Example Question - free fall
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Kinematics and Dynamics of Car Braking and Free Falling Object
<p>1) To find the total distance the car travels during braking and the time it takes:</p> <p>a) The deceleration \( a \) can be calculated using the relation \( a = \frac{{v^2 - u^2}}{{2s}} \), where initial velocity \( u = 50 \) m/s (since the car is initially moving at 50 km/h, we convert this to m/s by multiplying by \( \frac{{1000}}{{3600}} \)), final velocity \( v = 0 \) m/s (since the car stops), and \( s \) is the distance. But since we are given a 10% negative gradient, the effective deceleration is the sum of the deceleration due to braking and the deceleration due to the slope: \( a = a_{braking} + a_{slope} \).</p> <p>b) The deceleration due to the slope \( a_{slope} \) is given by \( a_{slope} = g\sin(\theta) \), where \( g = 9.81 \) m/s\(^2\) is the acceleration due to gravity, and \( \sin(\theta) \) can be approximated by the slope percentage over 100, thus \( \sin(\theta) \approx \frac{{10}}{{100}} = 0.1 \).</p> <p>c) Hence, \( a_{slope} = 9.81 \times 0.1 = 0.981 \) m/s\(^2\), and using \( a_{braking} = -0.3g = -0.3 \times 9.81 \) m/s\(^2\), we find \( a = a_{braking} + a_{slope} = -2.943 + 0.981 = -1.962 \) m/s\(^2\).</p> <p>d) Now, we can calculate the distance \( s \) using \( s = \frac{{v^2 - u^2}}{{2a}} = \frac{{0^2 - (50/3.6)^2}}{{2 \times (-1.962)}} \).</p> <p>e) The time \( t \) it takes to stop is given by \( t = \frac{{v - u}}{{a}} = \frac{{0 - (50/3.6)}}{{-1.962}} \).</p> <p>2) To find the speed at which the body impacts the water:</p> <p>a) Use the kinetic energy at impact \( K.E. = \frac{1}{2}mv^2 \) equating it to the potential energy at the start \( P.E. = mgh \).</p> <p>b) Since \( K.E. = P.E. \), we have \( \frac{1}{2}mv^2 = mgh \). After canceling mass \( m \), the equation simplifies to \( v^2 = 2gh \).</p> <p>c) Plug in \( g = 9.81 \) m/s\(^2\) and \( h = 10 \) m to find \( v = \sqrt{2 \times 9.81 \times 10} \).</p> <p>The calculations from the above steps will give you the distance the car travels during braking and the time it takes to stop, as well as the speed at which the object hits the water.</p>
Buoyancy and Free Fall in Fluids Problem
<p>1) Archimedes' principle states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. The volume of the car submerged, \( V_c \), can be calculated from the given density of the car, \( \rho_c = 50 \) kg/m³, and the weight of the car, \( W_c = 980 \) N. Using the equation \( W_c = \rho_c g V_c \), where \( g \) is the acceleration due to gravity (\( 9.8 \) m/s²), we can find the volume of the submerged part of the car:</p> <p>\[ V_c = \frac{W_c}{\rho_c g} = \frac{980}{50 \times 9.8} = 2 \text{ m}^3 \]</p> <p>Since 10% of the car is above water, the total volume of the car, \( V_t \), is \( V_t = \frac{V_c}{0.9} \). Therefore,</p> <p>\[ V_t = \frac{2}{0.9} \approx 2.22 \text{ m}^3 \]</p> <p>The buoyant force \( F_b \) on the car is equal to the weight of the water displaced, which is \( \rho_w g V_c \), where \( \rho_w = 1000 \) kg/m³ is the density of water. Thus:</p> <p>\[ F_b = \rho_w g V_c = 1000 \times 9.8 \times 2 = 19600 \text{ N} \]</p> <p>2) For the free-falling body, the velocity \( v \) at which it hit the bottom of the pond can be found using the kinematic equation \( v^2 = u^2 + 2as \), where \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the distance. In water, the acceleration is less due to upward buoyancy and resistance, \( a' = g - (\rho_w/\rho_s)g - 0.4 \) where \( \rho_s \) is the density of substance. If the body's density is \( \rho_b = 550 \) kg/m³, then:</p> <p>\[ a' = 9.81 - (\frac{1000}{550} \times 9.81) - 0.4 \]</p> <p>\[ a' = 9.81 - 17.84 - 0.4 \]</p> <p>\[ a' = -8.43 \text{ m/s}^2 \] (This is the effective acceleration considering buoyancy and resistance.)</p> <p>Using the kinematic equation with \( u = 0 \) (starting from rest), \( a' \) as the effective acceleration, and \( s = 100 \) m (distance to the bottom of the pond):</p> <p>\[ v^2 = 0 + 2(-8.43)(100) \]</p> <p>\[ v^2 = -1686 \]</p> <p>This yields an imaginary number which is not possible in real-life scenarios, which indicates that with the given parameters, the body would not reach the bottom due to the net upward acceleration. This suggests that an error might have been made in determining the resistance or the acceleration is not constant all the way down. In real-world scenarios, the object might eventually reach a terminal velocity where the net acceleration is zero.</p>
Calculating the Time for a Ball to Reach Zero Speed when Thrown Upward
<p>To calculate the time for the ball to reach zero speed, we can use the following kinematic equation for uniformly accelerated motion:</p> \[ v_f = v_i + a \cdot t \] <p>Where:</p> <p>\( v_f \) is the final velocity (0 m/s when the ball reaches its maximum height and stops momentarily)</p> <p>\( v_i \) is the initial velocity (4.5 m/s)</p> <p>\( a \) is the acceleration (acceleration due to gravity; for a ball thrown upward this is -9.8 m/s\(^2\), since it acts downward)</p> <p>\( t \) is the time in seconds</p> <p>We can set \( v_f \) to 0 and solve for \( t \):</p> \[ 0 = 4.5 \text{ m/s} - (9.8 \text{ m/s}^2) \cdot t \] \[ 9.8 \text{ m/s}^2 \cdot t = 4.5 \text{ m/s} \] \[ t = \frac{4.5 \text{ m/s}}{9.8 \text{ m/s}^2} \] \[ t \approx 0.459 \text{ s} \] <p>So, the ball will take approximately 0.459 seconds to reach zero speed.</p>
Understanding the Effects of Salt on Boiling Point and Exploring Free Fall
<p>No es una pregunta de matemáticas que requiera una solución paso a paso utilizando ecuaciones o fórmulas matemáticas. Es un resumen de los resultados de un experimento que examina el impacto de la sal en el punto de ebullición del agua y una declaración de intención para un experimento de caída libre. No se proporcionan datos o parámetros para resolver o calcular.</p>
Calculating Speed in Free Fall
Para resolver esta pregunta, necesitamos encontrar la velocidad de la pelota a los 6,5 segundos. Dado que tenemos una tabla de valores de tiempo y velocidad, parece sugerir que hay una relación lineal entre el tiempo y la velocidad. Sin embargo, al analizar las cifras, podemos recordar que la velocidad de un objeto en caída libre debido a la gravedad aumenta proporcionalmente con el tiempo. Esto sugiere que la relación es de hecho cuadrática, como lo demuestra la fórmula de la velocidad en caída libre sin resistencia del aire: v = g * t, donde g es la aceleración debida a la gravedad (aproximadamente 9,8 m/s^2) y t es el tiempo. Si observamos los valores proporcionados en la tabla, notamos que la velocidad se duplica cada segundo (aproximadamente). Por ejemplo, a los 2 segundos, la velocidad es aproximadamente el doble que a 1 segundo (9,8 m/s * 2 ≈ 19,6 m/s), lo que es coherente con la aceleración constante debida a la gravedad. Para obtener la velocidad a los 6,5 segundos, simplemente necesitamos multiplicar la aceleración de la gravedad por el tiempo: v = g * t v = 9,8 m/s^2 * 6,5 s v = 63,7 m/s Por lo tanto, la respuesta correcta es c) 63,7 m/s.
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