Help me solve this physics problem!

d = 1/2*a*t^2 + v*t

Eric Idle wrote:

d = 1/2*a*t^2 + v*t

What does v*t stand for?

fizyx wrote:

Eric Idle wrote:

d = 1/2*a*t^2 + v*t

What does v*t stand for?

Here's a hint - you can't print the actual word in a public forum so you just put the asterisk in as a place holder. Think about it. I can't really tell you more without putting myself at significant risk.

Depends how loose the girl is

177 meters or 547 ft

It's pretty easy

Change in position= Initial velocity(0m/s) times The amount of time (6s) plus 1/2 times acceleration (-9.8 m/s or -10 m/s (sig figs)) times the amount of time squared(36s)

So, basically it's:

delta x= 0 - 4.9 (36)

-120

-24

-27

-5.4

Or...

-176.4 meters

Eric Idle wrote:

d = 1/2*a*t^2 + v*t

that is correct..so around 180m if you use g=10m/s^2

Tommy2Nuttz wrote:

Eric Idle wrote:

d = 1/2*a*t^2 + v*t

that is correct..so around 180m if you use g=10m/s^2

Don't be a lazy bum. Use 9.8 m/s^2. It's still a fudge, but it's a more accurate fudge.

The airplane takes off.

MmmmkayChildren wrote:

It's pretty easy

Change in position= Initial velocity(0m/s) times The amount of time (6s) plus 1/2 times acceleration (-9.8 m/s or -10 m/s (sig figs)) times the amount of time squared(36s)

So, basically it's:

delta x= 0 - 4.9 (36)

-120

-24

-27

-5.4

Or...

-176.4 meters

Thanks. I have another one:

An astronaut on another planet drops a 1 kg rock from rest. The astronaut notices that the rock falls 2 meters straight down in one second. On this planet, how much does the rock weigh?

If you don't know what v*t stands for then you're not even ready to answer these questions.

Crap for brains wrote:

I'm going to assume you're from the US and cite this as another reason our country is headed into the shitter.

This is not a safe assumption. He could very well be from Cleveland.

fizyx wrote:

Thanks. I have another one:

An astronaut on another planet drops a 1 kg rock from rest. The astronaut notices that the rock falls 2 meters straight down in one second. On this planet, how much does the rock weigh?

You bet! Here's the relevant equation:

d = 1/2*a*t^2 + v*t

Use Newton's 2nd Law: F=ma, you know the mass of the rock is 1kg, then you can easily solve a [a=(delta)v/(delta)t].

Using a force diagram you know the only two forces that are going to be acting on the object in the vertical direction are the downward gravitational force, then acting against it is the normal force, or what we like to call weight. So you know that F(grav)=F(normal).

...I got 2 Newtons for an answer lol

fizyx wrote:

[quote]MmmmkayChildren wrote:

Thanks. I have another one:

An astronaut on another planet drops a 1 kg rock from rest. The astronaut notices that the rock falls 2 meters straight down in one second. On this planet, how much does the rock weigh?

1 Kg.

I got 4 Newtons.

s=.5at^2

2=.5(a)1^2

2=.5a

a=4 m/s/s

F=ma

F=(1kg)(4 m/s/s)

F=4 Newtons

Copy that, I misread it and thought 2 meters was 2 seconds, you're right at 4N

mmmmmmm .... Fig Newtons!

cookie monsters wrote:

mmmmmmm .... Fig Newtons!

That would be Four Fig Newtons, as in a car from Germany.