two forces act on a 8.50 - kg object. one of the forces is 14.0 n. if the object accelerates at 3.50 m/s²…

two forces act on a 8.50 - kg object. one of the forces is 14.0 n. if the object accelerates at 3.50 m/s², what is the greatest possible magnitude of the other force?\n10 n\n15.8 n\n43.8 n\n26.3 n\n73.5 n\nquestion 5\n10 pts\nif we knew that a nonzero net force is acting on an object, which of the following must we assume regarding the objects condition? the object is:\nat rest.\nmoving with a constant velocity.\nbeing accelerated.\nlosing mass.

two forces act on a 8.50 - kg object. one of the forces is 14.0 n. if the object accelerates at 3.50 m/s², what is the greatest possible magnitude of the other force?\n10 n\n15.8 n\n43.8 n\n26.3 n\n73.5 n\nquestion 5\n10 pts\nif we knew that a nonzero net force is acting on an object, which of the following must we assume regarding the objects condition? the object is:\nat rest.\nmoving with a constant velocity.\nbeing accelerated.\nlosing mass.

Answer

Question 1

Explanation:

Step1: Find the net force

According to Newton's second - law $F_{net}=ma$. Given $m = 8.50\ kg$ and $a=3.50\ m/s^{2}$, so $F_{net}=8.50\times3.50 = 29.75\ N$.

Step2: Determine the greatest possible magnitude of the second force

The two forces act on the object. Let $F_1 = 14.0\ N$ and $F_2$ be the other force. The net force $F_{net}=F_1 + F_2$ (when the two forces are in the same direction). Then $F_2=F_{net}-F_1$. Substitute $F_{net}=29.75\ N$ and $F_1 = 14.0\ N$ into the formula, we get $F_2=29.75 - 14.0=15.75\approx15.8\ N$.

Answer:

15.8 N

Question 2

Brief Explanations:

According to Newton's second - law $F = ma$, where $F$ is the net force, $m$ is the mass and $a$ is the acceleration. If a non - zero net force is acting on an object, then the object must have an acceleration. An object at rest or moving with a constant velocity has a net force of zero. There is no information to suggest the object is losing mass just because there is a non - zero net force.

Answer:

being accelerated.