PHYSICS QUESTION pretty easy..?

Question

a swimmer releases a 20kg rock from rest underwater. If the resistance and buoyancy provided by the water is 150N, how long will it take for the rock to fall 1m?

Answer 1

20 kg=200N
mass of the block under water=200-150
=50 N
time=1/5th of a second

Answer 2

Use the equation F=ma where F is force, m is mass and a is acceleration. Then you can find out how long it'll take to travel 1m easily enough, right?
To find the acceleration, divide the mass (20kg) by the force.
You can find the force by taking the using the mass and the distance (i think!) and then take away 150N from it to get the amount of force downward.
Sorry I can't tell you for sure how to get the Force, but I hope I've been able to be some help!

Answer 3

okay, remember that a Newton (N) is a kg-m/s^2

and F=ma or a=F/m

so the force on the bouyancy force on the rock will provide this acceleration:

a=(150 kg-m/s^2)/(20 kg)=7.5 m/s^2

this acceleration is in the opposite direction of gravity so the total acceleration

a=a(gravity)+a(bouyancy)=-9.8+7.5= -2.3 m/s^2

the next equation we need is:

d=1/2 at^2 or 2d/a=t^2 or t=sqrt(2d/a)

t=sqrt(2*-1m/-2.3m/s^2)
t=sqrt(.87 s^2)
t=.93 sec

you should check my math and my formulas but perhaps this will give you the idea

in a real situation there wouldn't be a constant bouyancy friction force because the friction gets bigger as the rock falls faster but it seems like this problem was ignoring that minor effect

Answer 4

easy indeed ;-)

net force on rock is its weight, less the buoyancy

so here, using 9.81 for g, you get a net force of

-20*9.81+150=-46.2, i.e. 46.2N, downwards

if you divide that force by the mass of the rock, you'll get the acceleration:
a=-46.2/20=-2.31m/s2

i.e. 2.31m/s2, downwards

now one of the equations for uniformly accelerated movement tells you that
x = (1/2)*a*t^2

which you can rearrange as:
t = sqrt(2*x/a)

using the a we just found, and x=1m, we get
t=sqrt(2*1/2.31)=0.93 seconds

whereas, out of the water, the rock would have needed just 0.45 seconds to fall 1m.



if you had used g=10 instead, you'd have found t=0.89 seconds.


hope this helps

Answer 5

i'm assuming it particularly is accelerating at a relentless fee and that relativity could properly be ignore approximately. i'm uncertain if those are lifelike assumptions. the customary speed is 25500 m/s + (one million/2) * 5.fifty seven * 10^6 m/s = 2.eighty one * 10^6 m/s d/v = t 0.15m/2.eighty one *10^6 m/s = 5.3 *10^-8 s

Answer 6

u made my calcalator blow up lol

Answer 7

0.2s

Answer 8

easy?!?!??!?