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Example Question - physics

Here are examples of questions we've helped users solve.

Calculating a Weighted Product

To solve the equation, first express the fraction: $\frac{9}{7} \times 32 \, \text{kg}$ Now, multiply the two values: $\frac{9 \times 32}{7} \, \text{kg}$ Calculating $9 \times 32 = 288$: $\frac{288}{7} \, \text{kg} \approx 41.14 \, \text{kg}$ Thus, the solution is approximately $41.14 \, \text{kg}$.

Determining the Volume of a Stone

This question seems to be asking about the method used to determine the volume of an irregularly shaped solid, such as a stone. The most commonly used method for this purpose is the displacement method, which is based on Archimedes' principle. To find the volume of the stone using the displacement method, you would: 1. Fill a graduated cylinder or beaker with water to a certain level and record the volume of water (V1). 2. Carefully place the stone into the water and make sure it is fully submerged without touching the sides of the container. Record the new volume of water (V2). 3. The volume of the stone (Vs) is equal to the difference in the volume of water before and after the stone was submerged: Vs = V2 - V1.

Applying Boyle's Law to Various Gas States

Para resolver este problema, aplicaremos la ley de Boyle-Mariotte, la cual establece que para una cantidad fija de gas a temperatura constante, el producto de la presión y el volumen es constante. $ P_1 \cdot V_1 = P_2 \cdot V_2 $. Esta relación se puede extender a todos los estados dados: $ P_0 \cdot V_0 = P_1 \cdot V_1 = P_2 \cdot V_2 = P_3 \cdot V_3 = P_4 \cdot V_4 $. Para hallar los valores faltantes, realizaremos las siguientes operaciones, asegurándonos de convertir todas las unidades a $ kg/cm^2 $ y $ cm^3 $ respectivamente, ya que las unidades deben ser consistentes a través de la igualdad: Del Estado 0 al Estado 1: $ P_0 = 1.5 \, kg/cm^2 $ $ V_0 = 20 \, L = 20,000 \, cm^3 $ $ P_0 \cdot V_0 = P_1 \cdot V_1 $ $ 1.5 \, kg/cm^2 \cdot 20,000 \, cm^3 = P_1 \cdot 1,500 \, cm^3 $ Ahora calculamos $ P_1 $: $ P_1 = \frac{1.5 \, kg/cm^2 \cdot 20,000 \, cm^3}{1,500 \, cm^3} $ $ P_1 = 20 \, kg/cm^2 $ Del Estado 0 al Estado 3: $ P_0 = 1.5 \, kg/cm^2 $ $ V_0 = 20,000 \, cm^3 $ $ P_0 \cdot V_0 = P_3 \cdot V_3 $ $ 1.5 \, kg/cm^2 \cdot 20,000 \, cm^3 = 1 \, atm \cdot V_3 $ Sabemos que $ 1 \, atm = 1.033 \, kg/cm^2 $, entonces: $ V_3 = \frac{1.5 \, kg/cm^2 \cdot 20,000 \, cm^3}{1.033 \, kg/cm^2} $ Resolviendo para $ V_3 $: $ V_3 = \frac{30,000 \, kg \cdot cm}{1.033 \, kg/cm^2} $ $ V_3 = 29,045.5 \, cm^3 $ Para el Estado 4: Conviene convertir la presión en Estado 4 a $ kg/cm^2 $, sabiendo que $ 1 \, bar $ es aproximadamente igual a $ 1.01972 \, kg/cm^2 $: $ P_4 = 0.017 \, bar \cdot 1.01972 \, kg/cm^2/bar $ $ P_4 = 0.01732 \, kg/cm^2 $ Ahora, podemos hallar $ V_4 $: $ P_0 \cdot V_0 = P_4 \cdot V_4 $ $ V_4 = \frac{1.5 \, kg/cm^2 \cdot 20,000 \, cm^3}{0.01732 \, kg/cm^2} $ $ V_4 = 1,732,899.08 \, cm^3 $ Finalmente, hemos encontrado los siguientes valores faltantes: - $ P_1 = 20 \, kg/cm^2 $ - $ V_3 = 29,045.5 \, cm^3 $ - $ V_4 = 1,732,899.08 \, cm^3 $

Pressure and Volume Relationships in Gas Laws

La imagen muestra una tabla relacionada con la Ley de Boyle-Mariotte, la cual establece que el volumen de un gas es inversamente proporcional a su presión a temperatura constante. La ley se puede expresar como $ P_1 V_1 = P_2 V_2 $, donde $ P_1 $ y $ V_1 $ son la presión y el volumen iniciales, respectivamente, y $ P_2 $ y $ V_2 $ son la presión y el volumen finales. Para resolver los problemas en la tabla, aplicamos esta ley a cada par de estados. La información inicial es un volumen de 20 L y una presión de 10 kg/cm² (Estado 0). Por ejemplo, para encontrar el volumen en el Estado 2 donde la presión es 20,000 Pa (o 0.2 kg/cm² ya que 1 Pa = 0.00001 kg/cm²), usamos la fórmula de la siguiente manera: $ V_2 = \frac{P_1 \cdot V_1}{P_2} = \frac{10 \cdot 20}{0.2} = 1000 $ cm³ Haciendo cálculos similares se pueden encontrar los valores de volumen o presión para los otros estados dadas las presiones o volúmenes. Nota: La imagen no proporciona suficiente información para resolver completamente el problema, ya que algunos datos están borrosos. Se necesita información adicional para proporcionar una solución completa.

Circuit Analysis of Currents in a Multi-Branch Circuit

Nesta análise de circuito, podemos aplicar a Lei de Ohm e as regras da série e paralelo para calcular a corrente em cada ramo. Primeiro, vamos encontrar a resistência total ($R_{total}$) do circuito: $R_{paralelo} = \left(\dfrac{1}{7\ \Omega} + \dfrac{1}{8\ \Omega}\right)^{-1} = \dfrac{7 \times 8}{7+8} = \dfrac{56}{15}\ \Omega $ $R_{total} = 2\ \Omega + R_{paralelo} = 2 + \dfrac{56}{15}\ \Omega = \dfrac{86}{15}\ \Omega$ Agora, vamos calcular a corrente total ($I_{total}$) usando a tensão fornecida pelo gerador (10V): $I_{total} = \dfrac{V}{R_{total}} = \dfrac{10\ V}{\dfrac{86}{15}\ \Omega} = \dfrac{150}{86} A = \dfrac{75}{43} A$ A corrente $I_{total}$ é a mesma que passa pela resistência de $2\ \Omega$, pois eles estão em série. Para descobrir a corrente que passa nos ramos de $7\ \Omega$ e $8\ \Omega$, que estão em paralelo: $I_{7\ \Omega} = \dfrac{V}{7\ \Omega} = \dfrac{10V}{7\ \Omega} = \dfrac{10}{7} A$ $I_{8\ \Omega} = \dfrac{V}{8\ \Omega} = \dfrac{10V}{8\ \Omega} = \dfrac{5}{4} A$ Contudo, o valor da tensão no nó entre as resistências de $7\ \Omega$ e $8\ \Omega$ não é o mesmo do gerador (10V), visto que há uma queda de tensão na resistência de $2\ \Omega$. Logo, precisamos calcular essa queda de tensão ($V_{2\ \Omega}$) e a nova tensão no nó ($V_{nó}$): $V_{2\ \Omega} = I_{total} \times 2\ \Omega = \dfrac{75}{43} A \cdot 2\ \Omega = \dfrac{150}{43} V$ $V_{nó} = V_{gerador} - V_{2\ \Omega} = 10V - \dfrac{150}{43} V = \dfrac{280}{43} V$ Agora, recalculamos as correntes $I_{7\ \Omega}$ e $I_{8\ \Omega}$ com a tensão $V_{nó}$: $I_{7\ \Omega} = \dfrac{V_{nó}}{7\ \Omega} = \dfrac{\dfrac{280}{43}V}{7\ \Omega} = \dfrac{40}{43} A$ $I_{8\ \Omega} = \dfrac{V_{nó}}{8\ \Omega} = \dfrac{\dfrac{280}{43}V}{8\ \Omega} = \dfrac{35}{43} A$ Com isso, temos os valores de corrente para cada ramo do circuito: Corrente no ramo de $2\ \Omega$: $I_{2\ \Omega} = \dfrac{75}{43} A$ Corrente no ramo de $7\ \Omega$: $I_{7\ \Omega} = \dfrac{40}{43} A$ Corrente no ramo de $8\ \Omega$: $I_{8\ \Omega} = \dfrac{35}{43} A$

Solving for Current in a Circuit with Resistors

Para resolver essa questão, é necessário aplicar análise de circuitos para encontrar a corrente em cada ramo. Usando a Lei de Ohm e as regras de circuitos em série e paralelo, fazemos os seguintes passos: 1. Primeiro, encontramos a resistência equivalente do circuito total. Temos dois resistores de $10 \Omega$ e $40 \Omega$ em série, que somam $50 \Omega$, e este conjunto está em paralelo com o resistor de $40 \Omega$. \[ R_{eq} = \frac{1}{\frac{1}{50 \Omega} + \frac{1}{40 \Omega}} \] \[ R_{eq} = \frac{1}{\frac{90}{2000 \Omega}} \] \[ R_{eq} = \frac{2000}{90 \Omega} \approx 22.22 \Omega \] 2. Com a resistência equivalente, podemos encontrar a corrente total ($I$) usando a tensão da fonte de $80V$: \[ I = \frac{V}{R_{eq}} \] \[ I = \frac{80V}{22.22 \Omega} \approx 3.6 A \] 3. A corrente que atravessa o resistor de $40 \Omega$ em paralelo é a mesma corrente total ($I \approx 3.6 A$), pois não há outros caminhos para a corrente fluir antes desse ponto. 4. Através do ramo que contém os resistores de $10 \Omega$ e $40 \Omega$ em série, a corrente é a mesma para ambos os resistores em série, e usamos a corrente total para encontrar essa corrente atravessando o nó entre os ramos paralelos: \[ I_1 = I \times \frac{R_{paralelo}}{R_{paralelo} + R_{serie}} \] \[ I_1 = 3.6 A \times \frac{40 \Omega}{40 \Omega + 50 \Omega} \] \[ I_1 = 3.6 A \times \frac{40}{90} \] \[ I_1 \approx 1.6 A \] 5. A corrente que atravessa o resistor de $10 \Omega$ em série é $I_1 \approx 1.6 A$, e essa será a mesma corrente através do resistor de $40 \Omega$ em série com este. Portanto, o valor da corrente em todos os ramos é aproximadamente $3.6 A$ no ramo com o resistor de $40 \Omega$ em paralelo, $1.6 A$ no ramo com o resistor de $10 \Omega$ em série, e $1.6 A$ no ramo com o resistor de $40 \Omega$ em série.

Circuit Analysis Question on Current Calculation

A imagem mostra um circuito com três resistores e uma fonte de tensão. Para encontrar a corrente em cada ramo, utilizaremos a Lei de Ohm e as regras de circuitos em série e paralelo. Passo 1: Determine a resistência equivalente do circuito. \[ R_{eq} = R_1 + \frac{1}{\left( \frac{1}{R_2} + \frac{1}{R_3} \right)} \] Onde $ R_1 = 1\Omega $, $ R_2 = 2\Omega $, e $ R_3 = 8\Omega $. \[ R_{eq} = 1\Omega + \frac{1}{\left( \frac{1}{2\Omega} + \frac{1}{8\Omega} \right)} \] \[ R_{eq} = 1\Omega + \frac{1}{\left( \frac{4+1}{8\Omega} \right)} \] \[ R_{eq} = 1\Omega + \frac{8\Omega}{5} \] \[ R_{eq} = 1\Omega + 1.6\Omega \] \[ R_{eq} = 2.6\Omega \] Passo 2: Calcule a corrente total fornecida pela fonte de tensão usando a Lei de Ohm. \[ I = \frac{V}{R_{eq}} \] A fonte de tensão é dada por $ V = 6V $. \[ I = \frac{6V}{2.6\Omega} \] \[ I = 2.30769231A \] (arredondando para quatro casas decimais). Passo 3: Calcule a corrente através dos resistores $ R_2 $ e $ R_3 $, que estão em paralelo. \[ I_{2,3} = \frac{V}{R_{2,3}} \] A tensão $ V $ no resistor $ R_2 $ é a mesma que no resistor $ R_3 $ porque eles estão em paralelo, então $ V = 6V $. A resistência equivalente $ R_{2,3} $ para os resistores em paralelo é: \[ R_{2,3} = \frac{1}{\left( \frac{1}{R_2} + \frac{1}{R_3} \right)} \] \[ R_{2,3} = \frac{1}{\left( \frac{1}{2\Omega} + \frac{1}{8\Omega} \right)} \] \[ R_{2,3} = \frac{8\Omega}{5} \] \[ R_{2,3} = 1.6\Omega \] \[ I_{2,3} = \frac{6V}{1.6\Omega} \] \[ I_{2,3} = 3.75A \] Passo 4: Use a divisão de corrente para encontrar as correntes em $ R_2 $ e $ R_3 $. \[ I_2 = I_{2,3} \times \frac{R_3}{R_2 + R_3} \] \[ I_2 = 3.75A \times \frac{8\Omega}{2\Omega + 8\Omega} \] \[ I_2 = 3.75A \times \frac{8}{10} \] \[ I_2 = 3A \] \[ I_3 = I_{2,3} - I_2 \] \[ I_3 = 3.75A - 3A \] \[ I_3 = 0.75A \] Passo 5: A corrente em $ R_1 $ é a mesma corrente total do circuito, pois está em série com o resto do circuito. \[ I_1 = I = 2.30769231A \] Portanto, as correntes são: \[ I_1 = 2.30769231A \] \[ I_2 = 3A \] \[ I_3 = 0.75A \] Note que os resultados finais devem ser arredondados com base no número de dígitos significativos desejados.

Circuit Analysis Problem Solving

\begin{align*} // Utilizando as Leis de Kirchhoff e a Lei de Ohm, temos (nomenclaturas i1, i2 e i3 para as correntes, R para as resistências e V para a força eletromotriz): \\ // 1. Aplicando a Lei de Kirchhoff para a malha à esquerda: \\ & -V + i_1 R + i_2 R = 0 \\ & -24 + 7i_1 + 3i_2 = 0 \quad (1) \\ // 2. Aplicando a Lei de Kirchhoff para a malha à direita: \\ & -i_2 R + i_3 R + V = 0 \\ & -3i_2 + 2i_3 + 24 = 0 \quad (2) \\ // 3. A terceira equação vem do nó entre as três resistências, utilizando a Lei de Kirchhoff para correntes (soma das correntes que chegam é igual a soma das correntes que saem): \\ & i_1 = i_2 + i_3 \quad (3) \\ // Resolvendo o sistema de equações: \\ // De (3), temos i1 em termos de i2 e i3 \\ & i_1 = i_2 + i_3 \\ // Substituímos i1 nas equações (1) e (2): \\ & -24 + 7(i_2 + i_3) + 3i_2 = 0 \\ & -24 + 10i_2 + 7i_3 = 0 \quad (4) \\ & -3i_2 + 2i_3 + 24 = 0 \quad (2) \\ // Multiplicamos (2) por 5 e somamos com (4): \\ & -15i_2 + 10i_3 + 120 + 10i_2 + 7i_3 = 0 \\ & 17i_3 + 120 = 0 \\ & i_3 = -\frac{120}{17} A \\ // Agora substituímos i3 em (2) para encontrar i2: \\ & -3i_2 + 2\left(-\frac{120}{17}\right) + 24 = 0 \\ & -3i_2 -\frac{240}{17} + \frac{408}{17} = 0 \\ & -3i_2 + \frac{168}{17} = 0 \\ & i_2 = \frac{168}{17 \times 3} A \\ & i_2 = \frac{56}{17} A \\ // Por fim, usamos i2 e i3 para encontrar i1 através de (3): \\ & i_1 = i_2 + i_3 \\ & i_1 = \frac{56}{17} - \frac{120}{17} \\ & i_1 = -\frac{64}{17} A \\ // Portanto, os valores das correntes são: \\ & i_1 = -\frac{64}{17} A \text{ (corrente no ramo da esquerda)} \\ & i_2 = \frac{56}{17} A \text{ (corrente no ramo do meio)} \\ & i_3 = -\frac{120}{17} A \text{ (corrente no ramo da direita)} \end{align*}

Determining the Number of Teachers Teaching Physics

Let $P$ be the number of teachers who teach physics and $M$ be the number of teachers who teach mathematics. We are given that there are 20 teachers in total, of which 12 teach mathematics, 4 teach both physics and mathematics. We need to find how many teach physics. We can use the principle of inclusion and exclusion to find the number of teachers who teach only physics. Number of teachers teaching only mathematics is $M - 4$. Number of teachers teaching physics, including those who teach both subjects, is $P$. The sum of teachers teaching only mathematics, only physics, and both is the total number of teachers: $(M - 4) + P = 20$ Given that $M = 12$, we substitute this into the equation: $(12 - 4) + P = 20$ $8 + P = 20$ $P = 20 - 8$ $P = 12$ There are 12 teachers who teach physics.

Vector Displacement and Distance Calculation

Ilustramos el problema con un diagrama de vectores: Punto inicial O. El coche se mueve 52 km al Este, llegando al punto A. Luego se mueve 27 km al Sur, llegando al punto B. \[\vec{OA} = 52 \text{ km Este}, \quad \vec{AB} = 27 \text{ km Sur}\] La distancia es la longitud total del camino recorrido, es decir: \[ \text{Distancia} = |\vec{OA}| + |\vec{AB}| = 52 \text{ km} + 27 \text{ km} = 79 \text{ km} \] El desplazamiento es el vector resultante desde el punto de inicio al punto final, es decir $\vec{OB}$. Calculamos su magnitud utilizando el teorema de Pitágoras, donde el desplazamiento es la hipotenusa de un triángulo rectángulo con lados 52 km y 27 km: \[ |\vec{OB}| = \sqrt{52^2 + 27^2} \] \[ |\vec{OB}| = \sqrt{2704 + 729} \] \[ |\vec{OB}| = \sqrt{3433} \] \[ |\vec{OB}| \approx 58.59 \text{ km} \] Por tanto, la distancia recorrida es de 79 km y el desplazamiento es aproximadamente de 58.59 km.

Calculating Distance and Displacement

La distancia es la longitud total del camino recorrido por el auto, que es igual a la suma de los desplazamientos en cada dirección: \[ \text{Distancia} = 52 \text{ km} + 27 \text{ km} = 79 \text{ km} \] El desplazamiento es el vector que va desde el punto inicial al final. Para calcular la magnitud del desplazamiento, usamos el teorema de Pitágoras, considerando un desplazamiento de 52 km hacia el Este y 27 km hacia el Sur, que forman un triángulo rectángulo. \[ \text{Desplazamiento} = \sqrt{(52 \text{ km})^2 + (27 \text{ km})^2} \] \[ \text{Desplazamiento} = \sqrt{2704 + 729} \] \[ \text{Desplazamiento} = \sqrt{3433} \] \[ \text{Desplazamiento} \approx 58.6 \text{ km} \]

Formula One Race Car Dynamics

La formule donnée pour calculer la variation de la vitesse est $\Delta v = \frac{\Delta p}{m}$, où $\Delta p$ est le changement de la quantité de mouvement et $m$ est la masse du système. La variation de vitesse, $\Delta t$ est le temps durant. Le problème demande de résoudre pour $\Delta t$. On peut réarranger l'équation pour isoler $\Delta t$ avec $\Delta t = \frac{\Delta p}{m \cdot \Delta v}$. Il est donné que $\Delta p = 6,53 \times 10^4 \ kg \cdot m/s$ et $m = 743 \ kg$, mais nous n'avons pas $\Delta v$ directement. On sait que la vitesse moyenne est directement liée à la distance parcourue par le temps pris, donc $v = \frac{d}{t}$. On sait que la distance parcourue $d = 5,8 \ km$ (qu'il faut convertir en mètres pour garder les unités constantes, donc $d = 5800 \ m$) et la vitesse moyenne $v = 230 \ km/h$ (qu'il faut aussi convertir en $m/s$ en utilisant le fait que $1 \ km/h = \frac{1}{3.6} \ m/s$, donc $v = \frac{230}{3.6} \ m/s$). Avec ces informations, on peut trouver le temps $\Delta t$ pris pour parcourir 5,8 km à cette vitesse. Calculons d'abord $\Delta v$: \[\Delta v = \frac{230}{3.6} \approx 63.89 \ m/s\] Maintenant, nous pouvons calculer $\Delta t$: \[\Delta t = \frac{6.53 \times 10^4}{743 \cdot 63.89} \approx 1,36 \ s\] Le temps $\Delta t$ est donc approximativement 1,36 secondes pour la distance donnée avec les changements spécifiés de la quantité de mouvement et de la masse.

Analysis of Harmonic Oscillation

Α. Η μέγιστη τιμή (στο σημείο όπου το ελατήριο είναι μέγιστα τεντωμένο ή συμπιεσμένο) της ταχύτητας είναι μηδέν. Όταν το σώμα είναι στη θέση ισορροπίας, έχει τη μέγιστη ταχύτητα. Η απόσταση από τη θέση ισορροπίας στο μέγιστο σημείο έκτασης είναι η πλάτος Α. Προκύπτει ότι A=0,5 m. Β. Η εξίσωση της ταχύτητας σε αρμονική ταλάντωση είναι v = Aωsin(ωt+φ), όπου ω η γωνιακή συχνότητα, φ η αρχική φάση και t ο χρόνος. Όταν το σώμα περνάει από τη θέση ισορροπίας για πρώτη φορά, φ=0 και sin(ωt)=1 (ωt=π/2+2πn, όπου n είναι ακέραιος). Βρίσκουμε τη μέγιστη ταχύτητα v_max ως v_max = Aω. Δεδομένου ότι η γονιακή ταχύτητα είναι σταθερή και ανεξάρτητη του χρόνου, προκύπτει ότι η μέγιστη ταχύτητα είναι η ταχύτητα που έχει το σώμα όταν περνά από τη θέση ισορροπίας. Γ. Η εξίσωση της επιτάχυνσης σε αρμονική ταλάντωση είναι a = -ω^2Acos(ωt+φ). Δεδομένου ότι αναφέρεται ότι η επιτάχυνση είναι a = -5kg⋅m/s^2 όταν η φάση είναι φ = -π/2 (cos(ωt - π/2) = sin(ωt)) και A = 0,5 m, βάζουμε στην εξίσωση και επιλύουμε για ω^2, έτσι ω^2 = (a/A) = (-5)/(0,5) = -10 s^-2. Εξαιτίας του ότι ω πρέπει να είναι θετικό, οπότε ω = √10 s^-1. Δ. Για να βρούμε τη φάση που συμβαίνει η επιτυχία με την ταχύτητα πέρασμα του σώματος από τη θέση ισορροπίας, χρησιμοποιούμε τη γνώση ότι το cos(φ) = 0,4 και ότι a = -ω^2Acos(ωt+φ). Επιβεβαιώνουμε ότι φ = arccos(0,4) ή φ = 2π - arccos(0,4).

Kinematics and Dynamics of Car Braking and Free Falling Object

1) To find the total distance the car travels during braking and the time it takes: 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} $. 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 $. 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$^2$. 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)}} $. e) The time $ t $ it takes to stop is given by $ t = \frac{{v - u}}{{a}} = \frac{{0 - (50/3.6)}}{{-1.962}} $. 2) To find the speed at which the body impacts the water: 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 $. 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 $. c) Plug in $ g = 9.81 $ m/s$^2$ and $ h = 10 $ m to find $ v = \sqrt{2 \times 9.81 \times 10} $. 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.

Buoyancy and Free Fall in Fluids Problem

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: \[ V_c = \frac{W_c}{\rho_c g} = \frac{980}{50 \times 9.8} = 2 \text{ m}^3 \] 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, \[ V_t = \frac{2}{0.9} \approx 2.22 \text{ m}^3 \] 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: \[ F_b = \rho_w g V_c = 1000 \times 9.8 \times 2 = 19600 \text{ N} \] 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: \[ a' = 9.81 - (\frac{1000}{550} \times 9.81) - 0.4 \] \[ a' = 9.81 - 17.84 - 0.4 \] \[ a' = -8.43 \text{ m/s}^2 \] (This is the effective acceleration considering buoyancy and resistance.) 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): \[ v^2 = 0 + 2(-8.43)(100) \] \[ v^2 = -1686 \] 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.

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Example Question - physics

Here are examples of questions we've helped users solve.

Calculating a Weighted Product

Determining the Volume of a Stone

Applying Boyle's Law to Various Gas States

Pressure and Volume Relationships in Gas Laws

Circuit Analysis of Currents in a Multi-Branch Circuit

Solving for Current in a Circuit with Resistors

Circuit Analysis Question on Current Calculation

Circuit Analysis Problem Solving

Determining the Number of Teachers Teaching Physics

Vector Displacement and Distance Calculation

Calculating Distance and Displacement

Formula One Race Car Dynamics

Analysis of Harmonic Oscillation

Kinematics and Dynamics of Car Braking and Free Falling Object

Buoyancy and Free Fall in Fluids Problem

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