Answer:
To solve this problem, you'll need to break the initial velocity of the seed into its horizontal and vertical components, then use the equations of motion to find the time of flight and horizontal distance.
The initial velocity (v) of the seed is 2.65 m/s. The angle it's launched at (θ) is 30.0 degrees below the horizontal. The height (h) it's launched from is 0.460 m.
First, calculate the horizontal (v_x) and vertical (v_y) components of the velocity. Because the seed is launched downward, the vertical component will be negative:
v_x = v * cos(θ) = 2.65 m/s * cos(30.0) = 2.29 m/s
v_y = v * sin(θ) = -2.65 m/s * sin(30.0) = -1.325 m/s
Next, use the equation of motion to find the time it takes for the seed to hit the ground:
h = v_y * t + 0.5 * g * t^2
Where g is the acceleration due to gravity, which is approximately 9.8 m/s². Solving the equation for t gives:
t = (-v_y - sqrt((v_y)^2 - 4 * 0.5 * g * (-h))) / (2 * 0.5 * g)
Plugging in the values:
t = (1.325 + sqrt((-1.325)^2 - 4 * 0.5 * 9.8 * (-0.460))) / (2 * 0.5 * 9.8)
t = 0.182 seconds
Finally, use the horizontal velocity and time of flight to find the horizontal distance the seed covers:
x = v_x * t = 2.29 m/s * 0.182 s = 0.417 m
So, the seed lands after approximately 0.182 seconds and travels approximately 0.417 meters horizontally.
High-power experimental engines are being developed by the Stevens Motor Company for use in its new sports coupe. The engineers have calculated the maximum horsepower for the engine to be 630HP
. Twenty five engines are randomly selected for horsepower testing. The sample has an average maximum HP of 650
with a standard deviation of 60HP
. Assume the population is normally distributed.
Step 1 of 2 : Calculate a confidence interval for the average maximum HP for the experimental engine. Use a significance level of α=0.01
. Round your answers to two decimal places.
The 99% confidence interval for the average maximum HP for the experimental engine is (610.12, 689.88).
To calculate the confidence interval for the experimental engines' average maximum HP, we can use the following formula:
To find the z-score for α=0.01, we can refer to a standard normal distribution table or use a calculator. The z-score is approximately 2.58.
Substituting the given values into the formula, we get:
CI = 650 ± 2.58*(60/√25) CI = 650 ± 30.96
Rounding to two decimal places, the confidence interval for the experimental engines' average maximum HP is:
CI = [619.04 HP, 680.96 HP]
Therefore, we can say with 99% confidence that the true average maximum HP for the experimental engines falls between 619.04 HP and 680.96 HP. Thus, we can conclude that the experimental engines' average maximum HP is likely to be within this range. However, note that this range does not include the manufacturer's claimed maximum HP of 630 HP, which may indicate that the engines are performing below expectations.
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How long does it take for the total energy stored in the circuit to drop to 10% of that value?
Express your answer with the appropriate units.A cylindrical solenoid with radius 1.00 cm
and length 10.0 cm
consists of 150 windings of AWG 20 copper wire, which has a resistance per length of 0.0333 Ω/m
. This solenoid is connected in series with a 10.0 μF
capacitor, which is initially uncharged. A magnetic field directed along the axis of the solenoid with strength 0.160 T
is switched on abruptly.
How long does it take for the total energy stored in the circuit to drop to 10% of that value?
Express your answer with the appropriate units.
The energy stored in the circuit at any time t is given by [tex]U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).[/tex]The units are in seconds.
The total energy stored in the circuit can be calculated using the formula: U = (1/2)L*I² + (1/2)Q²/C, where L is the inductance, I is the current, Q is the charge on the capacitor, and C is the capacitance.
Initially, the capacitor is uncharged, so the second term is zero.
Therefore, the initial energy stored in the circuit is U₀ = (1/2)L*I₀², where I₀ is the initial current, which is zero.
When the magnetic field is switched on, a current begins to flow in the solenoid.
This current increases until it reaches its maximum value, given by I = V/R, where V is the voltage across the solenoid and R is its resistance.
Since the solenoid is connected in series with the capacitor, the voltage across the solenoid is equal to the voltage across the capacitor, which is given by V = Q/C, where Q is the charge on the capacitor.
The charge on the capacitor is given by Q = C*V, where V is the voltage across the capacitor at any time t.
Therefore, we have I = V/R = Q/(R*C) = dQ/dt*(1/R*C), where dQ/dt is the rate of change of charge on the capacitor.
This is a first-order linear differential equation, which can be solved to give [tex]Q(t) = Q_{0} *(1 - e^{(-t/(R*C)}))[/tex], where Q₀ is the maximum charge on the capacitor, given by Q₀ = C*V₀, where V₀ is the voltage across the capacitor at t=0.
The current in the solenoid is given by I(t) = [tex]dQ/dt*(1/R*C) = (V_{0} /R)*e^{(-t/(R*C)}).[/tex]
The energy stored in the circuit at any time t is given by[tex]U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).[/tex]
The time t at which the energy stored in the circuit drops to 10% of its initial value can be found by solving the equation U(t) = U₀/10, or equivalently, [tex](1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C)}) + (1/2)C*V_{0} /R)^{2}*(1 - e^{(-2t/(R*C)})) = (1/20)L*I_{0} /R)^{2}.[/tex]
This equation can be solved numerically using a computer program, or graphically by plotting U(t) and U₀/10 versus t on the same axes and finding their intersection point.
The solution is t = 1.74 ms.
The units are in seconds.
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Two uncharged spheres are separated by 2.60 m. If 1.90 x 1012 electrons are removed from one sphere and placed on the other determine the magnitude of the Coulomb force (im N) an one of the spheres, treating the spheres as point charges.
The magnitude of the Coulomb force between the spheres, when 1.90 x [tex]10^{12[/tex] electrons are transferred, is 2.34 x [tex]10^{-4[/tex] Newtons.
To determine the magnitude of the Coulomb force between two uncharged spheres.
Given that 1.90 x [tex]10^{12[/tex] electrons are removed from one sphere and placed on the other, we need to calculate the charge on each sphere. The charge on a single electron is -1.6 x [tex]10^{-19[/tex] coulombs, so the charge transferred from one sphere to the other is:
Q = (1.90 x [tex]10^{12[/tex]) × (-1.6 x [tex]10^{-19[/tex]) = -3.04 x [tex]10^{-7[/tex] coulombs
Since one sphere loses electrons and becomes positively charged, while the other gains electrons and becomes negatively charged, the magnitude of the charge on each sphere is:
|Q| = 3.04 x [tex]10^{-7[/tex] coulombs
Now, we can calculate the magnitude of the Coulomb force using Coulomb's law:
F = k * (|Q1| * |Q2|) / [tex]r^2[/tex]
where k is the electrostatic constant (k = 8.99 x [tex]10^{9}[/tex] N [tex]m^{2}[/tex]/[tex]C^{2}[/tex]) and r is the distance between the centers of the spheres (r = 2.60 m).
Plugging in the values, we get:
F = (8.99 x [tex]10^{9}[/tex] N [tex]m^{2}[/tex]/[tex]C^{2}[/tex]) * (3.04 x [tex]10^{-7[/tex]C)² / [tex](2.60 m)^2[/tex]
Simplifying this expression, we find:
F ≈ 2.34 x [tex]10^{-4[/tex] N
Therefore, the magnitude of the Coulomb force between the spheres, when 1.90 x [tex]10^{12[/tex] electrons are transferred, is approximately 2.34 x [tex]10^{-4[/tex] Newtons.
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A Ferris wheel rotates three times each minute. It carries each car around a circle with the diameter of 19.0m. What force, magnitude (in N) and direction (in degrees), does the seat exert on a 35.0-kg child when the rider is halfway between the top and the bottom?
The forces acting on the child are the force of gravity and the normal force, which is the force exerted by the seat on the child. the force exerted by the seat on the child has a magnitude is Fn = 342.5N. the direction of the normal force is 45 degrees from the horizontal in the upward direction.
The Ferris wheel's speed can be calculated as follows: v = distance / time v = πd / tv = (3.14 * 19.0 m) / 1 min v = 60.66 m/min
The acceleration of the Ferris wheel can be calculated as follows: a = (v²) / ra = (60.66 m/min)² / (19.0 m / 2)a = 194.6 m/min²
The forces acting on the child are the force of gravity and the normal force, which is the force exerted by the seat on the child.
The magnitude of the normal force is equal to the sum of the weight of the child and the centripetal force acting on the child.
In this case, the child is halfway between the top and the bottom, so the angle between the normal force and the vertical direction is 45 degrees (90 degrees divided by 2).
As a result, the force exerted by the seat on the child has a magnitude of:
Fn = Fg + Fcf = maFn = (35.0 kg) * (9.8 m/s²) + (35.0 kg) * (194.6 m/min²)Fn = 342.5 N
The direction of the normal force is perpendicular to the seat, which is horizontal.
Since the angle between the normal force and the vertical direction is 45 degrees, the angle between the normal force and the horizontal direction is also 45 degrees.
Therefore, the direction of the normal force is 45 degrees from the horizontal in the upward direction.
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