The following four forces act on a 50.0 kg object:F₁ = 300 N east₹ 2 = 700 N north₹ 3 = 600 N westF 4 = 300 N southWhat is the acceleration of the object?14.0 m/s² in a direction 53.1º north of westO10.0 m/s2in a direction 53.1° north of west14.0 m/s2in a direction 36.9° north of westO 10.0 m/s² in a direction 36.9° north of west

Answered: Image /qna-images/answer/be9f598a-3936-47b5-85ea-09a1a4f1564e.jpg

The following four forces act on a 50.0 kg object: F₁ = 300 N east ₹ 2 = 700 N north ₹ 3 = 600 N west F 4 = 300 N south What is the acceleration of the object? 14.0 m/s² in a direction 53.1° north of west O 10.0 m/s2 in a direction 53.1° north of west 14.0 m/s2 in a direction 36.9° north of west 10.0 m/s² in a direction 36.9° north of west O

The following four forces act on a 50.0 kg object: F₁ = 300 N east ₹ 2 = 700 N north ₹ 3 = 600 N west F 4 = 300 N south What is the acceleration of the object? 14.0 m/s² in a direction 53.1° north of west O 10.0 m/s2 in a direction 53.1° north of west 14.0 m/s2 in a direction 36.9° north of west 10.0 m/s² in a direction 36.9° north of west O

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter4: The Laws Of Motion

Section: Chapter Questions

Problem 8P: A 3.00-kg object is moving in a plane, with its x and y coordinates given by x = 5t2 1 and y = 3t3...

See similar textbooks

Similar questions

The x and y coordinates of a 4.00-kg particle moving in the xy plane under the influence of a net force F are given by x = t4 6t and y = 4t2 + 1, with x and y in meters and t in seconds. What is the magnitude of the force F at t = 4.00 s?
There are two forces on the 1.19 kg box in the overhead view of the figure but only one is shown. For F1 = 15.7 N, a = 14.7 m/s2, and θ = 34.1°, find the second force (a) in unit-vector notation and as (b) a magnitude and (c) a direction. (State the direction as a negative angle measured from the +x direction.)
Two forces F1 = (55.6i - 26.83j) N and F2 = (27.41i - 80.52j) N are exerted on a particle. The particle's mass is 10.14 kg. (a) Find the particle's acceleration in component form (express your answer in vector form). (b) what are the magnitude (m/s2) and direction of the acceleration (counterclockwise from the +x axis). Use four decimal places.
A 1200 kg car goes from 10 m/s to 18 m/s in 5 s. what is the force acting on it?Required to answer. Single choice. a. 1600 kg m/s^2 b. 1920 kg m/s^2 c. 2000 kg m/s^2 d. 2400 kg m/s^2
If the only forces acting on a 2.0kg mass are F1=(3i-8j)N and F2=(5i+3j)N, what is the magnitude of the acceleration of the particle?
Consider 2 forces acting on a mass of 2.48 kg. F1 = 10.8 N @ 22.9 degrees and F2 = 8.29 N @ 70.6 degrees where the angles are measured relative to the +x direction. What is the magnitude of the acceleration produced considering only these 2 forces acting?
A freight train consists of two 8.00x104 kg engines and 45 cars with average masses of 5.50x104 kg. What force must each engine exert backward on the track to accelerate the train at a rate of 5.00x10-2 m/s2 if the force of friction is 7.50x105 N, assuming the engines exert identical forces? Group of answer choices 882 kN 441 kN 221 kN 110 kN
The launch thrust of the Space Shuttle’s rocket engines is approximately 3.47×107 N. If the Shuttle’s launch mass is about 2.00×106 kg, what is the acceleration at liftoff? I know that F net =F thrust, so the opposing but equal force is - mg F net = F thrust - mg m = 2.00 x 10^6 kg F net = N = 2.00 x 10^6 (9.8) =mg, expressed as N is 1.96x107 N = 3.47×107 N (up) - 1.96x107 N (down) then...
A 64 kg baseball player jumps into the air throws a 151 g baseball directly eastward with a force of 8.4 N. What is the acceleration of the player? 0.13 m/s2 to the east 0.13 m/s2 to the west 0.42 m/s2 to the west 0.42 m/s2 to the east
In the figure, a slab of mass m1 = 40 kg rests on a frictionless floor, and a block of mass m2 = 10 kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. A horizontal force F→ of magnitude 104 N begins to pull directly on the block, as shown. In unit-vector notation, what are the resulting accelerations of (a) the block and (b) the slab?
Horizontal force, F1=75 N, and a force, F2= 13.3 N acting at an angle of theta to the horizontal, or apply to a block of mass m= 2.4 kg. The coefficient of kinetic friction between the block in the surface is uk= 0.2. The block is moving to the right. F1= 75 N F2= 13.3 N M= 2.4 kg (A) all of numerically for the magnitude of the normal force, Fn in Newton’s, that acts on the block if theta =30 degrees. (B) solve numerically for the magnitude of acceleration of the block, a in m/s^2, if theta =30 degrees
According to a simplified model of a mammalian heart, at each pulse, approximately 20 g of blood is accelerated from 0.26 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle? Express your answer to two significant figures and include the appropriate units. F= ______ units I got 0.02N but it is wrong.