Net Force = mass x acceleration
3500=1,000a
So a= 3500/1000
a=35/10
a=3.5 m/s^2
<span>The bullfrog is sitting at rest on the log. The force of gravity pulls down on the bullfrog. We can find the weight of the bullfrog due to the force of gravity.
weight = mg = (0.59 kg) x (9.80 m/s^2)
weight = 5.782 N
The bullfrog is pressing down on the log with a force of 5.782 newtons. Newton's third law tells us that the log must be pushing up on the bullfrog with a force of the same magnitude. Therefore, the normal force of the log on the bullfrog is 5.782 N</span>
Answer:
= 0.5 m/s²
Explanation:
- According to Newton's second law of motion, the resultant force is directly proportion to the rate of change of linear momentum.
Therefore;<em> F = ma , where F is the Force, m is the mass and a is the acceleration.</em>
<em>Thus; a = F/m</em>
<em>but; F = 5 N, and m = 10 kg</em>
<em> a = 5 /10</em>
<u>= 0.5 m/s²</u>
The centripetal acceleration a is 4.32
10^-4 m/s^2.
<u>Explanation:</u>
The speed is constant and computing the speed from the distance and time for one full lap.
Given, distance = 400 mm = 0.4 m, Time = 100 s.
Computing the v = 0.4 m / 100 s
v = 4
10^-3 m/s.
radius of the circular end r = 37 mm = 0.037 m.
centripetal acceleration a = v^2 / r
= (4
10^-3)^2 / 0.037
a = 4.32
10^-4 m/s^2.
F = G*((m sub 1*m sub 2)/r^2)