Question

What happens to the acceleration if you triple the force that you apply to the painting with your hand? (Use the values from the example given in the previous part of the lecture.) Submit All Answers Answer: Not yet correct, tries 1/5 3. A driver slams on the car brakes, and the car skids to a halt. Which of the free body diagrams below best matches the braking force on the car. (Note: The car is moving in the forward direction to the right.] (A) (B) (C) (D) No more tries. Hint: (Explanation) The answer is A. The car is moving to the right and slowing down, so the acceleration points to the left. The only significant force acting on the car is the braking force, so this must be pointing left because the net force always shares the same direction as the object's acceleration. 4. Suppose that the car comes to a stop from a speed of 40 mi/hr in 24 seconds. What was the car's acceleration rate (assuming it is constant). Answer: Submit Al Answers Last Answer: 55 N Only a number required, Computer reads units of N, tries 0/5. 5. What is the magnitude (or strength) of the braking force acting on the car? [The car's mass is 1200 kg.) Answer: Submit Al Answers Last Answer: 55N Not yet correct, tries 0/5

Answer 1

Answer:

2) when acceleration triples force triples,  3) a diagram with dynamic friction force in the opposite direction of movement of the car

4)  a = 2.44 ft / s², 5)  fr = 894.3 N

Explanation:

In this exercise you are asked to answer some short questions

2)  Newton's second law is

         F = m a

when acceleration triples force triples

3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.

The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car

4) let's use the scientific expressions

          v = v₀ - a t

as the car stops v = 0

          a = v₀ / t

let's reduce the magnitudes

          v₀ = 40 mile / h ($\frac{5280 ft}{1 mile}$) ($\frac{1 h}{3600 s}$) = 58.667 ft / s

          a = 58.667 / 24

          a = 2.44 ft / s²

5) let's use Newton's second law

           fr = m a

We must be careful not to mix the units, we will reduce the acceleration to the system Yes

           a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²

           fr = 1200  0.745

           fr = 894.3 N