A 1540-kg car is traveling with a speed of 12.4 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 61.1 m? Number 236.40 Units N eTextbook and Media Hint

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A horizontal force, F1 = 55 N, and a force, F2 = 17 N acting at an angle of 0 to the horizontal, are applied to a block of mass m = 3.9 kg. The coefficient of kinetic friction between the block and the surface is uk 0.2. The block is moving to the right. = Randomized Variables F₁ = 55 N F₂=17 N m = 3.9 kg o vi ble. on d ed ed FN=46.72 (b) Solve numerically for the magnitude of acceleration of the block, a in m/s², if 0 = 30°. (a) Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if 0 = 30°. ✓ Correct! a = 15.487 tan() П acos() E 14 sinh() 1 * 1 tanh() cotanh() Degrees O Radians 2 0 VO BACKSPACE sin() cos() cotan() asin() atan() acotan() cosh() Submit Hint . Feedback 7 85 9 6 3 I give up! • DEL HOME ← END y. CLEAR X Hk m 0

A stockroom worker pushes a box with mass 11.6 kg on a horizontal surface with a constant speed of 3.40 m/s. The coefficient of kinetic Part A friction between the box and the surface is 0.21. What horizontal force must the worker apply to maintain the motion? Express your answer with the appropriate units. HA ? F = Value Units Submit Request Answer

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

A cart of mass 8.00 kg was moved by applying two constant forces. Force 1 is 28.0 N at 42.0°, and Force 2 is 13.0 N at 110°. Initially, the cart has a velocity of (3.50 i +2.20 j) m/s. d. What is the acceleration of the cart? Ans. ä = (2.05 i + 3.87 j) or 4.38 e. What is the cart's velocity after 5.00 s? Ans. i = (13.75 i + 21.55 j)" or 25.56" f. What is the position of the cart after 5.00 s? † = 43.125i + 59.375j or73.38 m

A 1.4 gram bullet is shot into a tree stump. It enters at a velocity of 341 m/sec and comes to rest after having penetrated the stump in a straight line. It takes 0.00032 sec to come to a stop after striking the stump. What was the force on the bullet during impact?  (answer in Newtons)

A 945.0kg car traveling rightward at 22.6 m/s slams on the brakes and skids to a stop (with locked wheels). If the coefficient of friction between tires and road is 0.9710, determine the force of friction. Use correct number of significant figures in answer Force =

Determine the force Q-> when the block moves with constant velocity. Express your answer in vector form.