D. They are heterotrophs that digest food internally.
It took so long because at the time there was no way for people to study the behavior formally. im not sure what helped it get recognized but i know wihelm wundt helped get it recongnized.
sorry i couldnt be much help
Ricks velocity would be zooomin out because it would fall off so strongly so it’d change and it’s weight too
Answer:
c. vf is greator than v2, but less than v1
Explanation:
The principle of conservation of linear momentum states that when two or more bodies act upon one another, their total momentum remains constant.
In a system of colliding bodies the total momentum of the system just before the collision is the same as the total momentum just after the collision.
Collisions in which the kinetic energy is conserved are called elastic collision.
Collisions in which the kinetic energy is not conserved are called inelastic collisions. If the two objects stick together after the collision and move with a common velocity, the collision is said to be perfectly inelastic.
<em>The above scenario is a perfectly inelastic collision. The initial velocity of particle 1 was greater than particle 2 before collision. After collision, its velocity will reduce to a final velocity vf as it transfers some of its kinetic energy to particle 2; whereas, the velocity of particle 2 will increase to a final velocity vf as it absorbs some of the kinetic energy of particle 1.</em>
Therefore,
a. vf = v2 is wrong because vf is greater than v2
b. vf is less than v2 is wrong because vf is greater than v2
c. vf is greater than v2, but less than v1 is correct.
d. vf = v1 is wrong because vf is less than v1
Answer:
B. 2 m/s
B. Acceleration = 4.05 m/s² and Tension = 297.5 N.
Explanation:
A force is applied on a mass m whose acceleration is 4 m/s
Force = mass × acceleration
a = F/m = 4 m/s
4 m/s = F/m
F = 4 m/s (m)
If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:
2F/4 m = F/ 2m
4m/s (m) / 2m = 2 m/s
a = 2 m/s
2.
Given that
mass 
= 30 kg
mass 
= 50 kg
= 0.1
From the question; we can arrive at two cases;
That :
----- equation (1)
---- equation (2)
50 a = 50 g - T
30 a = T - 30 g sin 30 - 4 × 30 g cos 30
By summation
80 a =![[ 50 - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]](https://tex.z-dn.net/?f=%5B%2050%20%20-%2030%20%2A%20%5Cfrac%7B1%7D%7B2%7D%20-%200.1%20%2A30%2A%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5D)
g
80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)
80 a = 324 m/s ²
a = 324/80
a = 4.05 m/s²
From equation , replace a with 4.05
50 × 4.05 = 50 × 10 - T
T = 500 -202.5
T =297.5 N
