<span>c. Mammal teeth do different jobs and are different sizes and shapes</span>
The answer would be B, George Darwin :)
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
Look up pulleys problem through Khan academy and a video should pop up with a problem similar and you should be able to walk through it .
Answer:
v ’= 21.44 m / s
Explanation:
This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s
f ’= f (v + v₀) / (v-
)
where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer
in this exercise both the source and the observer are moving, we will assume that both have the same speed,
v₀ = v_{s} = v ’
we substitute
f ’= f (v + v’) / (v - v ’)
f ’/ f (v-v’) = v + v ’
v (f ’/ f -1) = v’ (1 + f ’/ f)
v ’= (f’ / f-1) / (1 + f ’/ f) v
v ’= (f’-f) / (f + f’) v
let's calculate
v ’= (3400 -3000) / (3000 +3400) 343
v ’= 400/6400 343
v ’= 21.44 m / s