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Home Question Bank Physics Archive 2024 November 01

Physics Questions from Nov 01,2024

Browse the Physics Q&A Archive for Nov 01,2024, featuring a collection of homework questions and answers from this day. Find detailed solutions to enhance your understanding.

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La magnitud del peso de un cuerpo en la luna es \( P_{t}=320 \mathrm{~N} \). ¿Cuál es la magnitud \( P_{t} \), del peso del cuerpo en la tierra? 1. Any change in the speed or direction of an object's motion is Consider a system of three particles: One particle has mass \( m_{1}=3 \mathrm{~kg} \), position \( \vec{r}_{1}=<5,3,-5>\mathrm{m} \), and velocity \( \vec{v}_{1}=<2,-9,-5>\mathrm{m} / \mathrm{s} \). The second particle has mass \( m_{2}=8 \mathrm{~kg} \), position \( \vec{r}_{2}=<0,1,-3>\mathrm{m} \), and velocity \( \vec{v}_{2}=<4,-1,1>\mathrm{m} / \mathrm{s} \). The third particle has mass \( m_{3}=8 \mathrm{~kg} \), position \( \vec{r}_{3}=<4,3,-1>\mathrm{m} \), and velocity \( \vec{v}_{3}=<8,-1,0>\mathrm{m} / \mathrm{s} \). Just like we can figure out the momentum of the center of mass by multiplying the total mass of the system times the velocity of the center of mass, we can calculate the translational kinetic energy of the system using the total mass and center of mass velocity. This translational kinetic energy represents the energy associated with the center of mass motion of the whole system, but ignores any energy from motion within the system.] Find the translational kinetic energy of the system. encuentra conectado como se muestra en la figura. La diferencia de presione produce un desnivel en el líquido de 8 cm . E1 diámetro en la parte archa es d \( 2,5 \mathrm{~cm} \) y en la parte más angosta es de 3 mm . El liquido en el depósito tien ina densidad de \( 0,75 \mathrm{~g} / \mathrm{cm}^{3} \) mientras que la del aire en la bomba es \( 1,3 \mathrm{~kg} / \mathrm{m} \) Calcule: a) La diferencia de presiones entre las partes ancha y angosta de ombay y) Las velocidades del aire en cada rama de la bomba Consider a system of three particles: One particle has mass \( m_{1}=3 \mathrm{~kg} \), position \( \vec{r}_{1}=<5,3,-5>\mathrm{m} \), and velocity \( \vec{v}_{1}=<2,-9,-5>\mathrm{m} / \mathrm{s} \). The second particle has mass \( m_{2}=8 \mathrm{~kg} \), position \( \vec{r}_{2}=<0,1,-3>\mathrm{m} \), and velocity \( \vec{v}_{2}=<4,-1,1>\mathrm{m} / \mathrm{s} \). The third particle has mass \( m_{3}=8 \mathrm{~kg} \), position \( \vec{r}_{3}=<4,3,-1>\mathrm{m} \), and velocity \( \vec{v}_{3}=<8,-1,0>\mathrm{m} / \mathrm{s} \). Just like we can figure out the momentum of the center of mass by multiplying the total mass of the system times the velocity of the center of mass, we can calculate the translational kinetic energy of the system using the total mass and center of mass velocity. This translational kinetic energy represents the energy associated with the center of mass motion of the whole system, but ignores any energy from motion within the system.] Find the translational kinetic energy of the system. Consider a system of three particles: One particle has mass \( m_{1}=3 \mathrm{~kg} \), position \( \vec{r}_{1}=<5,3,-5>m_{1} \), and velocity \( \vec{v}_{1}=<2,-9,-5>\mathrm{m} / \mathrm{s} \). The second particle has mass \( m_{2}=8 \mathrm{~kg} \), position \( \vec{r}_{2}=<0,1,-3>m \), and velocity \( \vec{v}_{2}=<4,-1,1>\mathrm{m} / \mathrm{s} \). The third particle has mass \( m_{3}=8 \mathrm{~kg} \), position \( \vec{r}_{3}=<4,3,-1>m \), and velocity \( \vec{v}_{3}=<8,-1,0>\mathrm{m} / \mathrm{s} \). Find the translational kinetic energy of the system. Consider a system of three particles: One particle has mass \( m_{1}=3 \mathrm{~kg} \), position \( \vec{r}_{1}=<5,3,-5>\mathrm{m} \), and velocity \( \vec{v}_{1}=<2,-9,-5>\mathrm{m} / \mathrm{s} \). The second particle has mass \( m_{2}=8 \mathrm{~kg} \), position \( \vec{r}_{2}=<0,1,-3>\mathrm{m} \), and velocity \( \vec{v}_{2}=<4,-1,1>\mathrm{m} / \mathrm{s} \). The third particle has mass \( m_{3}=8 \mathrm{~kg} \), position \( \vec{r}_{3}=<4,3,-1>\mathrm{m} \), and velocity \( \vec{v}_{3}=<8,-1,0>\mathrm{m} / \mathrm{s} \). A motorcycle of mass 310 kg accelerates from rest to \( 46 \mathrm{~m} / \mathrm{s} \) in 6 seconds. Ignore air resistance. Assuming there's no slipping between the wheels and the paventent of the road Energy must always be conserved so find the sum of the translational kinetic energy and the relative kinetic energy. A motorcycle of mass 310 kg accelerates from rest to \( 46 \mathrm{~m} / \mathrm{s} \) in 6 seconds. Ignore air resistance. Assuming there's no slipping between the wheels and the pavement of the road (d) For the real system, how much work is done by the force of the road on the wheels? Consider a system of three particles: One particle has mass \( m_{1}=3 \mathrm{~kg} \), position \( \vec{r}_{1}=<5,3,-5>\mathrm{m} \), and velocity \( \vec{v}_{1}=<2,-9,-5>\mathrm{m} / \mathrm{s} \). The second particle has mass \( m_{2}=8 \mathrm{~kg} \), position \( \vec{r}_{2}=<0,1,-3>\mathrm{m} \), and velocity \( \vec{v}_{2}=<4,-1,1>\mathrm{m} / \mathrm{s} \). The third particle has mass \( m_{3}=8 \mathrm{~kg} \), position \( \vec{r}_{3}=<4,3,-1>\mathrm{m} \), and velocity \( \vec{v}_{3}=<8,-1,0>\mathrm{m} / \mathrm{s} \). The relative kinetic energy is due to the motion of the particle relative to the center of mass. Find the kinetic energy of the system relative to the center of mass.| density = b A block of aluminium has a volume of \( 5.0 \mathrm{~cm}^{3} \) and a mass of 13.5 g . Calculate the density of the aluminium. Give your answer in \( \mathrm{g} / \mathrm{cm}^{3} \). c Explain whether the block of aluminium will float or sink in water.
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