Solve 3 ,4,5 in english

Here are the solutions to questions 3, 4, and 5 from the image:

Question 3:

a) Calculate the time taken for the stone to reach the ground.

Step 1: Identify the knowns for vertical motion. Initial vertical velocity, $u_y = 0 \text{ m/s}$ (thrown horizontally). Vertical displacement, $s_y = 45 \text{ m}$. Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$.

Step 2: Use the kinematic equation for vertical motion. $s_y = u_y t + \frac{1}{2}gt^2$ Since $u_y = 0$: $s_y = \frac{1}{2}gt^2$ Rearrange to solve for $t$: $t^2 = \frac{2s_y}{g}$ $t = \sqrt{\frac{2s_y}{g}}$

Step 3: Substitute values and calculate $t$. $t = \sqrt{\frac{2 \times 45 \text{ m}}{9.8 \text{ m/s}^2}}$ $t = \sqrt{\frac{90 \text{ m}}{9.8 \text{ m/s}^2}}$ $t = \sqrt{9.1837 \text{ s}^2}$ $t \approx 3.030 \text{ s}$

The time taken for the stone to reach the ground is $\boxed{3.03 \text{ s}}$.

b) Calculate the horizontal distance from the base of the cliff where the stone lands.

Step 1: Identify the knowns for horizontal motion. Horizontal velocity, $v_x = 15 \text{ m/s}$ (constant). Time of flight, $t = 3.030 \text{ s}$ (from part a).

Step 2: Use the equation for horizontal distance. $s_x = v_x t$

Step 3: Substitute values and calculate $s_x$. $s_x = 15 \text{ m/s} \times 3.030 \text{ s}$ $s_x = 45.45 \text{ m}$

The horizontal distance is $\boxed{45.5 \text{ m}}$.

c) Calculate the velocity with which the stone hits the ground.

Step 1: Determine the horizontal and vertical components of velocity at impact. Horizontal velocity, $v_x = 15 \text{ m/s}$ (constant). Vertical velocity, $v_y = u_y + gt$. Since $u_y = 0$: $v_y = gt$ $v_y = 9.8 \text{ m/s}^2 \times 3.030 \text{ s}$ $v_y = 29.694 \text{ m/s}$

Step 2: Calculate the magnitude of the resultant velocity ($V$). $V = \sqrt{v_x^2 + v_y^2}$ $V = \sqrt{(15 \text{ m/s})^2 + (29.694 \text{ m/s})^2}$ $V = \sqrt{225 \text{ m}^2/\text{s}^2 + 881.73 \text{ m}^2/\text{s}^2}$ $V = \sqrt{1106.73 \text{ m}^2/\text{s}^2}$ $V \approx 33.27 \text{ m/s}$

Step 3: Calculate the angle ($\alpha$) the velocity makes with the horizontal. $\tan\alpha = \frac{v_y}{v_x}$ $\tan\alpha = \frac{29.694 \text{ m/s}}{15 \text{ m/s}}$ $\tan\alpha = 1.9796$ $\alpha = \arctan(1.9796)$ $\alpha \approx 63.2^\circ$

The velocity with which the stone hits the ground is $\boxed{33.3 \text{ m/s at } 63.2^\circ \text{ below the horizontal}}$.

Question 4:

a) Calculate the useful power developed by the pump.

Step 1: Identify given values. Mass of water, $m = 2000 \text{ kg}$. Vertical height, $h = 6.0 \text{ m}$. Time, $t = 10 \text{ s}$. Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$.

Step 2: Calculate the useful work done (potential energy gained). $W_{useful} = mgh$ $W_{useful} = 2000 \text{ kg} \times 9.8 \text{ m/s}^2 \times 6.0 \text{ m}$ $W_{useful} = 117600 \text{ J}$

Step 3: Calculate the useful power. $P_{useful} = \frac{W_{useful}}{t}$ $P_{useful} = \frac{117600 \text{ J}}{10 \text{ s}}$ $P_{useful} = 11760 \text{ W}$

The useful power developed by the pump is $\boxed{11760 \text{ W}}$.

b) Calculate the power input to the pump.

Step 1: Identify given values. Useful power, $P_{useful} = 11760 \text{ W}$ (from part a). Efficiency, $\eta = 80\% = 0.80$.

Step 2: Use the efficiency formula. $\eta = \frac{P_{useful}}{P_{input}}$ Rearrange to solve for $P_{input}$: $P_{input} = \frac{P_{useful}}{\eta}$

Step 3: Substitute values and calculate $P_{input}$. $P_{input} = \frac{11760 \text{ W}}{0.80}$ $P_{input} = 14700 \text{ W}$

The power input to the pump is $\boxed{14700 \text{ W}}$.

Question 5:

a) Calculate the angular frequency of the oscillation.

Step 1: Identify given values. Period, $T = 0.80 \text{ s}$.

Step 2: Use the formula for angular frequency. $\omega = \frac{2\pi}{T}$

Step 3: Substitute values and calculate $\omega$. $\omega = \frac{2\pi}{0.80 \text{ s}}$ $\omega = 2.5\pi \text{ rad/s}$ $\omega \approx 7.854 \text{ rad/s}$

The angular frequency is $\boxed{7.85 \text{ rad/s}}$.

b) Calculate the maximum velocity of the mass.

Step 1: Identify given values. Amplitude, $A = 0.10 \text{ m}$. Angular frequency, $\omega = 7.854 \text{ rad/s}$ (from part a).

Step 2: Use the formula for maximum velocity in SHM. $v_{max} = A\omega$

Step 3: Substitute values and calculate $v_{max}$. $v_{max} = 0.10 \text{ m} \times 7.854 \text{ rad/s}$ $v_{max} = 0.7854 \text{ m/s}$

The maximum velocity is $\boxed{0.785 \text{ m/s}}$.

c) Calculate the maximum kinetic energy of the mass.

Step 1: Identify given values. Mass, $m = 0.50 \text{ kg}$. Maximum velocity, $v_{max} = 0.7854 \text{ m/s}$ (from part b).

Step 2: Use the formula for kinetic energy. $KE_{max} = \frac{1}{2}mv_{max}^2$

Step 3: Substitute values and calculate $KE_{max}$. $KE_{max} = \frac{1}{2} \times 0.50 \text{ kg} \times (0.7854 \text{ m/s})^2$ $KE_{max} = 0.25 \text{ kg} \times 0.61685 \text{ m}^2/\text{s}^2$ $KE_{max} = 0.1542 \text{ J}$

The maximum kinetic energy is $\boxed{0.154 \text{ J}}$.

d) Calculate the maximum acceleration of the mass.

Step 1: Identify given values. Amplitude, $A = 0.10 \text{ m}$. Angular frequency, $\omega = 7.854 \text{ rad/s}$ (from part a).

Step 2: Use the formula for maximum acceleration in SHM. $a_{max} = A\omega^2$

Step 3: Substitute values and calculate $a_{max}$. $a_{max} = 0.10 \text{ m} \times (7.854 \text{ rad/s})^2$ $a_{max} = 0.10 \text{ m} \times 61.685 \text{ rad}^2/\text{s}^2$ $a_{max} = 6.1685 \text{ m/s}^2$

The maximum acceleration is $\boxed{6.17 \text{ m/s}^2}$.

e) Calculate the spring constant.

Step 1: Identify given values. Mass, $m = 0.50 \text{ kg}$. Period, $T = 0.80 \text{ s}$.

Step 2: Use the formula for the period of a mass-spring system. $T = 2\pi\sqrt{\frac{m}{k}}$ Rearrange to solve for $k$: $T^2 = (2\pi)^2 \frac{m}{k}$ $k = \frac{4\pi^2 m}{T^2}$

Step 3: Substitute values and calculate $k$. $k = \frac{4\pi^2 \times 0.50 \text{ kg}}{(0.80 \text{ s})^2}$ $k = \frac{4\pi^2 \times 0.50 \text{ kg}}{0.64 \text{ s}^2}$ $k = \frac{2\pi^2 \text{ kg}}{0.64 \text{ s}^2}$ $k = \frac{19.739 \text{ kg}}{0.64 \text{ s}^2}$ $k = 30.842 \text{ N/m}$

The spring constant is $\boxed{30.8 \text{ N/m}}$.