Answer:
The magnitude of the force exerted by the locomotive on the caboose is 3.08 N.
Explanation:
Given that,
Mass of a toy, m = 2.3 kg
Mass of caboose, m' = 1 kg
The frictional force of the track on the caboose is 0.48 N backward along the track.
The acceleration of train, ![a=2.6\ m/s^2](https://tex.z-dn.net/?f=a%3D2.6%5C%20m%2Fs%5E2)
The magnitude of the force exerted by the locomotive on the caboose is given by :
![F=m'a+f\\\\F=1\times 2.6+0.48\\\\F=3.08\ N](https://tex.z-dn.net/?f=F%3Dm%27a%2Bf%5C%5C%5C%5CF%3D1%5Ctimes%202.6%2B0.48%5C%5C%5C%5CF%3D3.08%5C%20N)
So, the magnitude of the force exerted by the locomotive on the caboose is 3.08 N.
An epiparasite is a parasite that parasitizes another organism that is also a parasite. Epiparasites are also called hyperparasites or secondary parasites. They feed on another organism, but they do not kill the organism they feed on.
Answer:
During its trajectory the cannon ball will have kinetic and potential energy
E = KE + PE
The maximum value of PE will occur when the cannon ball was fired straight up - for other trajectories PE will be less because of translational KE
So E = M g h = 1/2 M v^2 for maximum height
h = v^2 / (2 * g) = 400 / (2 * 9.8) = 20.4 m
Answer:
answer a: a large front gear with a small back gear
answer b: a small front gear with a large back gear
Explanation:
just simple gearing ratios
The period of the pendulum is given by the following equation
T = 2<span>π * sqrt (L/g)
Where g is the gravity (free fall acceleration)
L is the longitude of the pendulum
T is the period.
We find g.............> (T /2</span>π)<span>^</span><span>2 = L/g
g = L/(</span>T /2π)^2...........> g = 22.657 m/s^2