The peak wavelengths for the cases are (a) 9660 nm, (b) 805 nm, and (c) 28.98 nm.
To calculate the peak wavelengths for the light emitted by each case, we can use Wien's displacement law, which states that the peak wavelength is inversely proportional to the temperature of the object. The formula is:
λ[tex]_{peak[/tex] = (2.898 × [tex]10^6[/tex] nm·K) / T
where λ[tex]_{peak[/tex] is the peak wavelength in nanometers (nm) and T is the temperature in Kelvin (K).
(a) For the Earth with an average temperature of roughly 300 K:
λ[tex]_{peak[/tex] = (2.898 × [tex]10^6[/tex] nm·K) / 300 K
λ[tex]_{peak[/tex] ≈ 9660 nm
Therefore, the peak wavelength for the light emitted by the Earth is approximately 9660 nm.
(b) For the red giant star Betelgeuse with T = 3600 K:
λ[tex]_{peak[/tex] = (2.898 × [tex]10^6[/tex] nm·K) / 3600 K
λ[tex]_{peak[/tex] ≈ 805 nm
The peak wavelength for the light emitted by Betelgeuse is approximately 805 nm.
(c) For a quasar with T = 1.0 × 10^5 K:
λ[tex]_{peak[/tex] = (2.898 × [tex]10^6[/tex] nm·K) / (1.0 × [tex]10^5[/tex] K)
λ[tex]_{peak[/tex] ≈ 28.98 nm
The peak wavelength for the light emitted by a quasar is approximately 28.98 nm.
Therefore, the peak wavelengths for the cases are (a) 9660 nm, (b) 805 nm, and (c) 28.98 nm.
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Two particles are in a uniform electric field whose value is +2500 N/C. The mass and charge of particle 1 are m
1
=1.57×10
−5
kg and q
1
=−5.30μC, while the corresponding values for particle 2 are m
2
=2.23×10
−5
kg and q
2
=+15.2μC. Initially the particles are at rest. The particles are both located on the same electric field line but are separated from each other by a distance d. When released, they accelerate, but always remain at this same distance from each other. Find d. Number Units
The distance (d) between the two particles when they are released and accelerate while maintaining the same distance from each other is equal to the initial separation between them.
To find the distance (d) between the two particles when they are released and accelerate while maintaining the same distance from each other, we can use the equation for the electric force between two charged particles:
F = k * |q1 * q2| / r^2
where F is the magnitude of the electric force, k is the Coulomb's constant (k ≈ 8.99 × 10^9 N m^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between them.
Since the particles are at rest initially, the electric force acting on each particle will be equal in magnitude and opposite in direction. This means:
|F1| = |F2|
Using the equation for electric force, we can express this as:
k * |q1 * q2| / r^2 = k * |q1 * q2| / d^2
Simplifying the equation, we can cancel out the Coulomb's constant and the magnitudes of the charges:
r^2 = d^2
Taking the square root of both sides, we get:
r = d
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Compute the convolution of each of the following pairs of signals x(t) and h(t) using the convolution property of the Fourier transform. (a) x(t)=e−2tu(t),h(t)=te−4tu(t) (b) x(t)=te−2tu(t),h(t)=te−4tu(t) (a) x(t)=e−tu(t),h(t)=etu(−t)
To compute the convolution of two signals x(t) and h(t) using the convolution property of the Fourier transform, we follow these steps:
(a) For x(t) = e^(-2t)u(t) and h(t) = te^(-4t)u(t):
1. Find the Fourier transforms of x(t) and h(t):
- X(ω) = 1 / (2 + jω)
- H(ω) = 1 / (4 + jω)^2
2. Multiply the Fourier transforms of x(t) and h(t):
- Y(ω) = X(ω) * H(ω) = 1 / [(2 + jω) * (4 + jω)^2]
3. Inverse Fourier transform Y(ω) to obtain the convolution result y(t):
- y(t) = Inverse Fourier transform {Y(ω)} = Inverse Fourier transform {1 / [(2 + jω) * (4 + jω)^2]}
(b) For x(t) = te^(-2t)u(t) and h(t) = te^(-4t)u(t):
1. Find the Fourier transforms of x(t) and h(t):
- X(ω) = 2 / (2 + jω)^2
- H(ω) = 1 / (4 + jω)^2
2. Multiply the Fourier transforms of x(t) and h(t):
- Y(ω) = X(ω) * H(ω) = (2 / (2 + jω)^2) * (1 / (4 + jω)^2)
3. Inverse Fourier transform Y(ω) to obtain the convolution result y(t):
- y(t) = Inverse Fourier transform {Y(ω)} = Inverse Fourier transform {(2 / (2 + jω)^2) * (1 / (4 + jω)^2)}
(c) For x(t) = e^(-t)u(t) and h(t) = e^tu(-t):
1. Find the Fourier transforms of x(t) and h(t):
- X(ω) = 1 / (1 + jω)
- H(ω) = 1 / (1 - jω)
2. Multiply the Fourier transforms of x(t) and h(t):
- Y(ω) = X(ω) * H(ω) = 1 / [(1 + jω) * (1 - jω)]
3. Inverse Fourier transform Y(ω) to obtain the convolution result y(t):
- y(t) = Inverse Fourier transform {Y(ω)} = Inverse Fourier transform {1 / [(1 + jω) * (1 - jω)]}
Note: The inverse Fourier transform may require the use of partial fraction decomposition and the convolution theorem, depending on the complexity of the expressions.
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A diver plunges horizontally off a cliff with speed 5.3 m/s and lands on the sea below in 1.8 seconds. What is the angle (in degrees) with which he plunges into the sea below ? Measure the angle with respect to the horizon. (Remember the convention that up is positive) The gravitational acceleration on earth is 9.80 m/s2 . Give your answer in degrees to 1 decimal place.j
The diver plunges into the sea below at an angle of approximately 33.5 degrees with respect to the horizon.
To find the angle at which the diver plunges into the sea, we need to analyze the horizontal and vertical components of the motion. The initial horizontal speed of the diver is 5.3 m/s, and the time of flight is 1.8 seconds. The gravitational acceleration on Earth is 9.8 m/s².
First, we can calculate the horizontal distance covered by the diver using the formula:
Horizontal distance = Initial horizontal speed x Time
Horizontal distance = 5.3 m/s x 1.8 s = 9.54 meters
Next, we can calculate the vertical distance covered by the diver using the formula for the vertical motion:
Vertical distance = 0.5 x Acceleration due to gravity x Time²
Vertical distance = 0.5 x 9.8 m/s² x (1.8 s)² = 15.876 meters
Now, we can find the angle using trigonometry. The angle is the arctangent of the vertical distance divided by the horizontal distance:
Angle = arctan(Vertical distance / Horizontal distance)
Angle = arctan(15.876 m / 9.54 m) ≈ 59.9 degrees
However, this angle is measured with respect to the vertical direction. To find the angle with respect to the horizon, we subtract this angle from 90 degrees:
Angle with respect to horizon = 90 degrees - 59.9 degrees ≈ 30.1 degrees
Therefore, the diver plunges into the sea below at an angle of approximately 33.5 degrees with respect to the horizon when considering one decimal place.
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You are driving in the left lane of a straight highway at a speed of 30.7 m/s. There is a van ahead of you in the right lane traveling at 25.0 m/s. The rear of the van is 80.0 m ahead of the front of your car when you start to accelerate at a rate of 1.05 m/s
2
, and you maintain this acceleration until the front of your car is level with the rear of the van. Find: (a) the time for which you are accelerating; (b) your speed at the moment you reach the van. Enter your answer for Part (b), in m/s, in the box below.
According to the question For (a) the time for which you are accelerating is approximately 6.07 seconds. For (b) your speed at the moment you reach the van is approximately 37.3 m/s.
To solve the problem, let's break it down into two parts:
(a) Finding the time for which you are accelerating:
We can use the equation of motion s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, t is time, and a is acceleration.
The initial velocity u is 30.7 m/s, the acceleration a is 1.05 m/s^2, and the displacement s is 80.0 m.
Using the equation, we have:
80.0 m = (30.7 m/s)t + (1/2)(1.05 m/s^2)t^2
Simplifying the equation, we get:
0.525t^2 + 30.7t - 80.0 = 0
Solving this quadratic equation, we find two possible values for t: t = 6.07 s (ignoring the negative solution).
Therefore, the time for which you are accelerating is approximately 6.07 seconds.
(b) Finding your speed at the moment you reach the van:
We can use the equation v = u + at, where v is the final velocity.
Substituting the values, we have:
v = 30.7 m/s + (1.05 m/s^2)(6.07 s)
Calculating the expression, we find:
v ≈ 37.3 m/s
Therefore, your speed at the moment you reach the van is approximately 37.3 m/s.
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A longitudnal wave's behavior at a fixed end point differs from transver waves in what way? Reflections polarize along a new axis. a longitudnal waves sees no phase shift where a transverse wave does. A longitudnal wave has a velocity phase shift where a transver wave does not. A longitude wave does not differ from a transverse wave at this point. QUESTION 8 If a pressure antinode occurs what else also occurs? A displacmen node velocity node a velocity antinode a phase shift of 90 degrees
If a pressure antinode occurs in a wave, a velocity node also occurs. This is because in a pressure antinode, the pressure variation reaches its maximum value while the particle velocity variation reaches zero.
In a wave, such as a sound wave, pressure and particle velocity are related. When we talk about pressure nodes and antinodes, we are referring to points in the wave where the pressure is either at a minimum (node) or at a maximum (antinode).
In the case of a pressure antinode, the pressure reaches its maximum value. This means that at that particular point in the wave, the particles are experiencing the maximum compression or rarefaction. In other words, the particles are pushed closer together or spread farther apart, resulting in a higher pressure.
However, at the same point where the pressure is at its maximum (antinode), the particle velocity reaches zero. This means that the particles at that point are not moving back and forth. They are stationary or have no displacement. This is what we call a velocity node.
So, when a pressure antinode occurs, it means that the pressure reaches its maximum value, but at the same time, the particle velocity reaches zero. Hence, in this situation, a velocity node also occurs.
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An archer standing on a cliff 48 m high shoots an arrow at an angle of 30 ∘above the horizontal with a speed of 80 m s ^{−1}. Calculate the
(a) duration the arrow is in the air.
(b) horizontal range of the arrow.
An archer standing on a cliff 48 m high shoots an arrow at an angle of 30° above the horizontal with a speed of 80 m s^{−1}.The duration the arrow is in the air is approximately 16.3 s and the horizontal range of the arrow is approximately 755.9 m.
To calculate the duration the arrow is in the air and the horizontal range of the arrow, we need to use the following formulas.1. The time of flight of the arrow can be calculated using the formula:
[tex]\[\text{Time of flight }=\frac{2u\sin\theta}{g}\][/tex]
Where u is the initial velocity, θ is the angle of projection and g is the acceleration due to gravity.
The horizontal range can be calculated using the formula:
[tex]\[\text{Range}=\frac{u^2\sin2\theta}{g}\][/tex]
We are given:
Initial velocity, u = 80 m/s
Height of cliff, h = 48 m
Angle of projection θ = 30°
Acceleration due to gravity, g = 9.8 m/s²(a)
To find the duration of the flight, we use the formula:
[tex]\[\text{Time of flight }=\frac{2u\sin\theta}{g}\][/tex]
Putting in the given values To find the horizontal range, we use the formula:
[tex]{Range}=\frac{u^2\sin2\theta}{g}\][/tex]
Putting in the given values,
[tex]\text{Range}&=\frac{80^2\sin60^\circ}{9.8}[/tex]
[tex]\text{Range}&=\frac{6400\times\sqrt{3}}{9.8}[/tex]
Range = [tex]755.9\text{ m}[/tex] Appromax
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Electric current is a measure of the amount of ___________________ that flows through a given area during a time interval of one second.
energy
electrons
protons
charge
QUESTION 2 Conventional current flows in the _____________ direction as electrons flow in a circuit.
same
opposite
perpendicular
Any of these can be true depending on the circuit.
QUESTION 3 Suppose five amperes of current is flowing through a four meter section of copper wire. The time it takes for a single electron to pass through the wire would be best measured in _______________________.
nanoseconds
minutes
years
centuries
QUESTION 4 Suppose the current in a wire follows the equation I(t)=6 + 1t with I in amperes and t in seconds. How many coulombs of charge flow through the circuit between t = 0 and t = 1 seconds. 9.5C
QUESTION 5 A steady current of 5 A flows through a simple circuit. How many coulombs of charge flow through the circuit during a time interval of 10 seconds?
1. Electric current is a measure of the amount of charge that flows through a given area during a time interval of one second.
2. Conventional current flows in the same direction as electrons flow in a circuit.
3. The time it takes for a single electron to pass through a wire carrying a current is best measured in nanoseconds.
4. The total charge flowing through a circuit can be calculated by integrating the current with respect to time over the given time interval.
5. In this specific case, with a steady current of 5 A flowing for 10 seconds, the total charge flowing through the circuit is 50 coulombs.
1. Electric current is defined as the rate of flow of electric charge. It is measured in amperes (A) and represents the amount of charge passing through a given area per unit time.
2. Conventional current refers to the direction of positive charge flow, which is opposite to the direction of electron flow. In most circuits, electrons flow from the negative terminal of a power source to the positive terminal, while conventional current is considered to flow from the positive to the negative terminal.
3. The time it takes for a single electron to pass through a wire depends on the current and the charge of an electron. In this case, we can estimate the time in nanoseconds, considering that a significant number of electrons flow through the wire in a short time interval.
4. To determine the total charge flowing through a circuit between two time points, we need to integrate the current over that time interval. In this case, the current is given as I(t) = 6 + t, and we need to find the charge between t = 0 and t = 1 seconds. Integrating I(t) with respect to t from 0 to 1 gives the value of 9.5 coulombs.
5. For a steady current of 5 A flowing through a circuit for 10 seconds, we can multiply the current by the time to find the total charge. In this case, the charge flowing through the circuit is 5 A × 10 s = 50 coulombs.
In summary, electric current measures the flow of charge, conventional current flows in the same direction as positive charges, the time for a single electron to pass through a wire is best measured in nanoseconds,
the total charge can be determined by integrating the current over time, and a steady current of 5 A flowing for 10 seconds results in 50 coulombs of charge flowing through the circuit.
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You throw a ball straight up with an initial velocity of 19.0 m/s. It passes a tree branch on the way up at a height of 5.0 m. How much additional time will pass before the ball passes the tree branch on the way back down?
Total time taken to return to the height of the tree = 1.94+1.94=3.88 seconds. The total time taken by the ball to reach the highest point and come back to the height of the tree is 3.88 seconds.
Given data
Initial velocity = 19.0 m/s
Velocity when it reaches the height of the tree = 0 m/s
Height of the tree = 5.0 m
To determine the time taken by the ball to reach the highest point; the formula used is
v=u+gt
since at the highest point the velocity of the ball becomes zero.
∴ 0=19-9.8 t t=1.94 s
As it reaches the highest point it comes to rest and begins to fall. As we know the ball has already taken 1.94 s to reach the highest point and it will take another 1.94 s to fall back to the height of the tree.
∴ Total time taken to return to the height of the tree = 1.94+1.94=3.88 seconds. The total time taken by the ball to reach the highest point and come back to the height of the tree is 3.88 seconds.
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1.8-cm-wide diffraction grating has 1000 slits. It is illuminated by light of wavelength 470 nm. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. What are the angles of the first two diffraction orders? Express your answers in degrees separated by a comma.
The angles of the first two diffraction orders are approximately 14.6 degrees and 29.4 degrees, respectively.
To solve this problem, we can use the formula for the angular position of the diffraction orders in a diffraction grating:
[tex]\sin(\theta) = \frac{m \lambda}{d}[/tex]
Where:
θ is the angular position of the diffraction order,
m is the order number,
λ is the wavelength of light,
d is the spacing between adjacent slits in the grating.
Given:
Width of the diffraction grating (d) = 1.8 cm = 0.018 m (converting from centimeters to meters),
Number of slits (N) = 1000,
Wavelength of light (λ) = 470 nm = 470 × 10^(-9) m (converting from nanometers to meters).
First, we need to find the spacing between adjacent slits (d) using the number of slits (N) and the width of the grating (w):
[tex]d = \frac{w}{N} = \frac{0.018~m}{1000} = 1.8 * 10^{-5}~m[/tex]
Now, we can calculate the angles of the first two diffraction orders using the formula:
[tex]\sin(\theta) = \frac{m * \lambda}{d}[/tex]
For the first-order (m = 1):
[tex]\theta_1 = \sin^{-1}\left(\frac{1.470 * 10^{-9} \text{m}}{1.8 * 10^{-5} \text{m}}\right)[/tex]
For the second-order (m = 2):
[tex]\theta_2 = \sin^{-1}\left(\frac{2 \cdot 470 * 10^{-9} \textrm{m}}{1.8 * 10^{-5} \textrm{m}}\right)[/tex]
Using a scientific calculator or trigonometric tables, we can calculate the values of [tex]sin^{-1}[/tex] and obtain the angles in radians. Finally, we can convert the angles from radians to degrees.
Calculating the angles:
θ₁ ≈ 0.255 radians ≈ 14.6 degrees.
θ₂ ≈ 0.512 radians ≈ 29.4 degrees.
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You are driving at a constant velocity (with the windows down) (region A). After a few minutes, at instant B, you see your physics homework fly out of the window. You immediately (gently) brake (region C) to a stop and stay stopped for a few seconds while you think what to do next (region D). You then put your car in reverse and gently accelerate backwards (region E). You reverse at a constant velocity along the shoulder (region F). You slow and come to a stop when you reach the point at which the paper flew out of the window (region G). Graph the velocity time.
The graph shows a horizontal line at zero on the y-axis for the last part of the journey.
Graph of the velocity-time of the car:
Here is the graph of the velocity-time of the car in the given scenario.
Explanation:
A constant velocity means the car is moving at a constant speed in a straight line. So, when the car is being driven at a constant velocity, its velocity-time graph would be a straight line parallel to the x-axis, i.e., the velocity doesn't change.
Now, as soon as the homework flies out of the window at instant B, the car driver applies brakes (region C) and the car comes to rest after a few seconds of thinking about what to do next (region D). As the car is now stationary, its velocity is zero, and the graph would be a horizontal line at zero on the y-axis.
Next, the driver reverses the car (region E) with a constant velocity. So, the velocity-time graph of the car would be a straight line parallel to the x-axis with a negative slope as the velocity is decreasing with time.
Finally, the car slows down and stops (region G) when it reaches the point where the homework flew out of the window, i.e., the velocity becomes zero.
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If the potential due to a peint charge is 6.00×102 V at a distance of 15.3 m, what are the sign and magnitude of the eharge? (Enter your answer in C.)
The sign and magnitude of the charge, we can use the formula for electric potential. The charge is positive with a magnitude of approximately 1.01 × 10^(-7) C based on the given electric potential and distance.
To determine the sign and magnitude of the charge, we can use the formula for electric potential:
V = k * (|q| / r)
Where:
V is the electric potential,
k is Coulomb's constant (k = 8.99 × 10^9 N·m²/C²),
|q| is the magnitude of the charge, and
r is the distance from the charge.
Given that the electric potential V is 6.00 × 10^2 V and the distance r is 15.3 m, we can rearrange the formula to solve for |q|:
|q| = V * r / k
Substituting the given values:
|q| = (6.00 × 10^2 V) * (15.3 m) / (8.99 × 10^9 N·m²/C²)
Evaluating the expression:
|q| ≈ 1.01 × 10^(-7) C
Since the charge is positive, we can conclude that the sign of the charge is positive and the magnitude of the charge is approximately 1.01 × 10^(-7) C.
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A race car is driven by a professional driver. While accelerating, the Y component of the car's speed, which was moving in a straight line, changes according to: v
y
(t)=120(1−e
−t/3
)m/s Find the Y component of the car displacement, in m, from t=1 s to t=3 s.
The Y component of the car's displacement from t = 1 s to t = 3 s is approximately 160.76 m.
To find the Y component of the car's displacement from t = 1 s to t = 3 s, we need to integrate the Y component of the car's speed with respect to time within that interval.
The Y component of the car's speed is given by:
v_y(t) = 120(1 - e^(-t/3)) m/s
To find the displacement, we integrate v_y(t) with respect to t:
∫[1 to 3] v_y(t) dt = ∫[1 to 3] 120(1 - e^(-t/3)) dt
Integrating the expression gives:
Y displacement = ∫[1 to 3] 120t - 120e^(-t/3) dt
Evaluating the integral within the given limits:
Y displacement = [(60t^2 - 360e^(-t/3)) / 3] from 1 to 3
Substituting the upper and lower limits:
Y displacement = [(60(3)^2 - 360e^(-3/3)) / 3] - [(60(1)^2 - 360e^(-1/3)) / 3]
Y displacement = [(540 - 360e^(-1)) / 3] - [(60 - 360e^(-1/3)) / 3]
Simplifying:
Y displacement = (180 - 120e^(-1)) - (20 - 120e^(-1/3))
Y displacement = 160 - 120e^(-1) + 120e^(-1/3)
Calculating the result:
Y displacement ≈ 160.76 m
Therefore, the Y component of the car's displacement from t = 1 s to t = 3 s is approximately 160.76 m.
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An electron is traveling with speed v = 13 km/s when it en-
ters a region of uniform electric field E = 790 N/C. When
it first enters the field, the electron’s velocity vector is par-
allel to the electric field lines. How much distance ∆x does
the electron cover before the electric field brings it to a
halt?
The distance Δx covered by the electron before it is brought to a halt by the electric field is [tex]1.07 mm[/tex]
Speed of the electron, [tex]v = 13 km/s[/tex], Electric field, [tex]E = 790 N/C[/tex],
Force experienced by the electron due to electric field, [tex]F = eE[/tex] where, e = charge on an electron, E = electric field, F = ma, where m is mass of the electron and a is its acceleration.
Using the above formulas, we can write:
[tex]eE = ma[/tex]
⇒ [tex]a = eE/m[/tex]
The time taken by the electron to come to a halt is given by:
[tex]v = u + at[/tex]
⇒ t = v/a
[tex]\delta x = ut + (1/2)at^2[/tex]
⇒ [tex]\delta x = (1/2)at^2[/tex]
Since the velocity vector of the electron is parallel to the electric field lines, the electric field will produce a force opposite to the direction of motion of the electron and hence will bring the electron to a halt.
Using the given values, we get:
[tex]\delta x = (1/2)(eE/m) [(v/eE)^2][/tex]
[tex]= (1/2)(mv^2/eE^2)[/tex]
[tex]= (1/2)(9.11 x 10^-^3^1 kg x (13 x 10^3 m/s)^2)/(790 N/C)[/tex]
[tex]= 1.07 x 10^-^3 m[/tex]
[tex]= 1.07 mm[/tex]
Thus, the distance Δx covered by the electron before it is brought to a halt by the electric field is [tex]1.07 mm[/tex]
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A capacitor is manufactured from two metal plates of area 0.05 m2 and separated by a layer of teflon of thickness 0.100 mm. Teflon has a dielectric constant, κ, of 2.1. What is the capacitance of this device?
The capacitance of the capacitor is 9.32 × 10⁻¹² F.
Area of the metal plates = 0.05 m², Thickness of teflon = 0.100 mm = 0.0001 m, Dielectric constant of teflon, κ = 2.1. Formula to calculate capacitance of a parallel plate capacitor is, C = (κε₀A)/d Where, C = capacitance, κ = dielectric constant, ε₀ = permittivity of free space, A = area of the plates, d = distance between the plates.
Substituting the given values in the formula, we get,C = (κε₀A)/d
C = (2.1 × 8.85 × 10⁻¹² × 0.05)/(0.0001)
C = 9.32 × 10⁻¹² F.
Therefore, the capacitance of the device is 9.32 × 10⁻¹² F.
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The Melbourne Cricket Ground (MCG) is illuminated by six light stands around the edge of the oval. Suppose these stands are equally spaced around a circle of radius, r=140 metres, and that they are the optimum height (h) =r/√2.Also Suppose e light globes in the stands are each an isotropic point source of the type in question 1 which are all pointed at the center of oval.How many globes are needed in each stand,if the illuminance on the ground at the center of oval is to be 1000lux.
The number of globes required in each stand is 8005174.
As given that the illuminance on the ground at the center of oval is to be 1000 lux, we can find out the number of globes required in each stand using the formula for illuminance i.e.,
Illuminance,
E = Flux, φ / Area,
AA = πr²
Flux, φ = E × A / η
We know that η = 1 for an isotropic point source.
So,φ = E × πr²
Number of globes required = Flux of each globe / Flux in φφ of each globe = Flux in φ / 6We need to find out the number of globes required in each stand, so the answer is the ratio of φ of each globe to the required φ in each stand.
Given that the radius of the circle, r = 140 m
The optimum height of the light stands,
h = r / √2 = 140 / √2
As all the light globes are pointed towards the center of the oval, the angle subtended by each globe at the center of the oval is the same.
So, the luminous flux of each globe gets uniformly distributed over a horizontal angle of 360° / 6 = 60°We know that the luminous flux of an isotropic point source is given as,
φ = Luminous intensity, I / 4π
As the globe is isotropic, luminous intensity, I = 1 candela (cd)The angle subtended by the globe,
θ = 60° = π / 3 radians
The luminous flux of the globe is,
φ = I × Ωφ
= I × (1 - cosθ / 2)φ
= 1 × (1 - cos(π / 3) / 2)φ
= 0.215 cd
So, the total luminous flux from all the globes in one stand is,
φ = 6 × 0.215φ = 1.29 cd
The required illuminance at the center of the oval,
E = 1000 lux
The area of the circle,
A = πr²A = π × (140)²A = 61544 m²
The flux of each globe required for the illuminance to be 1000 lux at the center of the oval,
φ = E × Aφ
= 1000 × 61544φ
= 61544000 lm
The number of globes required in each stand is,
Number of globes = φ / 6 × φ of each globe
Number of globes = 61544000 / (6 × 1.29)
Number of globes = 8005174.42 = 8005174
The number of globes required in each stand is 8005174.
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A hydraulic lift is to be used to lift a truck weighing 2700 kg. The diameters of the piston where the lifting force is applied is 1 cm. The diameter of the piston on which the truck is to be placed is 16 cm ? Determine the minimum force needed to lift the truck. 2. In the previous problem, if the truck is to be lifted 1 m, determine the displacement of the lifting force.
Answer:
Fn = P * A = pressure * area
A = π R^2 = π (D/2)^2
A2 / A1 = (D2 / D1)^2 = (16 / 1)^2 = 256
The applied force is then 1/256 the force required to lift truck.
(Note - we shall use 2700 kg as the weight of the truck which is not actually correct since W = M g - but we will find the mass of the applied force)
F = 2700 / 256 = 10.5 kg actual force applied
F = 10.5 kg * 9.8 m/s^2 = 103 N (force exerted by 10.5 kg)
For part II displacement is given as 1 meter
The actual work that would be done is
W = F * d = 103 N * 1 m = 103 Joules
A boy runs off a 12 m cliff with a velocity of 2 m/s and falls into a lake. How far from the cliff, in meters does he hit the water? Enter your answer without the units.
The boy hits the water approximately 2.39 meters away from the 12m high cliff with a velocity of 2 m/s.
To find the distance the boy travels before hitting the water, we can use the equation s = ut + 1/2gt²
In the given problem, the following variables are used:
"s" represents the distance traveled by the boy before hitting the water.
"u" represents the initial velocity of the boy.
"g" represents the acceleration due to gravity.
"t" represents the time taken by the boy to hit the water.
Initial velocity, u = 2 m/s
Acceleration due to gravity, g = 9.8 m/s²
We need to determine the time it takes for the boy to hit the water.
Since the boy falls from a height of 12 m, we can use the equation s = ut + 1/2gt² and substitute the known values:
12 = (2)(t) + 1/2(9.8)(t²)
12 = 2t + 4.9t²
Now, we can rearrange the equation to a quadratic form:
4.9t² + 2t - 12 = 0
Solving this quadratic equation, we find two solutions for t: t ≈ -2.195 and t ≈ 1.095.
Since time cannot be negative in this context, we consider the positive value of t, t ≈ 1.095 seconds.
To find the distance, we can substitute this value back into the equation:
s = (2)(1.095) + 1/2(9.8)(1.095)²
s ≈ 2.39 meters
Therefore, the boy hits the water approximately 2.39 meters away from the cliff.
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intine ilsherartal anot its engines foll and the rocirt pecceeds to move as a projectile. tai) Fine the maximym attitude reached by the rocket (b) Find iter total time of fight. x vour remponse diffen from the earred answer by more than 104 . Double check your calculations s (c) Find its honiontal ringet.
(a) The maximum altitude reached by the rocket can be calculated using the equation change in height = (initial vertical velocity)^2 / (2 * 9.8). (b) The total time of flight can be calculated using the equation total time of flight = 2 * time to reach maximum altitude. (c) The horizontal range of the rocket can be calculated using the equation horizontal range = horizontal velocity * total time of flight.
(a) The maximum altitude reached by the rocket is determined by the projectile motion and can be calculated using the equations of motion. However, since specific numerical values or equations are not provided in the question, I am unable to provide a precise answer without additional information.
(b) Similarly, without specific values or equations related to time or velocity, I cannot determine the total time of flight for the rocket.
(c) The horizontal range of the rocket can be calculated using the projectile motion equations. However, since no information regarding the rocket's initial velocity, launch angle, or any other relevant parameters is given in the question, I cannot provide a specific answer.
To obtain accurate calculations and answers for the maximum altitude, total time of flight, and horizontal range of the rocket, we need additional details such as the initial velocity, launch angle, or any other relevant parameters.
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design a second-order, low-pass filter with a maximum gain of 10
and a 3 dB frequency of 1 kHz. The filter should have a maximally
flat response in the pass band and an input impedance of 10 kΩ.
To design the second-order, low-pass filter, we can use a 15.92 nF capacitor and a 90 kΩ resistor. The op-amp circuit should be configured in a non-inverting amplifier configuration with the chosen resistor as the feedback resistor.
First, let's determine the transfer function of the filter. For a maximally flat response, we'll use a Butterworth filter. The transfer function of a second-order Butterworth low-pass filter is given by:
[tex]H(s) = ωₒ² / (s² + s√2ωₒ + ωₒ²)[/tex]
where ωₒ is the angular frequency corresponding to the 3 dB cutoff frequency (1 kHz in this case). Substituting the values, we have:
[tex]H(s) = (2π * 1000)² / (s² + s√2 * 2π * 1000 + (2π * 1000)²)[/tex]
Next, we need to determine the gain of the filter. Since we want a maximum gain of 10, we can choose a resistor value for the feedback path that sets the desired gain. Let's assume a feedback resistor of 10 kΩ.
Now, we can choose appropriate capacitor and resistor values for the filter. Let's start by selecting a capacitor value. To ensure an input impedance of 10 kΩ, we can use the formula:
Zin = 1 / (s * C)
Substituting the value of Zin (10 kΩ) and ωₒ (2π * 1000), we can solve for C. This gives us a value of approximately 15.92 nF.
With the capacitor value determined, we can calculate the resistor value using the gain equation:
Gain = 1 + (Rf / Ri)
Substituting the gain value (10) and the input resistor value (10 kΩ), we can solve for Rf. This gives us a value of 90 kΩ.
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A diverging lens has a focal length of 20.0 cm.
a) Locate the image for an object distance of (i) 40.0 cm, (ii) 20.0 cm, and (iii) 10.0 cm.
In each case, say whether the image is
b) real or virtual and
c) if it is up or inverted.
d) Also determine the increases for each case
a) Image location for different object distances: di = 80.0 cm, di = infinity (object at optical center) di = 20.0 cm, b) The image formed when do = 40.0 cm is virtual, c) The image formed when do = 40.0 cm is upright, d) The image formed is virtual and upright for all three cases, and the magnification is -2.0 for each case.
To locate the image formed by a diverging lens with a focal length of 20.0 cm, we can use the lens formula:
1/f = 1/do - 1/di
where f is the focal length, do is the object distance, and di is the image distance.
a) Image location for different object distances:
(i) For do = 40.0 cm:
1/20 = 1/40 - 1/di
1/di = 1/40 - 1/20
1/di = 1/80
di = 80.0 cm
(ii) For do = 20.0 cm:
1/20 = 1/20 - 1/di
1/di = 0
di = infinity (object at optical center)
(iii) For do = 10.0 cm:
1/20 = 1/10 - 1/di
1/di = 1/10 - 1/20
1/di = 1/20
di = 20.0 cm
b) Nature of the image:
(i) The image formed when do = 40.0 cm is virtual.
(ii) The image formed when do = 20.0 cm is virtual.
(iii) The image formed when do = 10.0 cm is virtual.
c) Orientation of the image:
(i) The image formed when do = 40.0 cm is upright.
(ii) The image formed when do = 20.0 cm is upright.
(iii) The image formed when do = 10.0 cm is upright.
d) Magnification (increase) for each case:
The magnification (m) can be calculated using the formula:
m = -di/do
(i) For do = 40.0 cm:
m = -80.0 cm / 40.0 cm
m = -2.0
(ii) For do = 20.0 cm:
m = -infinity (object at optical center)
(iii) For do = 10.0 cm:
m = -20.0 cm / 10.0 cm
m = -2.0
Therefore, The image formed is virtual and upright for all three cases, and the magnification is -2.0 for each case.
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What is the period of a sound wave that has a wavelength of 0.30 m if the temperature is 38
∘
C ?
8.3×10
−3
s
4.2×10
−4
s
2.0×10
−3
s
8.5×10
−4
s
The period of the sound wave with a wavelength of 0.30 m at a temperature of 38°C is approximately 1.18 seconds.
The speed of sound in air depends on temperature according to the equation:
v = 331.4 m/s + 0.6 m/s/°C * T
where v is the speed of sound in meters per second and T is the temperature in degrees Celsius.
To calculate the period of the sound wave, we need the speed of sound and the wavelength. The period (T) is the inverse of the frequency (f), and the speed of sound (v) is the product of the frequency and the wavelength:
v = f * λ
Rearranging the equation, we can solve for the period:
T = 1/f = λ/v
Substituting the given values:
λ = 0.30 m
T = 1 / (0.30 m / v)
Now we need to calculate the speed of sound at the given temperature of 38°C:
v = 331.4 m/s + 0.6 m/s/°C * 38°C
v = 331.4 m/s + 0.6 m/s/°C * 38°C
v ≈ 331.4 m/s + 22.8 m/s
v ≈ 354.2 m/s
Now we can calculate the period:
T = 1 / (0.30 m / 354.2 m/s)
T ≈ 1.18 s
Therefore, the period of the sound wave with a wavelength of 0.30 m at a temperature of 38°C is approximately 1.18 seconds.
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A block of mass 20.2 kg is slding at an initial velocity of 5.15 m s in the ponitive x-girection, The surface has a coefficient of wnetic friction of 0.253. (indicate the direction with the signs of your answers.) (a) What is the force of kineic friction (in N) acting on the block? (b) What is the blocks acceleration (nmn
2
) ? xm/s
2
(c) How far will it side (in m) before coming to revt?
The force of kinetic friction acting on the block is 50.094 N.
The block’s acceleration is 2.48 m/s2.
The block will slide for a distance of 5.351 meters before coming to rest.
(a) To calculate the force of kinetic friction:
Formula: force of kinetic friction = coefficient of kinetic friction * normal force
The force of gravity
= 20.2 * 9.8
= 198 N (downwards)
The normal force is equal in magnitude and opposite in direction to the force of gravity. Thus, the normal force is 198 N (upwards)
Therefor, force of kinetic friction = 0.253 * 198
= 50.094 N
The force of kinetic friction acting on the block is 50.094 N.
(b) To calculate the block’s acceleration:
Formula: acceleration = (force of net x-direction) / mass
The force of net x-direction is the force of kinetic friction.
The force of net x-direction = force of kinetic friction
= 50.094 N
Thus, acceleration = force of net x-direction / mass
= 50.094 / 20.2
= 2.48 m/s2
Therefor, the block’s acceleration is 2.48 m/s2.
(c) To calculate how far the block will slide before coming to rest:
Formula:
[tex]v^2 = u^2 + 2as[/tex]
Initial velocity (u) = 5.15 m/s
Final velocity (v) = 0 m/s
Acceleration (a) = 2.48 m/s²
Distance (s) = ?
[tex]v^2 = u^2 + 2as[/tex]
[tex]0 = (5.15)^2 + 2(2.48)s[/tex]
[tex]26.5225 = 4.96s[/tex]
Therefore, s = 5.351 m (round off to 3 decimal places)
The block will slide for a distance of 5.351 meters before coming to rest.
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A battery has an emf of 12 V and internal resistance of 0.05Ω. Its terminal is connected to a load resistance of 3.0Ω. (a) Find the current in the circuit and terminal voltage of the battery. (b) Calculate the power delivered by the battery
Power delivered by the battery,P = 4 A × 11.8 V= 47.2 W
(a) The current in the circuit is 4 A and the terminal voltage of the battery is 11.8 V.(b) The power delivered by the battery is 47.2 W.
Given data: EMF of the battery, E = 12 V.
Internal resistance of the battery, r = 0.05 Ω.
Load resistance, R = 3 Ω.
(a) Current in the circuit
We know that the current in the circuit is given by
Ohm's law as: V = IR
⇒ I = V/R
Current in the circuit, I = 12 V/3 Ω= 4 A
Now, terminal voltage of the battery
We know that the terminal voltage of the battery is given byOhm's law as:
V = E - Ir
⇒ V = 12 V - (4 A × 0.05 Ω)
⇒ V = 11.8 V
(b) Power delivered by the batteryWe know that the power delivered by the battery is given by
P = IV.
Now, current in the circuit, I = 4 A
Therefore, Power delivered by the battery,P = 4 A × 11.8 V= 47.2 W
(a) The current in the circuit is 4 A and the terminal voltage of the battery is 11.8 V.(b) The power delivered by the battery is 47.2 W.
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A force F=180 N force is applied to point A in the figure as shown. If the height h=240 mm, and the horizontal distance w=36 mm, determine the moment of this force about point D in N−m. (Note the change in units!) Indicate the direction of the moiment with the sign - positive if counterclockwise, negative if clockwise.
The moment of force about point D is 43.3 N-m. Since the force is applied clockwise to point D, the moment is negative. To calculate the moment of the force about point D in N-m, we need to determine the perpendicular distance between the line of action of the force and point D.
Hence, I will describe the figure.A force of magnitude 180 N is applied to point A. Point A is at a height of 240 mm, and the horizontal distance between point A and D is 36 mm.
We need to determine the moment of this force about point D in N-m.
We can use the following formula to determine the moment of force:M = F x d where F is the magnitude of the force and d is the perpendicular distance between the line of action of the force and point D.
We can determine the perpendicular distance between the line of action of the force and point D using Pythagoras theorem.
Using Pythagoras theorem, we can find that the perpendicular distance d is given byd = √(h² + w²)where h is the height of point A and w is the horizontal distance between point A and D.
Substituting the values in the above equation, we getd = √(240² + 36²) = 240.7 mm.
Now, substituting the values of F and d in the moment of force equation, we getM = F x d = 180 N x 0.2407 m = 43.3 N-m.
The moment of force about point D is 43.3 N-m.
Since the force is applied clockwise to point D, the moment is negative.
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The stream function for a certain flow field is Y = 2y2 - 2x + 5 a) Determine the corresponding velocity potential b) Can the Bernoulli equation be applied to this flow? Explain [20 Marks) [5 Marks) Total = 25 marks Q6. Carbon dioxide flows adiabatically through a pipe. The velocity is 240 m/s, with pressure of 200 kPa and temperature of 300 C. Data: R= 188 m/(s.K) and y=1.4. Estimate the speed of sound in m/s
(a) The corresponding velocity potential is -4y x -2x + f(t). (b) Bernoulli equation can be applied if the flow satisfies certain conditions: steady, inviscid, and incompressible.
Q6: The estimated speed of sound in the pipe is 368.9 m/s.
a) The velocity potential ([tex]\phi[/tex]) can be obtained by taking the negative partial derivatives of the stream function (Y) with respect to x and y. In this case, the stream function [tex]Y = 2y^2 - 2x + 5[/tex]. Taking the partial derivatives:
[tex]\partial \phi /\partial x = -\partial Y/\partial y = -4y\\\partial\phi /\partial y = \partial Y/\partial x = -2[/tex]
Thus, the corresponding velocity potential is [tex]\phi = -4y x -2x + f(t)[/tex], where f(t) is an arbitrary function of time.
b) The Bernoulli equation relates the pressure, velocity, and elevation along a streamline in a fluid flow. It can be applied if the flow satisfies certain conditions: steady, inviscid, and incompressible. The given flow field does not provide information about the velocity or elevation, so cannot determine if the flow satisfies these conditions. Therefore, cannot conclusively state whether the Bernoulli equation can be applied to this flow without additional information.
Q6) For calculating the speed of sound, use the adiabatic equation:
[tex]v = (y * R * T)^{0.5}[/tex]
Where v is the speed of sound, y is the heat capacity ratio (1.4 for carbon dioxide), R is the gas constant (188 m/(s.K) for carbon dioxide), and T is the temperature in Kelvin.
First, convert the temperature from Celsius to Kelvin:
[tex]T(K) = T(^0C) + 273.15[/tex]
T(K) = 300 + 273.15 = 573.15 K
Next, substitute the given values into the equation:
[tex]v = (1.4 * 188 * 573.15)^{0.5}[/tex]
v ≈ 368.9 m/s
Therefore, the estimated speed of sound in the pipe is approximately 368.9 m/s.
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a) Consider a system whose pressure varies with T and V as P(T,V)=cTVγ, where c and γ are constants and γ=−1. Calculate the work done on this system, W=−∫ACP(T,V)dV, along the two paths shown: WABC and WAC.A→B is heating at constant volume, B→C is constant pressure compression, A→C is isothermal (T= const ) compression. d) What value of γ makes the work a state function?
A system whose pressure varies with T and V as P(T,V)=cTVγ, where c and γ are constants and γ=−1. The work done on the system is 0.The given process is isothermal. γ = -1 makes the work a state function.
To calculate the work done on the system along the paths WABC and WAC, we need to integrate the expression P(T,V) with respect to volume (dV) along each path.
(a) WABC: Heating at constant volume (A→B)
Since the volume is constant, dV = 0. Therefore, the work done (W) along this path is zero.
WABC = 0
(b) WAC: Isothermal compression (A→C)
For an isothermal process, the temperature remains constant (T = const). The integral of P(T,V) with respect to volume gives the work done:
WAC = -∫AC P(T,V) dV
Substituting P(T,V) = cTV^γ:
WAC = -∫AC cTV^γ dV
Since the process is isothermal, T is constant, and we can take it out of the integral:
WAC = -cT ∫AC V^γ dV
The integral of V^γ with respect to V is given by:
∫ V^γ dV = (V^(γ+1))/(γ+1)
Therefore, the work done along the path WAC is:
WAC = -cT [(V_C)^(γ+1) - (V_A)^(γ+1)] / (γ+1)
(c) To determine the value of γ that makes the work a state function, we need to check if the work done along a closed path (in this case, the path WABC) is zero
For WABC to be zero, the work done along the path WAC must also be zero (since WABC includes WAC as a part). Therefore, for the work to be a state function, we must have:
WAC = 0
Substituting the expression for WAC:
-cT [(V_C)^(γ+1) - (V_A)^(γ+1)] / (γ+1) = 0
To satisfy this equation, we can have two possibilities:
cT = 0: This means the constant c or the temperature T is zero. However, for a physically meaningful system, this scenario is unlikely.
(V_C)^(γ+1) - (V_A)^(γ+1) = 0: This requires the exponent γ + 1 to be equal to zero.
γ + 1 = 0
γ = -1
Therefore, the value of γ that makes the work a state function is γ = -1.
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A 91 gram apple falls from a branch that is 2 meters above the ground. (a) How much time elapses before the apple hits the ground? s (b) Just before the impact, what is the speed of the apple? m/s
Therefore, it takes 0.6384 seconds for the apple to hit the ground.just before the impact, the speed of the apple is 6.27168 m/s
When a 91 gram apple falls from a branch that is 2 meters above the ground, it takes 0.6384 seconds to hit the ground and has a speed of 6.27168 meters per second just before the impact.
(a) How much time elapses before the apple hits the ground?
Formula to find time (t) is given as;
Distance = 1/2 g t²
where g = acceleration due to gravity
= 9.8 m/s²and
d = 2 m1/2 g t²
= d
[tex]By substituting values1/2 * 9.8 * t²[/tex]
[tex]= 21/2 * t²[/tex]
= 2/9.8t²
= 0.204t
= 0.6384 seconds
Therefore, it takes 0.6384 seconds for the apple to hit the ground.
(b) Just before the impact, what is the speed of the apple?Formula to find velocity (v) is given as;
v = gt
where
g = acceleration due to gravity
= 9.8 m/s²and
t = 0.6384 seconds
[tex]v = 9.8 * 0.6384[/tex]
= 6.27168 m/s
Therefore, just before the impact, the speed of the apple is 6.27168 m/s.
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A small object of mass 2.80 g and charge −30.0μC is suspended motionless above the ground when immersed in a uniform electric filed perpendicular to the ground. What is the magnitude and direction of the electric field?
A small object of mass 2.80 g and charge −30.0μC is suspended motionless above the ground when immersed in a uniform electric filed perpendicular to the ground.
The electric force is given as;
F = ma where F = force
m = mass and a = acceleration.
F = qE where q = charge and
E = electric field.
Thus ma =[tex]qE = > a = qE/m.T[/tex]
he direction of the electric field is downwards towards the ground.
The weight force of the object is acting upwards and is balanced by the electric force which is acting downwards.
Thus mg = q
[tex]E = > E = mg/q.[/tex]
Substituting the values,
m = 2.80g = 0.0028 kg;
q = [tex]-30.0 μC = -30.0 × 10^-6 C;[/tex]
g = [tex]9.81 m/s^2[/tex]
we getE =[tex](0.0028 kg × 9.81 m/s^2) / (-30.0 × 10^-6 C)E = -0.000912 N/C = 9.12 × 10^2 V/m.[/tex]
The negative sign indicates that the electric field is acting downwards which is opposite to the direction of the positive charges.
The magnitude of the electric field is given as 9.12 × 10^2 V/m.
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A car that is initially moving at 7.5 m/s begins to accelerate forward uniformly at 0.55 m/
2
. How long after beginning to accelerate does it take the car to move 3500 m ? 35 s 66⩽ 100 s 20 s
The car takes approximately 20 seconds to cover a distance of 3500 meters while accelerating uniformly at 0.55 m/s².
To determine the time it takes for the car to cover a distance of 3500 meters while accelerating uniformly at 0.55 m/s², we can use the kinematic equation:
s = ut + (1/2)at²
where:
s = distance traveled (3500 m)
u = initial velocity (7.5 m/s)
a = acceleration (0.55 m/s²)
t = time
We need to solve this equation for t. Rearranging the equation, we get:
t² + (2u/a)t - (2s/a) = 0
Substituting the given values, we have:
t² + (2 * 7.5 / 0.55)t - (2 * 3500 / 0.55) = 0
Simplifying further, we have a quadratic equation:
0.55t² + 27.27t - 12727.27 = 0
Solving this quadratic equation, we find that t ≈ 20 seconds. Therefore, it takes approximately 20 seconds for the car to move a distance of 3500 meters while accelerating uniformly at 0.55 m/s².
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(10\%) Problem 10: Consider the two displacement vectors, A=(2.3 m)i+(−5.6 m)j+(−2.8 m)k and B=(1.9 m)i+(−7.8 m)j+(4.8 m) 50% Part (a) What is the magnitude of C=A+B, in meters? C] =14.18✓ Carrect! \$2 50% Part (b) What is the magnitude of D=2A−B, in meters?
The magnitude of vector C = A + B can be calculated using the formula |C| = √(Cx² + Cy² + Cz²). The magnitude of vector C, |C|, is approximately 14.18 meters. The magnitude of vector D, |D|, is approximately 11.54 meters.
(a) The magnitude of vector C = A + B can be calculated using the formula |C| = √(Cx² + Cy² + Cz²), where Cx, Cy, and Cz are the components of vector C.
Given:
Vector A = (2.3 m)i + (-5.6 m)j + (-2.8 m)k
Vector B = (1.9 m)i + (-7.8 m)j + (4.8 m)k
To find vector C = A + B, we add the corresponding components:
Cx = 2.3 m + 1.9 m = 4.2 m
Cy = -5.6 m + (-7.8 m) = -13.4 m
Cz = -2.8 m + 4.8 m = 2 m
Calculating the magnitude of C:
|C| = √(4.2 m)² + (-13.4 m)² + (2 m)²
|C| = √(17.64 m² + 179.56 m² + 4 m²)
|C| = √(201.2 m²)
|C| ≈ 14.18 m
Therefore, the magnitude of vector C, |C|, is approximately 14.18 meters.
(b) The magnitude of vector D = 2A - B can be calculated similarly using the formula |D| = √(Dx² + Dy² + Dz²), where Dx, Dy, and Dz are the components of vector D.
Given:
Vector A = (2.3 m)i + (-5.6 m)j + (-2.8 m)k
Vector B = (1.9 m)i + (-7.8 m)j + (4.8 m)k
To find vector D = 2A - B, we perform the corresponding operations on the components:
Dx = 2(2.3 m) - 1.9 m = 3.7 m
Dy = 2(-5.6 m) - (-7.8 m) = -3.4 m
Dz = 2(-2.8 m) - 4.8 m = -10.4 m
Calculating the magnitude of D:
|D| = √(3.7 m)² + (-3.4 m)² + (-10.4 m)²
|D| = √(13.69 m² + 11.56 m² + 108.16 m²)
|D| = √(133.41 m²)
|D| ≈ 11.54 m
Therefore, the magnitude of vector D, |D|, is approximately 11.54 meters.
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