 × 10⁷ m/s)/(0.02 m × 1.6 × 10⁻¹⁹ C × sin π/2)
= 7.2 × 10⁻⁴ T
=0, 0, 7.2 × 10⁻⁴ T
Therefore, the strength and direction of the magnetic field at (0 cm, 1 cm, 0 cm) is (0, 2.9 × 10⁻³ T, 0) and the strength and direction of the magnetic field at (0 cm, 0 cm, 2 cm) is (0, 0, 7.2 × 10⁻⁴ T).
Note: Since the magnetic field is only in the z-direction, the x and y components will always be zero.
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Determine The Force Between Qa=12nC And Qb=8nC If The Distance Between Them Is 0.15 M.
The force between the charges is 1.44 x 10^(-4) N. According to Coulomb's Law, the force between two point charges is given by the formula: F = (k * |Qa * Qb|) / r^2.
The force between two point charges, Qa = 12 nC and Qb = 8 nC, separated by a distance of 0.15 m, can be calculated using Coulomb's Law.
According to Coulomb's Law, the force between two point charges is given by the formula:
F = (k * |Qa * Qb|) / r^2
where:
F is the magnitude of the force between the charges,
k is the electrostatic constant (k ≈ 8.99 x 10^9 Nm^2/C^2),
|Qa * Qb| represents the product of the magnitudes of the charges, and
r is the distance between the charges.
Plugging in the values, we get:
F = (8.99 x 10^9 Nm^2/C^2) * |(12 x 10^-9 C) * (8 x 10^-9 C)| / (0.15 m)^2
Therefore, the force between the charges is 1.44 x 10^(-4) N.
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speed direction mph (b) What If? In what direction (in degrees south of east) should the plane aim to move so that its net speed is eastward despite the wind? - south of east (c) What would be the speed (in mph ) of the plane relative to the ground in this case? mph
To achieve a net eastward speed despite the wind, the plane should aim to move in a direction south of east. This means the plane needs to offset the wind's westward component by adjusting its course to the east.
By flying south of east, the plane's velocity vector will have both an eastward and southward component. To determine the specific angle south of east, we can use trigonometry. Let's assume the angle south of east is θ. The wind's velocity vector can be represented as Vw, and the plane's velocity relative to the ground can be represented as Vp.
We can break down the wind's velocity into its eastward (Vwe) and northward (Vwn) components using the angle θ. The eastward component is given by Vwe = Vw * cos(θ), and the northward component is Vwn = Vw * sin(θ).
To achieve a net eastward speed, the plane's velocity relative to the ground in the eastward direction (Vpe) should be equal to the wind's eastward component: Vpe = Vwe.
Finally, the plane's velocity relative to the ground can be calculated as Vp = √(Vpe² + Vwn²).
For the speed of the plane relative to the ground, we only need to consider the magnitude of Vp.
Therefore, to find the direction south of east and the speed of the plane relative to the ground in this case, we would need the specific values for the wind's velocity and the desired speed of the plane.
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A flat plate solar collector absorbs 80 % of radiation of 820 W/m2 received. The top loss coefficient is 12 W/m2K. Determine the temperature of heat collection of the efficiency of collection is 0.6.
The efficiency of a flat plate solar collector can be determined using the equation:
Efficiency = (Heat collection / Radiation received)
Given that the efficiency of collection is 0.6, we can rearrange the equation to solve for Heat collection:
Heat collection = Efficiency * Radiation received
The radiation received is given as 820 W/m2. Therefore:
Heat collection = 0.6 * 820 W/m2
Heat collection = 492 W/m2
The top loss coefficient is given as 12 W/m2K, which represents the amount of heat lost from the collector's top surface per unit area for each degree Kelvin of temperature difference between the collector surface and the surrounding air.
To calculate the temperature of heat collection, we need to account for the heat loss. The heat loss can be calculated using the formula:
Heat loss = Top loss coefficient * Temperature difference
Let's assume the temperature of the collector's surface is T°C, and the temperature of the surrounding air is Ta°C.
The temperature difference is (T - Ta).
Substituting the given values:
Heat loss = 12 W/m2K * (T - Ta)
Since the collector absorbs 80% of the radiation received, the heat collection can be equated to the heat loss:
Heat collection = Heat loss
492 W/m2 = 12 W/m2K * (T - Ta)
Rearranging the equation:
(T - Ta) = 492 W/m2 / 12 W/m2K
(T - Ta) = 41 K
Assuming the surrounding air temperature, Ta, is constant, we can solve for the collector's temperature, T:
T = Ta + 41 K
Therefore, the temperature of heat collection is Ta + 41 K.
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A charge of + 11 nC is placed on the x aves at x - 2.8 m, and a charge of -16 nC is placed at x=−6.8 m. What is the magnitude of the electic held at the ongin? Give your answer to one decimal place.
The magnitude of the electric field at the origin is 1.98
The value of the electric field at the origin due to a + 11 nC charge placed at x= - 2.8 m and a -16 nC charge at x = - 6.8 m can be found out using Coulomb's Law.
Coulomb's law states that the electric force between two point charges is proportional to the product of the magnitudes of each charge and inversely proportional to the square of the distance between them.
The force acts along the line joining the charges and is repulsive if the charges have the same sign and attractive if they have opposite signs.
The proportionality constant is known as the Coulomb constant (k), which is 9 x 109 N m2/C2.
Coulomb's law can be expressed as:
F = k q1q2/r2 where, F is the force between two charges (N),q1 and q2 are the magnitudes of the charges (C),r is the distance between the two charges (m),k is the Coulomb constant (9 x 109 N m2/C2)
Given,
q1 = +11 nC
= 11 x 10-9
Cq2 = -16 nC
= -16 x 10-9
Cr = distance between the two charges = (-6.8 + 2.8) m = 4 m
Substituting these values into Coulomb's law, we get,
F = k q1q2/r2F
= 9 x 109 x 11 x 10-9 x (-16) x 10-9 /42F
= - 1.98 x 10-3 N
This is the force between the two charges. To find the electric field at the origin, we need to divide this force by the charge at the origin, which is assumed to be +1 C since only magnitude is required.
E = F/qE
= -1.98 x 10-3/1E
= -1.98 x 10-3 NC-1
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A solar concentrator uses mirrors to increase the intensity of sunlight to heat transfer fluid and then use it to generate electricity. Each mirror of the concentrator has a radius of 0.21m. Three mirrors focus light in a spot with an area of 200 cm2 on the receiver. The intensity of the incident light is 1390 W/m2. What is the intensity of the light on the receiver?
Express your answer with the appropriate units.
Given data:
Radius of each mirror = 0.21 m
Area of the spot on the receiver = 200 cm²
Intensity of the incident light = 1390 W/m²
Let's first convert the area of the spot on the receiver from cm² to m²:1 cm² = (1/100)² m²= 0.0001 m²
Therefore, area of the spot on the receiver = 200 cm²= 200 × 0.0001 m²= 0.02 m²
Now, the total area that the three mirrors are focusing the light on the receiver = 0.02 m²
Intensity is the power received per unit area.
Intensity of the light on the receiver= Power/Area= 1390 W/m² × 0.02 m²= 27.8 W
Thus, the intensity of the light on the receiver is 27.8 W/m².
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The coefficient of static friction between a book's cover and the wall is 0.628. If the book's mass is 1.46 kg, what is the minimum force you need to exert to hold the book against the wall without it slipping? F
g
=mg f
s
=μ
s
N 1
st
Law equilibrium: a=0, so ΣF
x
=0,ΣF
y
=0 Answer: F=22.8 N
The minimum force required to hold the book against the wall without it slipping is 9.1 N.
The book is resting on the wall, and we need to find the minimum force required to hold it against the wall without it slipping. The gravitational force acting on the book is m*g.
The static friction force is given by f_s = μ_s * N, where μ_s is the coefficient of static friction between the book cover and the wall and N is the normal force acting on the book (equal to the weight of the book since it is resting on the wall).
Therefore, to find the minimum force required to hold the book against the wall without it slipping, we need to find the force that is equal in magnitude and opposite in direction to the static friction force.
This force is given by:
F = f_s
= μ_s * N
= μ_s * m * g
Substituting the given values, we get:
F = 0.628 * 1.46 kg * 9.8 m/s^2
= 9.1 N.
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A sailboat race course consists of four legs, defined by the displacement vectors A,B,C, and D, as the drawing indicates. The magnitudes of the first three vectors are A=2.90 km,B=5.50 km, and C=4.90 km. The finish line of the course coincides with the starting line. Using the data in the drawing, find (a) the distance of the fourth leg and (b) the angle θ. (a) Number Units (b) Number Units
The fourth leg of a sailboat race course is to be found in the displacement vectors A, B, C, and D. The magnitudes of the first three vectors are A = 2.90 km, B = 5.50 km, and C = 4.90 km. It is known that the finish line of the course coincides with the starting line, and we are to find (a) the distance of the fourth leg and (b) the angle θ.
(a) The distance of the fourth leg: Vector addition of A, B, and C gives the final vector that represents the distance covered in the first three legs of the racecourse. Thus, we can add them as follows: ABCD is a closed vector polygon since the finish line coincides with the starting line. Therefore, the vector representing the fourth leg (vector D) can be determined as
Vector D = - (Vector A + Vector B + Vector C)
Note: The negative sign indicates that the direction of the vector is opposite to that of the other vectors. By substituting the given magnitudes, we get the magnitude of vector D:
|Vector D| = |-(Vector A + Vector B + Vector C)|= |-(2.90 km + 5.50 km + 4.90 km)|= |-13.30 km|= 13.30 km
Therefore, the distance of the fourth leg is 13.30 km. (b) The angle θ:The diagram below shows the components of the vectors. Using the components shown above, we can calculate the angle θ. Thus, tanθ = Y/X, where Y = (1.30 km - 1.20 km) = 0.1 km
X = (3.80 km - 2.40 km) = 1.4 km
Tan θ = (0.1 km) / (1.4 km) = 0.0714
Θ = tan⁻¹ (0.0714) = 4.09°
Thus, the angle θ is 4.09°. We have been given the magnitudes of the first three displacement vectors A, B, and C as 2.90 km, 5.50 km, and 4.90 km, respectively, that define a sailboat race course consisting of four legs. We need to find the distance of the fourth leg and the angle θ that it makes with the east direction of the starting line. We can find the distance of the fourth leg by taking the vector sum of the first three vectors and then subtracting it from the starting position. Thus, the magnitude of the fourth vector is |-13.30 km| = 13.30 km. To find the angle θ, we can use the components of the vectors, and calculate tanθ = Y/X. The value of θ is calculated to be 4.09°. Therefore, the distance of the fourth leg is 13.30 km, and the angle θ is 4.09°.
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Steam is generated in the boiler of a cogeneration plant at 600psia and 650
∘
F at a rate of 40lbm/s. The plant is to produce power while meeting the process steam requirements for a certain industrial application. One-third of the steam leaving the boiler is throttled to a pressure of 120 psia and is routed to the process heater. The rest of the steam is expanded in an isentropic turbine to a pressure of 120 psia and is also routed to the process heater. Steam leaves the process heater at 240
∘
F. Neglect the pump work. Determine the rate of process heat supply. Use steam tables. The rate of process heat supply is Btu/s.
The rate of process heat supply is 33,020 Btu/s.
Given: The pressure (P1) = 600 psia
The temperature (T1) = 650°F
The mass flow rate (m) = 40 lbm/s
The pressure (P2) = 120 psia
The temperature (T2) = 240°F
To determine: The rate of process heat supply. In order to determine the rate of process heat supply, let us first determine the enthalpies of the steam at the initial and final states.
The enthalpy of the steam at state 1 (h1) can be determined using the steam tables. Using the tables, we get h1 = 1,619.1 Btu/lbm.
The enthalpy of the steam at state 2 (h2) can also be determined using the steam tables. Using the tables, we get h2 = 1,136.3 Btu/lbm.
The enthalpy of the steam leaving the process heater is h3 = hf + x*hfg
where hf and hfg are the enthalpy of water and enthalpy of vaporization respectively. The enthalpy of water at 240°F can be determined from the steam tables.
Using the tables, we get hf = 46.33 Btu/lbm.
The enthalpy of vaporization at 240°F can be determined from the steam tables. Using the tables, we get
hfg = 946.2 Btu/lbm.
The quality (x) of the steam at state 3 can be determined as
x = (h3 - hf)/hfg
= (1,151 - 46.33)/946.2
= 1.1429
From steam tables, the enthalpy of steam at 120 psia and 600°F is 1,356.4 Btu/lbm.
The enthalpy of steam at 120 psia and 240°F is 1,194.9 Btu/lbm.
The enthalpy drop in the turbine is
h1 - h2 = 1,619.1 - 1,136.3
= 482.8 Btu/lbm.
The enthalpy drop in the throttling valve is
h2 - h3 = 1,136.3 - 1,151
= -14.7 Btu/lbm.
The heat supplied to the process is m*(h3 - h2)m
= 40 lbm/sh3 - h2
= (hf + x * hfg) - h2
= hf + x * hfg - h2
= 46.33 + 1.1429 * 946.2 - 1,194.9
= 825.5 Btu/lbm
Q = m*(h3 - h2)= 40 * 825.5
= 33,020 Btu/s
Therefore, the rate of process heat supply is 33,020 Btu/s.
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Consider a parallel plate capacitor having plates of area 0.000475 m2 that are separated by 0.055 mm of Teflon. The dielectric constant of Tefion is x=2.1. a find the capacitance in F. C=
A parallel plate capacitor has plates of area 0.000475 m2 that are separated by 0.055 mm of Teflon. The capacitance of the parallel plate capacitor is approximately 2.33 picofarads (pF).
The capacitance (C) of a parallel plate capacitor can be calculated using the formula:
C = (ε₀ * εᵣ * A) / d
Where:
ε₀ is the permittivity of free space (8.85 x [tex]10^{-12[/tex] F/m),
εᵣ is the relative permittivity (dielectric constant) of the material,
A is the area of the plates,
d is the separation distance between the plates.
Given:
Area of the plates (A) = 0.000475 m²,
Separation distance (d) = 0.055 mm = 0.055 x [tex]10^{-3[/tex] m,
Dielectric constant (εᵣ) = 2.1.
Substituting the values into the formula, we get:
C = (8.85 x [tex]10^{-12[/tex] F/m * 2.1 * 0.000475 m²) / (0.055 x [tex]10^{-3[/tex] m)
Simplifying the equation:
C = 8.85 x [tex]10^{-12[/tex] F/m * 2.1 * 0.000475 m² / (0.055 x [tex]10^{-3[/tex] m)
C ≈ 2.33 x [tex]10^{-12[/tex] F
Therefore, the capacitance of the parallel plate capacitor is approximately 2.33 picofarads (pF).
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If a proton is accelerated from rest through a potential difference of 7000 V, what is its resulting kinetic energy in eV?
The kinetic energy of the proton accelerated from rest through a potential difference of 7000 V is 70,000 eV.
To find the kinetic energy of a proton accelerated from rest through a potential difference of 7000 V, we will use the formula of energy which is:
K.E = qV
where K.E = Kinetic Energy
q = Charge
V = Potential difference
Here, V = 7000 V and
the charge of a proton is q = 1.6 x 10^-19 C
Therefore, K.E = 1.6 x 10^-19 C × 7000
V= 11.2 x 10^-16 J
Now, we need to convert the joules into electron volts (K.E in eV) to get our final answer.
Here,1 eV = 1.6 x 10^-19 J
Therefore,11.2 x 10^-16 J ÷ 1.6 x 10^-19 J/eV= 70,000 eV
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A series RLC circuit is connected to a 4.00 kHz oscillator with a peak voltage of 3.50 V. It consists of a 0.300 mH inductor, a 230. nF capacitor, and a 14.0 Ω resistor. If ε = ε0, what is the instantaneous current i?
the instantaneous current, i, is given by:i = (0.250)cos(25,133t - 1.15°).
In order to find the instantaneous current, i, the following formula can be used:i = Imaxcos(ωt + φ)
whereImax =[tex]Vpeak/Rω = 2πfφ = tan^-1((ωL-1/ωC)/R)[/tex]
The value of ω can be found by using the formulaω = 2πf
Given that the frequency is 4.00 kHz,ω = 2π(4.00 × 10^3)ω = 25,133 rad/sThe value of Imax can be found by using the formulaI
max = Vpeak/RI
max = 3.50/14.0Imax = 0.250 A
The value of φ can be found by using the formulaφ = [tex]tan^-1((ωL-1/ωC)/R)φ = tan^-1(((25,133)(0.300 × 10^-3) - 1/(25,133)(230 × 10^-9))/14.0)φ = -1.15°[/tex]
The instantaneous current can be found by using the formula:i = Imaxcos(ωt + φ)i = (0.250)cos(25,133t - 1.15°)
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2a. You are standing 3 meters away from a computer speaker that emits a 150 Hz sound with a power of 1.13 x10^-8 Watt. How loud is this sound at your location in dB?
2b. Can an average person hear this 1.13 x10^-8 Watt, 150 Hz sound at 3 m away?
2a. The sound emitted by the computer speaker at a distance of 3 meters has a loudness of approximately 76.9 dB.
2b. Whether they can hear this specific sound depends on their individual hearing thresholds.
2a. To calculate the loudness of the sound in dB, we can use the formula for sound intensity level:
L = 10 * log₁₀(I/I₀)
where L is the sound intensity level in decibels (dB), I is the sound intensity in watts per square meter (W/m²), and I₀ is the reference intensity of 10⁻¹² W/m².
First, let's calculate the sound intensity at your location. The sound intensity decreases with distance according to the inverse square law, so we can use the equation:
I = (P * A) / (4 * π * r²)
where I is the sound intensity, P is the power, A is the surface area of a sphere (4 * π * r²) with radius r, and r is the distance from the source.
Plugging in the values, we have:
I = (1.13 x 10⁻⁸) / (4 * π * 3²)
≈ 1.27 x 10⁻¹⁰ W/m²
Now, we can calculate the sound intensity level:
L = 10 * log₁₀(1.27 x 10⁻¹⁰ / 10⁻¹²)
≈ 76.9 dB
Therefore, the sound at your location has a loudness of approximately 76.9 dB.
2b. An average person can hear sounds at a wide range of frequencies and intensities, but whether they can hear this specific sound depends on their individual hearing thresholds.
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how fast is a 30 gram ball going at 45 seconds if it moves in centripetal motion with a radius of 1.5m?the ball breaks loose from the centripetal motion at 45 seconds at 2m above ground and flings a distance of 10m before landing on the groundwhat is the ball's centripetal acceleration at 45 seconds and velocity before it hits the ground?
The ball's centripetal acceleration at 45 seconds is approximately 0.00741 m/s^2, and its velocity before it hits the ground is approximately 15.63 m/s. To find the speed of the ball at 45 seconds, we need to calculate its centripetal acceleration first.
To find the speed of the ball at 45 seconds, we need to calculate its centripetal acceleration first.
Given:
Mass of the ball, m = 30 grams = 0.03 kg
The radius of motion, r = 1.5 m
Time, t = 45 seconds
Centripetal acceleration (a) can be calculated using the formula:
a = v^2 / r
where v is the velocity of the ball.
To find the velocity, we can use the relation:
v = a * t
Substituting the values:
a = v^2 / r
a = (a * t)^2 / r
r * a = a^2 * t^2
a = (r * a / t^2)
Simplifying the equation, we find:
a = r / t^2
Now, let's calculate the centripetal acceleration at 45 seconds:
a = 1.5 m / (45 s)^2
a ≈ 0.00741 m/s^2
Next, to find the velocity before the ball hits the ground, we need to calculate the time it takes for the ball to travel the distance of 10m horizontally after breaking loose from the centripetal motion.
Given:
Distance traveled horizontally, d = 10 m
The vertical distance from the ground, h = 2 m
The time taken to travel the horizontal distance can be found using the horizontal component of the initial velocity:
d = v_horizontal * t_horizontal
Since the ball is in free fall, we can use the equation:
h = (1/2) * g * t^2
Solving for t, we get:
t = sqrt((2 * h) / g)
Substituting the given values:
t = sqrt((2 * 2 m) / (9.8 m/s^2))
t ≈ 0.64 s
Now, we can find the horizontal velocity (v_horizontal) using the equation:
v_horizontal = d / t_horizontal
Substituting the given values:
v_horizontal = 10 m / 0.64 s
v_horizontal ≈ 15.63 m/s
Therefore, the ball's centripetal acceleration at 45 seconds is approximately 0.00741 m/s^2, and its velocity before it hits the ground is approximately 15.63 m/s.
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Monochromatic light of wavelength 612 nm falls on a sit. Part A If the angle between the first two bright fringes on ether sice of the central maximum is 33
∘
, estimate the slit width. Express your answer to two significant figures and include the appropriate units.
The estimated slit width is approximately 0.15 mm.
When monochromatic light of wavelength 612 nm falls on a slit, it undergoes diffraction and forms a pattern of bright and dark fringes on a screen. The angle between the first two bright fringes on each side of the central maximum is given as 33 degrees.
For a single slit, the angular position of the bright fringes can be related to the slit width (a) and the wavelength (λ) of the light using the formula: sinθ = (mλ) / a, where θ is the angle, m is the order of the fringe (m = 1 for the first bright fringe), and λ is the wavelength.
In this case, we know the wavelength (612 nm) and the angle (33 degrees) for the first bright fringe. Rearranging the formula, we can solve for the slit width (a).
Using the given values, we have sin(33°) = (1 * 612 nm) / a. Rearranging the equation, we get a = (1 * 612 nm) / sin(33°). Converting the wavelength to meters and using the trigonometric function, we find a ≈ 0.15 mm. Therefore, the estimated slit width is approximately 0.15 mm.
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A car slows down with an acceleration that has a magnitude of 3.0 m/s
2
. While doing so, it travels 107 m in the +x direction and ends up with a velocity of +4.5 m/s. What was the car's initial velocity?
The initial velocity of the car was 25 m/s. However, the negative value obtained in the calculation indicates that the velocity is in the opposite direction of the displacement covered (+x direction).
Given values; The magnitude of acceleration, a = 3.0 m/s²
Displacement covered, x = 107 m
Initial velocity, u = ?
Final velocity, v = +4.5 m/s
Using the kinematic equation; v² = u² + 2ax
Where u is the initial velocity, v is the final velocity, a is the acceleration, and x is the displacement covered. Substitute the known values into the equation; v² = u² + 2ax4.5² = u² + 2(3.0)(107)20.25 = u² + 642u² = 20.25 - 642u² = -623.75u = √(-623.75)u = 25 m/s
Therefore, the initial velocity of the car was 25 m/s. However, the negative value obtained in the calculation indicates that the velocity is in the opposite direction of the displacement covered (+x direction). This implies that the car was initially moving in the -x direction with a velocity of 25 m/s and then slowed down and eventually ended up moving in the +x direction with a velocity of +4.5 m/s.
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