Studyit

Mechanics - Scalars and Vectors

Need urgent help on the following, thanks to whoever answers it!

Bob is at the start of his 100m sprint race. He is at rest and, at the firing of the starting gun he accelerates for 6.0 s up to his sprinting speed. He travels 25 m while accelerating.

a. Calculate Bob’s acceleration. Give your answer to the correct number of significant figures.

Please answer the following:

Jason drops a leaf into the water. (The leaf is lighter than the stone.) How will the velocity of the leaf compare with the velocity of the stone when it hits the water? Compare the acceleration of the leaf and the stone as they fall from rest.

For this question you have the distance and time and you want to find the acceleration. You also know the intial velcoity since he started from rest so you will use d=vit+1/2at2 in this situation and solve for acceleration.

d= 25m vi=0 t=6.0s
25=0×6+a×62
25=36a
a=25/36
a=0.6944
a=0.69ms-2 (2sf)

The acceleartion of the leaf and the stone will be the same. On Earth’s surface all objects regardless of their mass accelerate at 9.8ms-2. When they hit the water, if we consider friction to be neglegible (no air resistance and we assume that the only force acting on the objects is gravity) and we also assume they are dropped from the same height then the velocity of the leaf and the stone will be the same when they hit the water.

Do you think you can answer an atomic nuclear physics question?

I can have a go, I think I remember most of it from last year haha.

The half-life of Strontium 90 is about 29 years and it decays into Yttrium 90.

Yttrium 90 is also radioactive and is a beta particle emitter with a half-life of 2.67 days.

Discuss the amount of Yttrium that would be present in the above example after the

Strontium had been decaying for the first 8 days

You need to know the intial amount of stronitum 90, did the question say that?


This is my attempt at a general solution when x=the intial amount of Strontium 90.

I was thinking the same thing, unfortunetly no