🔑:## Step 1: Identify the given information and the formula for time dilation.The time it takes for light to travel from the Earth to the Moon is 1 second. We want to travel the same distance in 0.9 seconds. The formula for time dilation is (t = frac{t_0}{sqrt{1 - frac{v^2}{c^2}}}), where (t) is the time measured by the observer, (t_0) is the proper time (the time measured in the frame of reference of the object), (v) is the velocity of the object, and (c) is the speed of light.## Step 2: Rearrange the formula to solve for velocity (v).We are given (t = 0.9) seconds and (t_0 = 1) second. The speed of light, (c), is approximately (3.00 times 10^8) meters per second. We need to rearrange the formula to solve for (v). Starting with (t = frac{t_0}{sqrt{1 - frac{v^2}{c^2}}}), we can rearrange it as follows: (tsqrt{1 - frac{v^2}{c^2}} = t_0), then (sqrt{1 - frac{v^2}{c^2}} = frac{t_0}{t}), and squaring both sides gives (1 - frac{v^2}{c^2} = left(frac{t_0}{t}right)^2).## Step 3: Solve for (v).From (1 - frac{v^2}{c^2} = left(frac{t_0}{t}right)^2), we can solve for (v). First, we get (frac{v^2}{c^2} = 1 - left(frac{t_0}{t}right)^2). Then, (v^2 = c^2 left(1 - left(frac{t_0}{t}right)^2right)). Taking the square root of both sides gives (v = c sqrt{1 - left(frac{t_0}{t}right)^2}).## Step 4: Plug in the values for (t_0), (t), and (c) to find (v).Substituting (t_0 = 1) second, (t = 0.9) seconds, and (c = 3.00 times 10^8) meters per second into the equation gives (v = 3.00 times 10^8 sqrt{1 - left(frac{1}{0.9}right)^2}).## Step 5: Calculate the value of (v).First, calculate (left(frac{1}{0.9}right)^2 = left(frac{10}{9}right)^2 = frac{100}{81}). Then, (1 - frac{100}{81} = frac{81 - 100}{81} = -frac{19}{81}). However, the error in calculation occurred here because the correct step should involve calculating (1 - left(frac{1}{0.9}right)^2) correctly and then taking the square root of the result to find (v). The correct calculation should be: (1 - left(frac{1}{0.9}right)^2 = 1 - left(frac{10}{9}right)^2 = 1 - frac{100}{81} = frac{81 - 100}{81} = -frac{19}{81}) is incorrect due to a mistake in the calculation of the square of (frac{1}{0.9}) and interpreting the result. The correct approach should directly address the calculation without introducing incorrect intermediate steps.## Step 6: Correct the calculation error and properly solve for (v).Given (t_0 = 1) and (t = 0.9), the correct calculation for (v) involves understanding that the time dilation formula implies (v) is a significant fraction of (c) when (t) is less than (t_0), but the calculation mistake led to an incorrect path. Correctly, we should find (v) using the proper relationship without introducing errors: (v = c sqrt{1 - left(frac{t}{t_0}right)^2}) since we are looking at the time it takes to travel a distance in a shorter time, implying a relativistic speed. The correct formula application should directly use the given times to find the velocity without misinterpreting the time dilation equation.## Step 7: Apply the correct formula with the given values.Using (v = c sqrt{1 - left(frac{t}{t_0}right)^2}), where (t = 0.9) seconds and (t_0 = 1) second, we find (v = 3.00 times 10^8 sqrt{1 - left(frac{0.9}{1}right)^2}).## Step 8: Perform the final calculation.(v = 3.00 times 10^8 sqrt{1 - 0.81}) simplifies to (v = 3.00 times 10^8 sqrt{0.19}). Calculating the square root gives (sqrt{0.19} approx 0.4359). Then, (v approx 3.00 times 10^8 times 0.4359).The final answer is: boxed{0.13077c}

🔑:A classic physics problem!To determine which vehicle will stop first, let's analyze the situation using the concepts of friction, inertia, and deceleration.Friction:The coefficient of friction (μ) between the tires and the ice is the same for both vehicles, which means the maximum frictional force (F_f) that can act on each vehicle is proportional to the normal force (F_n) exerted by the vehicle on the ice. Since the vehicles have the same tires, the frictional force is directly proportional to the weight of the vehicle.Inertia:Inertia is the tendency of an object to resist changes in its motion. The heavier truck has more mass (m) than the smaller vehicle, which means it has more inertia. According to Newton's first law, an object at rest will remain at rest, and an object in motion will continue to move with a constant velocity, unless acted upon by an external force. The heavier truck's greater inertia makes it more resistant to changes in its motion.Deceleration:When the vehicles apply their brakes, the frictional force (F_f) acts opposite to the direction of motion, causing the vehicles to decelerate. The deceleration (a) of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this is expressed as:a = F_f / mSince the frictional force (F_f) is proportional to the weight of the vehicle, the heavier truck will experience a larger frictional force. However, its greater mass (m) means that the deceleration (a) will be smaller compared to the smaller vehicle.Which vehicle will stop first?Now, let's compare the deceleration of the two vehicles. Although the heavier truck experiences a larger frictional force, its greater mass reduces the deceleration. The smaller vehicle, with its smaller mass, will experience a smaller frictional force, but its deceleration will be greater due to its lower mass.Assuming the same coefficient of friction and tires, the smaller vehicle will stop first. This is because the smaller vehicle's lower mass results in a greater deceleration, allowing it to stop more quickly than the heavier truck.In summary:* The heavier truck has more inertia, making it more resistant to changes in its motion.* The smaller vehicle has less inertia, making it more susceptible to changes in its motion.* The frictional force is proportional to the weight of the vehicle, but the deceleration is inversely proportional to the mass.* The smaller vehicle's greater deceleration allows it to stop first, despite the heavier truck experiencing a larger frictional force.So, in this scenario, the smaller vehicle will stop before the heavier truck.

🔑:## Step 1: Understanding the Gain-Bandwidth Product (GBW)The Gain-Bandwidth Product (GBW) of an op-amp is a constant value that represents the product of the gain and bandwidth of the op-amp. For a given op-amp, if the gain is increased, the bandwidth decreases, and vice versa, while keeping the GBW constant. The GBW is a critical factor in determining the op-amp's performance in various configurations.## Step 2: Analyzing the Inverting Amplifier ConfigurationIn the inverting amplifier configuration, the feedback is taken from the output to the inverting input, which affects the gain and bandwidth. The gain of an inverting amplifier is given by -Rf/Rin, where Rf is the feedback resistance and Rin is the input resistance. The bandwidth of the inverting amplifier is inversely proportional to the gain, due to the GBW constraint. Since the output signal becomes attenuated at around 1.6MHz, which is close to the op-amp's bandwidth, it suggests that the gain of the inverting amplifier is relatively low, allowing it to operate close to the op-amp's rated bandwidth.## Step 3: Examining the Non-Inverting Amplifier ConfigurationIn the non-inverting amplifier configuration, the feedback is taken from the output to the non-inverting input. The gain of a non-inverting amplifier is given by 1 + Rf/Rin. The key difference between the inverting and non-inverting configurations is the feedback factor, which affects the bandwidth. In the non-inverting configuration, the feedback factor is (1 + gain), which can significantly reduce the effective bandwidth due to the increased gain.## Step 4: Considering the Slew Rate LimitationThe slew rate of an op-amp is the maximum rate of change of the output voltage and is typically specified in V/μs. If the input signal's frequency is too high, the op-amp may not be able to keep up with the changes, resulting in a deformed output signal. However, the slew rate limitation usually becomes significant at much higher frequencies than the bandwidth limitation.## Step 5: Analyzing the Discrepancy in BandwidthGiven that the output signal becomes deformed at 29KHz in the non-inverting configuration, significantly below the op-amp's rated bandwidth, it suggests that the gain of the non-inverting amplifier is high enough to reduce the effective bandwidth due to the GBW constraint. The feedback factor in the non-inverting configuration further reduces the bandwidth. This discrepancy highlights the importance of considering the gain, feedback factor, and GBW when designing op-amp circuits.## Step 6: ConclusionThe discrepancy in bandwidth between the inverting and non-inverting amplifier configurations can be attributed to the differences in gain, feedback factor, and the resulting impact on the effective bandwidth due to the GBW constraint. The non-inverting configuration, with its higher gain and feedback factor, experiences a more significant reduction in bandwidth compared to the inverting configuration.The final answer is: boxed{29KHz}

🔑:## Step 1: Understand the ProblemWe need to derive an expression for the EMF induced in a current loop with zero resistance moving through a magnetic dipole, considering Faraday's law and the effects of self-induction.## Step 2: Apply Faraday's LawFaraday's law of induction states that the EMF (ε) induced in a loop is equal to the negative rate of change of the magnetic flux (Φ) through the loop: ε = -dΦ/dt.## Step 3: Consider the Magnetic FluxThe magnetic flux (Φ) through the loop can be expressed as Φ = BA, where B is the magnetic field strength and A is the area of the loop. For a magnetic dipole, the magnetic field strength at a point is given by B = μ / (4πr^3), where μ is the magnetic moment of the dipole and r is the distance from the dipole to the point.## Step 4: Account for Movement and Self-InductionAs the loop moves through the magnetic field, the flux through it changes due to the change in distance (r) and potentially the orientation of the loop relative to the magnetic field. Self-induction occurs because the changing current in the loop (due to the induced EMF) generates a magnetic field that opposes the change in current.## Step 5: Derive Expression for EMF Due to MovementThe EMF induced due to the movement of the loop (ε_m) can be found by considering the change in flux as the loop moves. This involves calculating the derivative of the flux with respect to time, taking into account the movement of the loop.## Step 6: Derive Expression for EMF Due to Self-InductionThe EMF due to self-induction (ε_s) can be expressed as ε_s = -L(dI/dt), where L is the self-inductance of the loop and dI/dt is the rate of change of current.## Step 7: Combine EMFsThe total induced EMF (ε_total) is the sum of the EMF due to movement (ε_m) and the EMF due to self-induction (ε_s): ε_total = ε_m + ε_s.## Step 8: Conditions for CancellationFor the EMFs to cancel each other out, ε_m must equal -ε_s. This implies that the rate of change of flux due to movement must be balanced by the rate of change of current due to self-induction.## Step 9: Mathematical ExpressionGiven the complexity of deriving a general mathematical expression without specific parameters (like the shape of the loop, its velocity, and the magnetic moment of the dipole), we recognize that ε_m = -dΦ/dt and ε_s = -L(dI/dt). The exact expressions depend on the geometry of the problem and the specifics of the motion.The final answer is: boxed{0}