Answer:
v₃ = 3.33 [m/s]
Explanation:
This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.
In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.
where:
m₁ = mass of the car = 1000 [kg]
v₁ = velocity of the car = 10 [m/s]
m₂ = mass of the truck = 2000 [kg]
v₂ = velocity of the truck = 0 (stationary)
v₃ = velocity of the two vehicles after the collision [m/s].
Now replacing:
![(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]](https://tex.z-dn.net/?f=%281000%2A10%29%2B%282000%2A0%29%3D%281000%2B2000%29%2Av_%7B3%7D%5C%5Cv_%7B3%7D%3D3.33%5Bm%2Fs%5D)
Answer:
Nucleus And electron cloud
Explanation:
Hope this helps
Answer:
568.18 N
Explanation:
From the question,
The formula for gravitational potential is given as
Ep = mgh........................ Equation 1
Where Ep = Gravitational potential, m = mass of the diver,h = Height.
But,
W = mg.................... Equation 2
Where W = weight of the diver.
Substitute equation 2 into equation 1
Ep = Wh
Make W the subject of the equation
W = Ep/h................... Equation 3
Given: Ep = 25000 J, h = 44 m
Substitute into equation 3
W = 25000/44
W = 568.18 N.
Hence the weight of the diver = 568.18 N
It is gaining potental energy which will then transfer to knetic energy as it falls
