Question

3 multiple choice 0.5 points my car ran out of gas and now has to be pushed to the gas station over on the left side of the road. i can apply 45 n of force while i push and steer. luckily, i had the help of my friends. chad was able to push with 85 n and karen was able to push with 90 n of force. what is the net force applied to the car? 220 newtons - 130 newtons 130 newtons -220 newtons 4 multiple choice 0.5 points two friends are having an argument to see who is stronger. they decide to push on a refrigerator to see who can overtake the other. one pushes left on it with 250 n of force while the other pushes to the right with 265 n of force. what is the net force on the refrigerator? -15 newtons 15 newtons 515 newtons -515 newtons

Explanation:

Question 3

Step1: Identify forces direction

All forces (my 45 N, Chad's 85 N, Karen's 90 N) are in the same direction (pushing to the gas station on the left, so we can consider them as positive in that direction).

Step2: Sum the forces

Net force $F_{net}=45 + 85+90$
$F_{net}=220$ N

Step1: Assign signs to directions

Let right be positive and left be negative. So the force to the left is $- 250$ N and the force to the right is $265$ N.

Step2: Calculate net force

Net force $F_{net}=265+( - 250)$
$F_{net}=265 - 250 = 15$ N (positive means direction is right, but the magnitude is 15 N. If we took left as positive, it would be - 15 N, but typically we can also just calculate the magnitude difference with sign based on direction. Since right force is larger, net force is 15 N in right direction, or + 15 N. But let's check the options: 15 Newtons is an option. Wait, wait, no: wait, if left is negative and right is positive, then $F_{net}=265-250 = 15$ N (positive, so right direction). But let's re - check: force1 (left) = - 250 N, force2 (right)=265 N. Net force $F = F_1+F_2=-250 + 265=15$ N.

Answer:

220 Newtons

Question 4