ESAT Physics Mock 4

At $\displaystyle I=0.30\,\mathrm{A}$ , the graph gives $\displaystyle V_{\rm lamp}=4.0\,\mathrm{V}$ . The resistor voltage is $\displaystyle IR=3.0\,\mathrm{V}$ , so this operating point uses the full $\displaystyle 7.0\,\mathrm{V}$ . Thus $\displaystyle P_{\rm lamp}=VI=4.0\times0.30=1.2\,\mathrm{W}$ . The total supply power is not the lamp power.

For any electromagnetic wave, the speed of light $\displaystyle c$ , frequency $\displaystyle f$ , and wavelength $\displaystyle \lambda$ are related by the wave equation:This can be rearranged to find the frequency for a given wavelength:This shows that frequency is inversely proportional to wavelength. A shorter wavelength corresponds to a higher frequency, and a longer wavelength corresponds to a lower frequency.

From the graph and the stem, we are given that $\displaystyle \lambda_B$ is a longer wavelength than $\displaystyle \lambda_A$ , so $\displaystyle \lambda_B > \lambda_A$ . Applying the inverse relationship, the corresponding frequencies must satisfy $\displaystyle f_B < f_A$ . So, the peak that was at the shorter wavelength $\displaystyle \lambda_A$ will be at the higher frequency $\displaystyle f_A$ .

Re-plotting the data against a different variable on the x-axis does not change the intensity values on the y-axis. The intensity of each component of the signal remains the same. The stem and graph state that the peak at $\displaystyle \lambda_A$ is taller than the peak at $\displaystyle \lambda_B$ . Therefore, on the new graph, the peak at frequency $\displaystyle f_A$ will be taller than the peak at frequency $\displaystyle f_B$ .

Combining these two points, the peak from $\displaystyle \lambda_A$ is at a higher frequency ( $\displaystyle f_A > f_B$ ) and is the taller peak. This corresponds to option A.

The original period is $\displaystyle 8.0\,\mathrm{ms}$ , so the first positive maximum is at $\displaystyle 2.0\,\mathrm{ms}$ . Doubling the rotation speed halves all corresponding times, giving $\displaystyle 1.0\,\mathrm{ms}$ . Faster cutting of the magnetic field produces a larger peak voltage.

For a fixed mass of an ideal gas at constant temperature, Boyle's Law states that pressure $\displaystyle P$ is inversely proportional to volume $\displaystyle V$ . We can write this as $\displaystyle P = \dfrac{k}{V}$ , where $\displaystyle k$ is a constant.

The problem states that the volume increases at a constant rate. We can express this as $\displaystyle V(t) = V_0 + rt$ , where $\displaystyle V_0$ is the initial volume and $\displaystyle r$ is a constant positive rate.

Substituting this into the pressure equation gives the pressure as a function of time:As time $\displaystyle t$ increases, the denominator $\displaystyle (V_0 + rt)$ increases. Since the numerator $\displaystyle k$ is a positive constant, the pressure $\displaystyle P(t)$ must decrease. This eliminates options A, B, and C.

Now we must consider the rate of decrease. Let's look at the change in pressure over two consecutive, equal time intervals. In the first interval, the volume increases by a fixed amount $\displaystyle \Delta V$ . In the second interval, it increases by the same amount $\displaystyle \Delta V$ .

The change in pressure is given by $\displaystyle \Delta P = P_{final} - P_{initial}$ . Because the graph of $\displaystyle P$ vs $\displaystyle V$ ( $\displaystyle P=k/V$ ) is a convex curve, for a given change in volume $\displaystyle \Delta V$ , the corresponding change in pressure $\displaystyle |\Delta P|$ is larger at smaller volumes and smaller at larger volumes.

Since the volume is continuously increasing, the pressure drop for each successive time interval becomes smaller. Therefore, the pressure is decreasing, and the rate at which it decreases is also decreasing.

At terminal velocity, the net force on the sphere is zero. The forces acting are Weight ( $\displaystyle W$ ), Upthrust ( $\displaystyle U$ ), and Drag ( $\displaystyle F$ ).Using $\displaystyle W = V \rho_{\text{sphere}} g$ and $\displaystyle U = V \rho_{\text{fluid}} g$ , where volume $\displaystyle V = \dfrac{4}{3}\pi r^3$ :We are given the drag law $\displaystyle F = k r v$ . Equating the expressions for force:Rearranging for velocity $\displaystyle v$ :Now we calculate the value for each sphere:

For sphere P ( $\displaystyle r, 3\rho$ ):For sphere Q ( $\displaystyle 2r, 2\rho$ ):Taking the ratio:

The efficiency, $\displaystyle \beta$ , is defined as the ratio of the useful power output, $\displaystyle P_{\text{out}}$ , to the total electrical power input, $\displaystyle P_{\text{in}}$ .The input power is given as $\displaystyle P_{\text{in}} = 160$ W.

The useful power output is the power used to lift the mass. Since the mass is lifted at a constant speed, the upward lifting force, $\displaystyle F$ , provided by the motor must be equal in magnitude to the weight of the mass, $\displaystyle W = mg$ .The useful power is the rate at which work is done by this force, which is given by $\displaystyle P_{\text{out}} = Fv$ .Now, we can calculate the efficiency:Expressed as a percentage, the efficiency is $\displaystyle 0.75 \times 100\% = 75\%$ .

Both microwave and infrared signals travel at the same speed in vacuum. Add the two path lengths, rather than their difference: $\displaystyle t=\dfrac{(36000+45000)\times1000}{3.0\times10^8}=0.27\,\mathrm{s}.$

The displacement of particle $\displaystyle \text{P}$ as a function of time $\displaystyle t$ is given by the kinematic equation: $\displaystyle s_{\text{P}}(t) = s_0 + ut + \dfrac{1}{2}at^2$ Substituting $\displaystyle s_0 = 9.0\text{ m}$ , $\displaystyle u = 1.0\text{ m s}^{-1}$ , and $\displaystyle a = 2.0\text{ m s}^{-2}$ : $\displaystyle s_{\text{P}}(t) = 9.0 + 1.0t + \dfrac{1}{2}(2.0)t^2 = t^2 + t + 9.0$ The displacement of particle $\displaystyle \text{Q}$ moving with constant speed $\displaystyle v$ from $\displaystyle s = 0$ is: $\displaystyle s_{\text{Q}}(t) = vt$ Since the line representing $\displaystyle \text{Q}$ is tangent to the curve representing $\displaystyle \text{P}$ , they touch at exactly one time $\displaystyle t$ .

Method 1 (Equating velocities and positions):
At the point of tangency, the instantaneous velocity of $\displaystyle \text{P}$ equals the speed of $\displaystyle \text{Q}$ : $\displaystyle v_{\text{P}}(t) = \dfrac{\mathrm{d}s_{\text{P}}}{\mathrm{d}t} = 2t + 1 = v$ Since their positions are also equal at this instant: $\displaystyle t^2 + t + 9.0 = vt$ Substitute $\displaystyle v = 2t + 1$ into the position equation: $\displaystyle t^2 + t + 9.0 = (2t + 1)t = 2t^2 + t$ $\displaystyle t^2 = 9.0 \implies t = 3.0\text{ s} \quad (\text{since } t > 0)$ Thus, the speed of $\displaystyle \text{Q}$ is: $\displaystyle v = 2(3.0) + 1 = 7.0\text{ m s}^{-1}$ Method 2 (Using the discriminant):
Equating displacements: $\displaystyle t^2 + t + 9.0 = vt \implies t^2 + (1 - v)t + 9.0 = 0$ For tangency, there must be a unique solution (repeated root), so the discriminant must be zero: $\displaystyle \Delta = (1 - v)^2 - 4(1)(9.0) = 0$ $\displaystyle (1 - v)^2 = 36$ $\displaystyle 1 - v = -6 \implies v = 7.0\text{ m s}^{-1}$

Convection is driven by buoyancy. We analyse the forces on the parcel:

1. **Weight ( $\displaystyle W$ ):** The mass $\displaystyle m$ of the parcel is constant, so its weight $\displaystyle W = mg$ is constant.
2. **Upthrust ( $\displaystyle U$ ):** By Archimedes' principle, the upward buoyant force is equal to the weight of the fluid displaced: $\displaystyle U = \rho_{\text{fluid}} V_{\text{parcel}} g$ .

When the parcel is heated, its temperature increases and it expands (volume $\displaystyle V_{\text{parcel}}$ increases). Since mass is constant, its density $\displaystyle \rho_{\text{parcel}} = \dfrac{m}{V_{\text{parcel}}}$ decreases.

The expansion increases the volume of fluid displaced, increasing the upthrust $\displaystyle U$ . Since the surrounding fluid is cooler and denser than the heated parcel ( $\displaystyle \rho_{\text{fluid}} > \rho_{\text{parcel}}$ ), the upthrust $\displaystyle U$ becomes larger than the parcel's weight $\displaystyle W$ .Thus, the direct cause of the upward force is the decrease in density (relative to the surroundings), which creates a buoyant imbalance.

1. **Displacement in the positive direction ( $\displaystyle 0\text{ s} \le t \le 30\text{ s}$ ):**
The area under the graph from $\displaystyle t = 0$ to $\displaystyle t = 30\text{ s}$ forms a trapezium with parallel sides of length $\displaystyle 30\text{ s}$ and $\displaystyle (20 - 10) = 10\text{ s}$ , and height $\displaystyle 8.0\text{ m s}^{-1}$ : $\displaystyle s_1 = \dfrac{1}{2} \times (30 + 10) \times 8.0 = 160\text{ m}$ 2. **Displacement in the negative direction ( $\displaystyle 30\text{ s} \le t \le 50\text{ s}$ ):**
The area between the graph and the time axis from $\displaystyle t = 30$ to $\displaystyle t = 50\text{ s}$ forms a triangle with base $\displaystyle 20\text{ s}$ and height $\displaystyle -6.0\text{ m s}^{-1}$ : $\displaystyle s_2 = \dfrac{1}{2} \times 20 \times (-6.0) = -60\text{ m}$ 3. Total distance travelled:
Distance is a scalar quantity equal to the sum of the magnitudes of each segment's displacement: $\displaystyle d = |s_1| + |s_2| = 160 + 60 = 220\text{ m}$ 4. Distance from starting position:
The net displacement is: $\displaystyle \Delta x = s_1 + s_2 = 160 - 60 = 100\text{ m}$ Therefore, the vehicle is $\displaystyle 100\text{ m}$ from its starting position.

5. Magnitude of average velocity:
Average velocity is total displacement divided by total time: $\displaystyle |\bar{v}| = \dfrac{|\Delta x|}{\Delta t} = \dfrac{100\text{ m}}{50\text{ s}} = 2.0\text{ m s}^{-1}$ *(Note: Average speed would be $\displaystyle \dfrac{220}{50} = 4.4\text{ m s}^{-1}$ , but average velocity is requested).*

Thus, the correct row is E.

The principle of conservation of energy states that the total power input from the heater must equal the sum of the power used to melt the ice and the power lost to the surroundings.First, we must calculate the power required to melt the ice, $\displaystyle P_{\text{melt}}$ . Power is the rate of energy transfer, and the energy required for a phase change is given by $\displaystyle E = mL_f$ . Therefore, the power required for melting is the rate of mass change multiplied by the specific latent heat of fusion, $\displaystyle L_f$ .

The rate of melting is given in g s⁻¹, which must be converted to kg s⁻¹:Now, calculate $\displaystyle P_{\text{melt}}$ :This is the 'useful' power. The power lost to the surroundings is the difference between the input power and the useful power:

For P, $\displaystyle T_P=4\times0.50=2.0\,\mathrm{ms}$ . For Q, $\displaystyle T_Q=2\times0.50=1.0\,\mathrm{ms}$ , so its frequency is doubled. Both voltage amplitudes are $\displaystyle 3\times0.20=2\times0.30=0.60\,\mathrm{V}$ ; hence the sound amplitudes are equal.

Half-life describes the probability of decay, not an exact schedule for a small sample. Each surviving nucleus still has a one-half probability of surviving the next half-life. The expected number remaining is $\displaystyle 72/2=36$ , with random variation around that value.

There are two equivalent ways to solve this problem:

Method 1: Geometric optics property of successive reflections
When a light ray undergoes two successive reflections in a single plane, the total angular deflection of the ray depends only on the angle $\displaystyle \phi$ between the planes of the two mirrors (the total deflection is $\displaystyle 2\phi$ ).

Since the direction of the incident ray is fixed, the direction of the emerging ray will remain unchanged if and only if the angle $\displaystyle \phi$ between mirror $\displaystyle M_1$ and mirror $\displaystyle M_2$ remains unchanged.

Therefore, if mirror $\displaystyle M_1$ is rotated clockwise through an angle $\displaystyle \alpha$ , mirror $\displaystyle M_2$ must also be rotated clockwise through the same angle $\displaystyle \alpha$ so that the relative angle between them is preserved.

Method 2: Direct angle tracing
1. When mirror $\displaystyle M_1$ is rotated clockwise by $\displaystyle \alpha$ , the ray reflected from $\displaystyle M_1$ rotates clockwise by $\displaystyle 2\alpha$ .
2. If mirror $\displaystyle M_2$ is rotated clockwise by an angle $\displaystyle \beta$ , its normal also rotates clockwise by $\displaystyle \beta$ .
3. The incoming ray to $\displaystyle M_2$ has shifted by $\displaystyle 2\alpha$ clockwise, while the normal to $\displaystyle M_2$ has shifted by $\displaystyle \beta$ clockwise. The angle of the reflected ray from $\displaystyle M_2$ changes by $\displaystyle 2\beta - 2\alpha$ relative to its initial direction.
4. For the emerging ray to remain strictly vertical, the net change in its final direction must be zero: $\displaystyle 2\beta - 2\alpha = 0 \implies \beta = \alpha$ Thus, mirror $\displaystyle M_2$ must be rotated clockwise through an angle $\displaystyle \alpha$ .

The work done by air resistance is given by the integral of the drag force over the distance travelled:Mechanical energy is continuously dissipated by air resistance throughout the flight. Therefore, at any specific height $\displaystyle h$ , the mechanical energy of the ball is lower on the way down than on the way up. Since the potential energy $\displaystyle mgh$ is the same at height $\displaystyle h$ in both phases, the kinetic energy (and thus speed) must be lower on the way down:Since the force of air resistance increases with speed, the drag force at any height $\displaystyle h$ is smaller during the descent than during the ascent. Consequently, the integral of the force over the height $\displaystyle H$ is smaller for the descent:

First, we determine the wavelength $\displaystyle \lambda$ of the wave. The wavelength is the distance traveled by the wave in one period ( $\displaystyle T$ ). Using the wave speed $\displaystyle v$ :Substituting the given values:The number of complete wavelengths $\displaystyle N$ that fit into a total distance $\displaystyle d = 120 \text{ m}$ is found by dividing the distance by the wavelength:

The power $\displaystyle P$ dissipated in a resistor with resistance $\displaystyle R$ and a constant voltage $\displaystyle V$ across it is given by the formula:We are told that $\displaystyle V$ is constant and $\displaystyle R$ decreases at a constant rate. Let the initial resistance be $\displaystyle R_0$ and the constant rate of decrease be $\displaystyle k$ . Then the resistance at time $\displaystyle t$ is $\displaystyle R(t) = R_0 - kt$ .

Substituting this into the power equation gives:As time $\displaystyle t$ increases, the denominator $\displaystyle (R_0 - kt)$ decreases. Since the numerator $\displaystyle V^2$ is constant, the value of the fraction $\displaystyle P(t)$ must increase.

Now we must determine the rate of this increase. Let's consider the change in power over two consecutive, equal time intervals, $\displaystyle \Delta t$ . In each interval, the resistance decreases by the same amount, $\displaystyle \Delta R = k \Delta t$ .

Consider the change in power when the resistance changes from $\displaystyle R$ to $\displaystyle R - \Delta R$ . The change is $\displaystyle \Delta P = \dfrac{V^2}{R - \Delta R} - \dfrac{V^2}{R}$ .

The graph of $\displaystyle y = 1/x$ is convex (it curves upwards). This means that for a fixed change in the input (a fixed $\displaystyle \Delta R$ ), the change in the output is larger when the input value is smaller.

As time passes, $\displaystyle R$ gets smaller. Therefore, the change in power, $\displaystyle \Delta P$ , for each subsequent time interval $\displaystyle \Delta t$ becomes larger. This means the power is increasing, and the rate of that increase is also increasing.

Sound in air is longitudinal: particles move back and forth parallel to the propagation direction. Energy moves along the tube without the bulk transport of air with the wave.

Let each spring constant be $\displaystyle k$ . In X, each spring carries $\displaystyle W$ , so the total energy is $\displaystyle 2[W^2/(2k)]=W^2/k$ . In Y, each carries $\displaystyle W/2$ , so the total energy is $\displaystyle 2[(W/2)^2/(2k)]=W^2/(4k)$ . The ratio is $\displaystyle 4:1$ .

Overall the object has lost $\displaystyle 3.0\times10^{10}$ electrons, so it is positive. The charge is $\displaystyle (3.0\times10^{10})(1.6\times10^{-19})=4.8\times10^{-9}\,\mathrm{C}=+4.8\,\mathrm{nC}$ .

Only a changing current in P produces a changing magnetic field and an induced voltage in Q. The current increases in I and decreases in III, so the induced voltages are non-zero with opposite signs. The constant currents in II and IV give zero voltage, even though the current in II is non-zero.

For a fixed resistance $\displaystyle R$ , the electrical power dissipated is given by $\displaystyle P = \dfrac{V^2}{R}$ .

This implies that power is proportional to the square of the voltage ( $\displaystyle P \propto V^2$ ).

The voltage changes by a factor of $\displaystyle 0.8$ (or $\displaystyle \dfrac{4}{5}$ ). The power therefore changes by a factor of:The new power is:

The period of the wave is $\displaystyle T = \dfrac{1}{f} = \dfrac{1}{5} = 0.2 \text{ s}$ .

The wave travels from P to Q. The time taken for a specific wavefront to travel this distance, $\displaystyle t_{\text{travel}}$ , relates to the observed time difference $\displaystyle \Delta t_{\text{obs}} = 0.10 \text{ s}$ by:where $\displaystyle n$ is a non-negative integer ( $\displaystyle 0, 1, 2, \dots$ ). The possible travel times are:

- If $\displaystyle n=0$ : $\displaystyle t = 0.10 \text{ s} \implies v = \dfrac{6}{0.1} = 60 \text{ m/s}$ .
- If $\displaystyle n=1$ : $\displaystyle t = 0.30 \text{ s} \implies v = \dfrac{6}{0.3} = 20 \text{ m/s}$ .
- If $\displaystyle n=2$ : $\displaystyle t = 0.50 \text{ s} \implies v = \dfrac{6}{0.5} = 12 \text{ m/s}$ .

We calculate the corresponding wavelengths using $\displaystyle \lambda = \dfrac{v}{f}$ :

- For $\displaystyle v = 60 \text{ m/s}$ , $\displaystyle \lambda = 12 \text{ m}$ .
- For $\displaystyle v = 20 \text{ m/s}$ , $\displaystyle \lambda = 4 \text{ m}$ .
- For $\displaystyle v = 12 \text{ m/s}$ , $\displaystyle \lambda = 2.4 \text{ m}$ .

The constraint $\displaystyle 3 \text{ m} < \lambda < 5 \text{ m}$ is only satisfied by $\displaystyle \lambda = 4 \text{ m}$ , so the speed is $\displaystyle 20 \text{ m/s}$ .

We determine the scaling factors based on the radius ratio of $\displaystyle 3$ ( $\displaystyle r_Y = 3r_X$ ).

1. Energy Absorbed: The rate of heat absorption is proportional to surface area ( $\displaystyle A$ ).
For a sphere, area scales with the square of the radius ( $\displaystyle r^2$ ).Therefore, Y absorbs $\displaystyle 9$ times as much energy as X ( $\displaystyle 9:1$ ).

2. Heat Capacity: Heat capacity is proportional to mass ( $\displaystyle M$ ).
Assuming constant density, mass scales with volume ( $\displaystyle V$ ), which scales with the cube of the radius ( $\displaystyle r^3$ ).Therefore, Y has a heat capacity $\displaystyle 27$ times that of X.

3. Rate of Temperature Rise: The rate of temperature change ( $\displaystyle \dfrac{\Delta T}{t}$ ) is determined by the energy input divided by the heat capacity:Comparing Y to X:Thus, Y absorbs $\displaystyle 9$ times the energy but heats up at $\displaystyle \dfrac{1}{3}$ the rate of X ( $\displaystyle 1:3$ ).

First, calculate the mass and weight of the cuboid.Next, find the areas of the faces. Let the side lengths of the cuboid be $\displaystyle x, y, z$ . The face areas are $\displaystyle A_1 = xy$ , $\displaystyle A_2 = yz$ , and $\displaystyle A_3 = zx$ . The volume is $\displaystyle V = xyz$ . The product of the face areas is:The areas are in the ratio $\displaystyle 1:2:8$ . Let them be $\displaystyle A_1 = k$ , $\displaystyle A_2 = 2k$ , and $\displaystyle A_3 = 8k$ . Their product is $\displaystyle (k)(2k)(8k) = 16k^3$ .

We know the volume $\displaystyle V = 0.0040\ \text{m}^3 = 4.0 \times 10^{-3}\ \text{m}^3$ . So, $\displaystyle V^2 = (4.0 \times 10^{-3})^2 = 16 \times 10^{-6}\ \text{m}^6$ .

Equating the two expressions for the product of the areas:The three face areas are $\displaystyle A_1 = 0.01\ \text{m}^2$ , $\displaystyle A_2 = 0.02\ \text{m}^2$ , and $\displaystyle A_3 = 0.08\ \text{m}^2$ .

Maximum pressure, $\displaystyle P_{\text{max}}$ , is exerted when the cuboid rests on its smallest face, $\displaystyle A_{\text{min}} = A_1 = 0.01\ \text{m}^2$ .

Let $\displaystyle P(t)$ be the pressure at time $\displaystyle t$ , and let $\displaystyle P_f$ be the final equilibrium pressure. The excess pressure is $\displaystyle \Delta P(t) = P(t) - P_f$ .

From the problem statement, the rate of cooling follows Newton's Law of Cooling, which implies an exponential decay for the temperature difference. Since $\displaystyle P \propto T$ for an ideal gas at constant volume, the excess pressure $\displaystyle \Delta P(t)$ also decays exponentially.

Let $\displaystyle \Delta P_0$ be the excess pressure at $\displaystyle t=0$ . The excess pressure at a later time $\displaystyle t$ is given by $\displaystyle \Delta P(t) = \Delta P_0 \left(\dfrac{1}{2}\right)^{t/T_{1/2}}$ , where the half-life is $\displaystyle T_{1/2} = 10\,\text{minutes}$ .

The total pressure is $\displaystyle P(t) = P_f + \Delta P(t)$ .

We are given two data points:
1. At $\displaystyle t=0$ : $\displaystyle P(0) = 240\,\text{kPa}$ .(Equation 1)

2. At $\displaystyle t=20\,\text{minutes}$ : $\displaystyle P(20) = 120\,\text{kPa}$ .
The number of half-lives passed is $\displaystyle 20/10 = 2$ .(Equation 2)

We now have a system of two linear equations. Subtracting Equation 2 from Equation 1:Substitute $\displaystyle \Delta P_0$ back into Equation 1 to find $\displaystyle P_f$ :So, the final equilibrium pressure is $\displaystyle 80\,\text{kPa}$ .

Now we need to find the pressure at $\displaystyle t=30\,\text{minutes}$ .
The number of half-lives passed is $\displaystyle 30/10 = 3$ .The final equilibrium pressure is $\displaystyle 80\,\text{kPa}$ and the pressure at $\displaystyle t=30\,\text{minutes}$ is $\displaystyle 100\,\text{kPa}$ .