The free electric charge per unit area on the boundary that makes the total charge per unit area zero is [tex]-(ϵ2 - ϵ1) · ∇ · E / ϵ0.[/tex]
In the given question, we have two dielectric mediums separated by a boundary at the x=2 plane. Let's denote dielectric medium 1 as D1 (where x<2) and dielectric medium 2 as D2 (where x>2).
In D1, the electrostatic field intensity is given as E1 = max + 2ay + 3az. In D2, the electrostatic field intensity is given as E2 = nax + pay + qaz.
To determine the polarization charge per unit area on the boundary when the free electric charge per unit area on the boundary is zero, we need to consider the electric permittivity of each medium. Let's denote the electric permittivity of D1 as ϵ1 and the electric permittivity of D2 as ϵ2.
The polarization charge density, ρp, is related to the electric field intensity by the equation:
[tex]ρp = -(ϵ2 - ϵ1) · ∇ · E[/tex]
Since the electric field intensity is constant in both mediums, [tex]∇ · E = 0[/tex], and therefore the polarization charge per unit area on the boundary is zero when [tex]ϵ2 = ϵ1.[/tex]
Now, to determine the free electric charge per unit area on the boundary that makes the total charge per unit area on the boundary zero, we need to consider the Gauss's law in differential form:
[tex]∇ · E = (ρf + ρp) / ϵ0[/tex]
where [tex]ρf[/tex]is the free electric charge density and ϵ0 is the permittivity of free space.
Since ∇ · E = 0 (from the previous analysis), we have[tex]ρf = -ρp = -(ϵ2 - ϵ1) · ∇ · E / ϵ0.[/tex]
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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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The central ideas in this problem are reviewed in Mutiple-Concept Example 9. One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and its magnitude is slowly increased. When the force reaches 49.2 N, the upper block just begins to slide. The force is then removed from the upper block. and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block?
The magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block, is 24.6 N.
To solve this problem, we can apply the concept of static friction and the condition for impending motion.
Given:
Force applied to the upper block = 49.2 N
Let's assume:
Mass of each block = m (since they are identical)
To find the magnitude of the horizontal force required to slide the lower block, we need to consider the maximum static friction force acting between the lower block and the horizontal surface. This maximum static friction force can be determined using the equation:
Maximum static friction force = coefficient of static friction * normal force
The normal force acting on the lower block is equal to the weight of the upper block plus the weight of the lower block:
Normal force = (m * g) + (m * g) = 2mg
where g is the acceleration due to gravity.
When the upper block just begins to slide, the maximum static friction force is equal to the applied force:
Maximum static friction force = 49.2 N
Substituting the values into the equation:
coefficient of static friction * (2mg) = 49.2 N
Simplifying the equation:
coefficient of static friction = 49.2 N / (2mg)
Now, let's consider the scenario where we want to determine the magnitude of the horizontal force required to make the lower block slide out from under the upper block. At this point, the static friction force between the blocks and the coefficient of static friction remain the same.
Using the condition for impending motion, the magnitude of the horizontal force required on the lower block is equal to the maximum static friction force between the blocks:
Force on the lower block = coefficient of static friction * normal force
Substituting the value of the coefficient of static friction:
Force on the lower block = (49.2 N / (2mg)) * (m * g)
Simplifying:
Force on the lower block = 24.6 N
Therefore, the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block, is 24.6 N.
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Drag-race tires in contact with an asphalt surface have a very high coefficient of static friction.
Assuming a constant acceleration and no slipping of tires, estimate the coefficient of static friction needed for a drag racer to cover 1.9 km in 13 s , starting from rest.
The coefficient of static friction needed for a drag racer to cover 1.9 km in 13 s, starting from rest, is approximately 0.008. This low value is expected since the drag racer is on an asphalt surface, which provides a high coefficient of static friction.
To estimate the coefficient of static friction needed for a drag racer to cover 1.9 km in 13 s, starting from rest, we can make use of the following formula:
$$s=\frac{1}{2}at^2$$
$$a=\frac{2s}{t^2}$$
Where s is the distance travelled, a is the acceleration, and t is the time taken. We are given that s=1.9 km and t=13s. We are to find the value of a, and we will assume that there is no slipping of tires.Let's solve for a first:
$$a=\frac{2s}{t^2}$$
$$a=\frac{2(1.9\text{ km})}{(13\text{ s})^2}$$
$$a=0.0802\text{ km/s}^2$$
Now we can estimate the coefficient of static friction needed for this drag racer. We can make use of the following formula that relates acceleration and coefficient of static friction:
$$a=g\mu$$
$$\mu=\frac{a}{g}$$
$$\mu=\frac{0.0802\text{ km/s}^2}{9.81\text{ m/s}^2}$$
$$\mu=0.008$$
Therefore, the coefficient of static friction needed for a drag racer to cover 1.9 km in 13 s, starting from rest, is approximately 0.008. This low value is expected since the drag racer is on an asphalt surface, which provides a high coefficient of static friction.
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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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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 1500 kg car is moving at 10 m/s when the driver applies the brakes. The car slows to 5 m/s while covering a distance of 30 m.
a. How much work was done on the car?
b. How much force was applied to the car?
By substituting the values into the equations, we can calculate the work done on the car and the force applied to the car.
a. To calculate the work done on the car, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. The work done (W) can be calculated using the formula:
W = ΔKE = KE_final - KE_initial,
where KE is the kinetic energy.
The initial kinetic energy (KE_initial) of the car is given by:
KE_initial = (1/2) * m * v_initial^2,
where m is the mass of the car and v_initial is the initial velocity.
Substituting the given values:
KE_initial = (1/2) * 1500 kg * (10 m/s)^2.
The final kinetic energy (KE_final) of the car is given by:
KE_final = (1/2) * m * v_final^2,
where v_final is the final velocity.
Substituting the given values:
KE_final = (1/2) * 1500 kg * (5 m/s)^2.
Now we can calculate the work done:
W = KE_final - KE_initial.
b. To calculate the force applied to the car, we can use Newton's second law of motion, which states that the force (F) is equal to the rate of change of momentum. The force applied can be calculated using the formula:
F = Δp / Δt,
where Δp is the change in momentum and Δt is the time interval.
The momentum (p) of the car is given by:
p = m * v,
where v is the velocity.
The initial momentum (p_initial) of the car is given by:
p_initial = m * v_initial.
The final momentum (p_final) of the car is given by:
p_final = m * v_final.
Now we can calculate the change in momentum:
Δp = p_final - p_initial.
Finally, we can calculate the force:
F = Δp / Δt.
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Radio station WCCO in Minneapolis broadcasts at a frequency of 8.30×10^5Hz. At a point some Find the wavelength. distance from the transmitter, the magnetic-field Express your answer in meters. amplitude of the electromagnetic wave from WCCO is 4.12×10^−11T λ=[
Find the wave number. Express your answer in radians per meter. Radio station WCCO in Minneapolis broadcasts at a frequency of 8.30×10^5Hz. At a point some Calculate the angular frequency. distance from the transmitter, the magnetic-field Express your answer in radians per second. amplitude of the electromagnetic wave from WCCO Express your answer in radians per second. is 4.12×10^−11T. Part D Calculate the electric-field amplitude. Express your answer in volts per meter.
The wavelength of the electromagnetic wave from radio station WCCO is 361.45 meters.
The wave number is 0.0174 radians per meter.
The angular frequency is 5.21 × 10⁶ radians per second.
The electric-field amplitude of the electromagnetic wave is 1.24 volts per meter
Radio station WCCO in Minneapolis broadcasts at a frequency of 8.30 × 10⁵Hz.
At a point some distance from the transmitter,
the magnetic-field wavelength of the electromagnetic wave from WCCO is 4.12 × 10⁻¹¹ T.
We have to find the wavelength of the wave in meters.λ = v/f
Where f = 8.30 × 10⁵ Hz. v is the speed of light (c)
which is 3 × 10⁸ m/s.λ = 3 × 10⁸/8.30 × 10⁵λ = 361.45 meters
The wavelength of the electromagnetic wave is 361.45 meters.
The wave number is given by:k = 2π/λk = 2π/361.45k = 0.0174 radians per meter
The angular frequency is given by:ω = 2πfω = 2π × 8.30 × 10⁵ω = 5.21 × 10⁶ radians per second
The electric-field amplitude is given by:B = E/cwhere B = 4.12 × 10⁻¹¹ T and c = 3 × 10⁸ m/sE = B × cE = 4.12 × 10⁻¹¹ × 3 × 10⁸E = 1.24 volts per meter
The electric-field amplitude is 1.24 volts per meter.
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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 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 small glass bead has been charged to +60.0nC. A small metal ball bearing 2.60 cm above the bead feels a 1.10×10
−2
N downward electric force. What is the charge on the ball bearing? Express your answer with the appropriate units.
The charge on the ball bearing is approximately 2.04 × 10^(-8) C.
To find the charge on the ball bearing, we can use Coulomb's Law, which states that the electric force between two charged objects is proportional to the product of their charges and inversely proportional to the square of the distance between them.
Charge on the glass bead (Q1) = +60.0 nC
Distance between the bead and the ball bearing (r) = 2.60 cm = 0.0260 m
Electric force (F) = 1.10 × 10^(-2) N
Using Coulomb's Law, we can express the relationship as:
F = k * |Q1 * Q2| / r^2
where k is the electrostatic constant (k ≈ 8.99 × 10^9 N·m^2/C^2), Q2 is the charge on the ball bearing.
Rearranging the equation to solve for the charge on the ball bearing:
|Q2| = (F * r^2) / (k * |Q1|)
Substituting the given values:
|Q2| = (1.10 × 10^(-2) N * (0.0260 m)^2) / (8.99 × 10^9 N·m^2/C^2 * 60.0 × 10^(-9) C)
Simplifying the expression:
|Q2| ≈ 2.04 × 10^(-8) C
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Suppose the moon of a planet has a mass of 1/8th the mass of the planet it is orbiting (note: the moons shown above actually are even a smaller fraction than that!). What is the ratio of the force the moon applies to the planet compared to the force the planet applies to the moon? (Express your answer as a number--don't enter anything like A:B or A/B, just the single number you get by dividing A by B.)
The ratio of the force the moon applies to the planet compared to the force the planet applies to the moon is 1.
The force of gravity between two objects can be calculated using Newton's law of universal gravitation, which states that the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
In this case, let's consider the force exerted by the moon on the planet (F_moon) and the force exerted by the planet on the moon (F_planet).
According to Newton's law of universal gravitation, the force between two objects is given by:
F = G * (m1 * m2) / r^2,
where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.
Given that the moon's mass (m_moon) is 1/8th the mass of the planet (m_planet), we can express it as m_moon = (1/8) * m_planet.
The ratio of the force the moon applies to the planet compared to the force the planet applies to the moon can be calculated as:
F_moon / F_planet = (G * (m_moon * m_planet) / r^2) / (G * (m_planet * m_moon) / r^2).
Simplifying the equation, we find:
F_moon / F_planet = (m_moon * m_planet) / (m_planet * m_moon) = 1.
Therefore, the ratio of the force the moon applies to the planet compared to the force the planet applies to the moon is 1.
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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) 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 speedboat moving at 40.6 m/s approaches a Norwalk buoy marker 100 m ahead. The pilot slows the boat with a constant acceleretion of -3.70m/s² by reducing the throttle (a) How long does it take the boat to reach the buoy?
b). What is the velocity of the boat when it reaches the buoy?
a) It takes 10.97 seconds for the speedboat to reach the buoy.
b) The velocity of the boat when it reaches the buoy is 0.08 m/s.
a) The equation we can use here is
vf = vi + at
where
vf = final velocity,
vi = initial velocity,
a = acceleration,
t = time taken for the object to reach its final velocity
Initial velocity vi = 40.6 m/s
Acceleration, a = -3.7 m/s²
Distance, d = 100 m
Velocity when it reaches the buoy, vf = 0 (since it stops)
Using vf = vi + at, we can solve for t:
vf = vi + at
0 = 40.6 + (-3.7)t3.7
t = 40.6t = 40.6 / 3.7
t = 10.97 seconds
Therefore, it takes 10.97 seconds for the speedboat to reach the buoy.
b) Since we now know the time it takes the boat to reach the buoy (t = 10.97 s), we can use the equation
vf = vi + at to find its velocity when it reaches the buoy:
vf = vi + att = 10.97 seconds
Initial velocity, vi = 40.6 m/s
Acceleration, a = -3.7 m/s²
vf = 40.6 + (-3.7 × 10.97)
vf = 0.08 m/s
Therefore, the velocity of the boat when it reaches the buoy is 0.08 m/s.
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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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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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people are talking a distance of 3.0m away from where you are and
you measure the sound intensity as 1.1x10^-7 W/m^2. another student
is 5.0 m away from the talkers. what is the sound intensity?
The sound intensity at a distance of 5m from the talkers is 4.888 × 10⁻¹⁰ W/m².
We are given the distance between the talkers and the sound intensity. We need to find the sound intensity at another student’s position who is at a distance of 5m from the talkers. We can use the inverse square law of sound to solve the problem.
Inverse square law states that the intensity of sound at any point is inversely proportional to the square of the distance from the source of the sound.
So, the formula for the intensity of sound is:
I ∝ 1/d²
where,
I is the intensity of sound
d is the distance from the source of the sound.
Solving the above equation, we get:
I = K/d²
where K is the constant of proportionality.
To find the value of K, we can use the values of distance and sound intensity for a particular point. Let’s assume that the value of K is I1d1² = I2d2², where I1 is the intensity of sound at a distance of d1 from the source and I2 is the intensity of sound at a distance of d2 from the source.
Substituting the given values, we get:
I1 (3)² = 1.1 × 10⁻⁷
I1 = 1.1 × 10⁻⁷ / 9
I1 = 1.222 × 10⁻⁸
Now, using this value of K, we can find the sound intensity at a distance of 5m from the talkers.
I2 = K/d²
I2 = (1.222 × 10⁻⁸)/5²
I2 = 4.888 × 10⁻¹⁰ W/m²
Therefore, the sound intensity at a distance of 5m from the talkers is 4.888 × 10⁻¹⁰ W/m².
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A coil is in a magnetic field that points parallel to the coil's axis and that is described by the expression B = 0.0400t + 0.0400t2. The 2.60 cm diameter coil has 11 turns and a resistance of 0.990 Ω. What is the induced current at time t = 5.00 s?
Magnitude:
The magnitude of the induced current at time t = 5.00 s is 0.0017 A (rounded to three significant figures).
Given the following values:
B = 0.0400t + 0.0400t²
Radius, r = 2.60/2 = 1.30 cm = 0.0130 m
Number of turns, N = 11
Resistance, R = 0.990 Ω
We know that the magnitude of the induced emf is given by:
ε = -N(dΦ/dt)
Where N is the number of turns and Φ is the magnetic flux.
If we assume the area of the coil to be perpendicular to the magnetic field, then the flux, Φ = BA, where B is the magnetic field intensity and A is the area of the coil (πr²).
Let's calculate the magnetic field at time t = 5.00 s:
B = 0.0400t + 0.0400t² = 0.0400(5.00) + 0.0400(5.00)² = 1.00 + 10.00 = 11.00 T
The radius of the coil, r = 0.0130 m
Number of turns, N = 11
The magnetic field at the coil, B = 11.00 T
The area of the coil, A = π(0.0130)² = 0.0005309 m²
The flux, Φ = BA = 11.00 x 0.0005309 = 0.005848 Tm
The induced emf is given by:
ε = -N(dΦ/dt)
Therefore, ε = -N(d/dt)(BA) = -NAdB/dt
The magnetic field, B = 0.0400t + 0.0400t²
Differentiating with respect to time, we get:
dB/dt = 0.0400 + 2(0.0400)t
Substituting the values, we get:
dB/dt = 0.0400 + 2(0.0400)(5.00) = 0.280 Vm⁻¹
The induced emf is given by:
ε = -NAdB/dt
ε = -11 x 0.0005309 x 0.280
ε = -0.001658 V
The induced current is given by:
I = ε/R
I = -0.001658/0.990
I = -0.0017 A
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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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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 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 plane has a velocity of 520.9 km/h at 16 degrees south of east relative to the
wind. The plane is flying in a wind that is 215.2 km/h at 65 degrees south of east. Find the
magnitude of the plane relative to the earth in km/h using the method of components.
The magnitude of the plane's velocity relative to the earth is approximately 553.1 km/h.
To find the magnitude of the plane's velocity relative to the earth, we can use the method of vector addition. We need to add the vector representing the plane's velocity relative to the wind to the vector representing the wind's velocity.
Given the plane's velocity of 520.9 km/h at 16 degrees south of east relative to the wind, we can break it down into its horizontal and vertical components. The horizontal component is 520.9 km/h * cos(16°), and the vertical component is 520.9 km/h * sin(16°).
Similarly, for the wind's velocity of 215.2 km/h at 65 degrees south of east, we can determine its horizontal and vertical components using the same method.
Next, we add the horizontal components of both vectors together and the vertical components together. This gives us the horizontal and vertical components of the plane's velocity relative to the earth.
Finally, we can use these components to calculate the magnitude of the plane's velocity relative to the earth using the Pythagorean theorem:
Magnitude = [tex]\sqrt(horizontal ^2 + vertical ^2)[/tex]
After performing the calculations, we find that the magnitude of the plane's velocity relative to the earth is approximately 553.1 km/h.
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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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\$ 25% Part (d) How far, in meters, has the elevator moved above its original starting point? y=
In order to determine how far the elevator has moved above its original starting point, we need to analyze the equation representing the elevator's vertical position.
Unfortunately, the equation representing the vertical position of the elevator, denoted as y, has not been provided in the question. Without this equation, it is not possible to calculate the exact displacement or distance traveled by the elevator.
To determine how far the elevator has moved above its original starting point, we would need the specific equation or additional information regarding the elevator's motion, such as its initial position or velocity. With these details, we could calculate the displacement by evaluating the change in position from the starting point to a given time or position.
Please provide the equation or additional information related to the elevator's vertical position, and I would be happy to assist you further in calculating the displacement.
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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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Consider a 4 mm long linear conductor carrying a current of 2 A. The conductor is parallel to the x-axis, it's base is positioned at the point (0,4,−2), and is surrounded by air. Determine the magnetic field at the origin.
The magnetic field at the origin, due to the current in the linear conductor,
is [tex]10^(-7) / √5 T[/tex].
To determine the magnetic field at the origin, we can use the Biot-Savart law
which states that the magnetic field created by a current-carrying wire is directly proportional to the current, length of the wire, and inversely proportional to the distance from the wire.
First, let's find the distance from the origin to the conductor. The base of the conductor is positioned at (0, 4, -2).
Since we're looking for the magnetic field at the origin, The distance is simply the magnitude of this position vector.
which is [tex]√(0^2 + 4^2 + (-2)^2)[/tex]
=[tex]√20 = 2√5.[/tex]
Next, we can calculate the magnetic field using the formula:
B = (μ0 * I * L) / (2π * r),
where μ0 is the permeability of free space[tex](4π × 10^(-7) T·m/A),[/tex]
I is the current (2 A)
L is the length of the conductor (4 mm = 0.004 m)
r is the distance from the conductor[tex](2√5 m).[/tex]
Plugging in the values, we get:
B = [tex](4π × 10^(-7) * 2 * 0.004) / (2π * 2√5)[/tex]
=[tex](8π × 10^(-7)) / (4π * 2√5)[/tex]
= [tex](2 × 10^(-7)) / (2√5)[/tex]
= [tex]10^(-7) / √5[/tex].
The magnetic field at the origin is 10^(-7) / √5 T.
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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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2.1 Distinguish between the following:
(a) beam, diffuse, and total radiation. (3)
(b) extra-terrestrial and terrestrial solar radiation. (3)
(c) solar irradiance and solar irradiation. (2)
2.2 Explain why it is more difficult to predict diffuse irradiance than beam irradiance. (5)
2.1a Beam radiation is directional radiation and is expressed in watts per square meter
Diffuse radiation is expressed in watts per square meter
Total radiation is expressed in watts per square meter
2.1b Extra-terrestrial Solar Radiation called space radiation
Terrestrial Solar Radiation is received on the earth's surface after atmospheric absorption and scattering.
2.1c Solar Irradiance is expressed in watts per square meter
Solar Irradiation is expressed in Joules per square meter
2.2 Diffuse irradiance is received from many directions, whereas beam irradiance is directional, making it easier to predict.
2.1 (a) Beam Radiation: This is a form of radiation that includes solar radiation that reaches the earth's surface without having been diffused or scattered by the atmosphere. It is directional radiation and is expressed in watts per square meter (Wm-2).
Diffuse Radiation: It refers to the radiation that reaches the earth's surface after it has been scattered by the atmosphere. The scattered radiation is not directional and can be received from different points of the sky. It is expressed in watts per square meter (Wm-2).
Total Radiation: It is the summation of beam and diffuse radiation that is received on the earth's surface. It is expressed in watts per square meter (Wm-2).
2.1 (b) Extra-terrestrial Solar Radiation: This is the amount of solar radiation that is received on the outermost layer of the earth's atmosphere. It is also called space radiation.
Terrestrial Solar Radiation: This refers to the amount of solar radiation that is received on the earth's surface after atmospheric absorption and scattering.
2.1 (c) Solar Irradiance: It is the amount of solar radiation that is received on the earth's surface per unit area. It is expressed in watts per square meter (Wm-2).
Solar Irradiation: It is the amount of solar radiation that is absorbed per unit area of the earth's surface. It is expressed in Joules per square meter (Jm-2).
2.2 It is more difficult to predict diffuse irradiance than beam irradiance because diffuse radiation results from multiple scattering events in the atmosphere and is dependent on cloud cover, atmospheric aerosols, and the amount of water vapor in the atmosphere, among other factors. These variables make it more difficult to predict the amount of diffuse irradiance than beam irradiance, which is only dependent on the position of the sun in the sky. Additionally, diffuse irradiance is received from many directions, whereas beam irradiance is directional, making it easier to predict.
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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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Consider the circuit shown in the diagram below. Switch 1 has been closed for a long time. The capacitor is initially uncharged. a. Now switch \( \mathrm{S}_{2} \) is closed. What happens to the brigh
(a). Initially, the bulbs will have equal brightness when switch S2 is closed.
(b). After switch S2 has been closed for a long time, the brightness of the bulbs will remain constant.
(c). If switch S1 is opened after switch S2 has been closed for a long time, the brightness of the bulbs will gradually decrease as the charged capacitor discharges through them.
(a). When switch S2 is closed, the brightness of the bulbs will initially be equal. This is because the uncharged capacitor acts like a short circuit when first connected. The current flows through the bulbs in parallel, and since they are identical, they will have the same brightness.
(b). After switch S2 has been closed for a long time, the brightness of the bulbs will not change. This is because the capacitor will become fully charged, and it will block the flow of direct current (DC) through the circuit. Since the capacitor blocks the flow of DC, the bulbs will not receive any current and their brightness will remain constant.
(c). If switch S1 is opened after switch S2 has been closed for a long time, the brightness of the bulbs will gradually decrease over time. This is because the charged capacitor will start discharging through the bulbs. Initially, the brightness will be high, but it will decrease as the charge on the capacitor decreases. Eventually, the brightness will become zero as the capacitor discharges completely.
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Complete question is,
Consider the circuit shown in the diagram below. Switch 1 has been closed for a long time. The capacitor is initially uncharged. TE e e B
a. Now switch S2 is closed. What happens to the brightness of (current through) each of the bulbs immediately after switch S2 is closed? Explain your reasoning.
b. Compare the brightness of the bulbs after switch S2 has been closed for a long time. Explain your reasoning.
c. If, after the switch S2 has been closed for a long time, switch S1 is then opened, how would the brightness other bulbs compare over time? (Switch S2 remains closed.) Explain your reasoning.
