You are given two masses, m₁ and m2, strapped on a pulley with a rope. m₁-13 kg and m2-17 kg. What is the equation to be used to find the system's acceleration? a= = (m₂g + m1g)/(m₂ - m₁) a= = (m2g - m₁g)/(m₂ - m₁) a = (m2g + mig)/(m2+m1) O a = (m2g - m_g)/(m2+m1)

You are given two masses, m₁ and m2, strapped on a pulley with a rope. m₁-13 kg and m2-17 kg. What is the equation to be used to find the system's acceleration? a= = (m₂g + m1g)/(m₂ - m₁) a= = (m2g - m₁g)/(m₂ - m₁) a = (m2g + mig)/(m2+m1) O a = (m2g - m_g)/(m2+m1)

Name: You are given two masses, m₁ and m2, strapped on a pulley with a rope
Uploaded: 2024-03-20T06:56:46.000Z
Channel: Bartleby
Description: You are given two masses, m₁ and m2, strapped on a pulley with a rope. m₁-13 kg andm2-17 kg. What is the equation to be used to find the system's acceleration?O a (m₂g + m₁g)/(m₂ - m₁)=a== (m2g - m1g)/(m₂ - m₁)a = (m₂g + mig)/(m2 + m₁)O a (m₂g-m₁g)/

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

The graph shows the speed of from breaking at 10 s. a 2360.0 kg car as it slows down after the breaks are applied. At a time of 30.0 s the car comes to a complete stop. Assuming other forces are small, what is the magnitude of the force applied to the car 30 25 20 Tries 0/10 Submit Answer 15 10 10 15 20 25 30 Time (s) Speed (m/s)
A 1,910 kg African bush elephant (Loxodonta africana) is pushing a 2,890 kg car. The elephant and the car move forward together with a maximum forward force of 8,750 N. Assuming friction is zero, what is the magnitude of the force (in kN) of the rhinoceros on the car? Enter only the numerical part of your answer to three significant figures.
During your winter break, you enter a sled race. You push the sled with a force of 210 N at 30.0° below the horizontal as shown. The mass of the sled (and your nephew sitting on it) is 35.0 kg. The coefficients of friction between the sled and icy surface are us = 0.35 and uk = 0.20. What is the magnitude of the acceleration of the sled, in m/s2? Use g = 10.0 m/s?. Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement.
The 1.5 Mg = 1,500 kg sports car has a tractive force of F = 4.5 kN = 4,500 N. If it produces the velocitydescribed by v-t graph shown, plot the air resistance R versus t for this time period.
The acceleration of a particle moving along the = -7 t + 4 t* + 3 t2, 7. x-axis is given by the function where a is in meters per second squared and t in seconds. What is the particle's velocity at t = 2 s if the object's mass is 11.3 kg and has an initial velocity of 1.6 m/s?
G=25
An object of mass 3.1 kg with a certain initial speed on a horizontal surface comes to rest after traveling a distance of 15.9 m. If the coefficient of kinetic friction between the object and the horizontal surface is 0.27, what is the initial speed of the object? Express your answer in m/s, to at least one digit after the decimal point. plz use q =9.8
In the rollercoaster shown in the figure, the height of the first hill is 24.5 meters, and the height of the second hill is 12.8 meters. The track of the rollercoaster is frictionless. If the rollercoaster car’s velocity is 20.4 m/s, what was its velocity at the top of the first hill? v = m/s
m = 8 kg v0 = 6 m/s F = 12 N the box reaches 10 m in 1.4 seconds If you wanted the box to reach a distance of x = 10 m in 0.761 seconds, to what value would you change the force (in N)?
A physics student, in a stationary elevator, places a 240.0 g mass on an electronic balance. The student hits the elevator button and it begins to move upwards. While the elevator is moving, the student records this information Time = 0 to 3.0 sec the balance reads 278 grams Time = 3.0 - 12 sec the balance reads 240. grams Time = 12.0 to 15.0 sec the balance reads 192 gram How far up (in meters) did the elevator travel? Consider: UP to be (+), DOWN to be (-) and g =9.8 m/s2 . ROUND YOUR ANSWER TO A WHOLE NUMBER of meters (no decimal places) HINT: This problem requires finding THREE different displacements and adding them up! 2nd HINT: You will need to (A) use your equations of motion or (B) make a v-t graph to find those displacements!
Determine the force Q-> when the block moves with constant velocity. Express your answer in vector form.
Answer to 5)1.96