The solution for this problem is:
Let u denote speed.
Equating momentum before and after collision:
= 0.060 * 40 = (1.5 + 0.060) u
= 2.4 = 1.56 u
= 2.4 / 1.56 = 1.56 u / 1.56
= 1.6 m / s is the answer for this question. This is the speed after the collision.
A. 0.5kg
To get this answer you need to follow the equation of KE=0.5*mv^2
But we don't have the m part in the equation. So just plug in the numbers to see which works best, though I can tell you before we do that the answer would be a.
As you may know, gravity, is a force of 9.8 m/s. And we want to get 9.8 Joules. So if we take a half a kg stone, release it at one meter, we get half of the normal gravity pull, 4.90 Joules. That means if we take half a kg stone and drop it at a doubled height, we get 9.8 Joules.
That is also to say that if we have a 1kg stone and drop it at one meter you will get the normal pull of gravity in Joules, 9.8J.
Be careful though, this does not mean if you drop a 1kg stone and a .5 kg stone the 1kg will hit first. This simply means that the 1kg stone will have twice the Joules that the .5kg stone has.
Answer:
First one, third one, and fourth one
Answer:
Here's what I got:
Let's assume that N and E are + directions while S and W are - directions.
Wind is blowing from SW; thus, it is blowing towards NE (or at 45 deg N of E).
Dividing the wind's speed into components:y-component: +70.71 km/h; x-component: +70.71 km/h
Dividing the airplane's speed into components:y-component: -600 km/h; x-component: 0 km/h
Adding the components to get the resulting components:y-component: -529.29 km/h; x-component: +70.71
Using the Pythagorean Theorem to find the resulting speed:v^2 = y^2 + x^2 so v = 533.99 km/h
To find the angle of direction, use arctan (y/x):arctan (529.29/70.71) = 82.39 deg
ANSWER: velocity = 533.99 km/h at 82.39 deg S of E
Explanation:
