Answer:
The increase in gravitational potential energy is the same in both cases
Explanation:
It is easier to climb a mountain in a zigzag way rather than climbing on a straight line but since the distance is the same ( vertical height ) , mass and gravity is the same. Hence the increase in gravitational potential energy is the same in both cases.
gravitational potential energy = mgh ( same in both cases )
m = mass
g = acceleration due to gravity
h = distance ( vertical height )
The rock it traveling really, really fast.
It is hard to exactly determine how fast bc u need the height of the cliff and how big the rock is.
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Answer:
The example of the center of the gravity is the middle of a seesaw
Explanation:
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Answer:
The position of the car at t = 1.5 s is at -8.1625 meters
Explanation:
The initial position of the car is 3.2 meters
The initial velocity is -8.4 m/s
The constant acceleration is 1.1 m/s²
We need to find the final position of the car at the time t = 1.5 seconds
The displacement <em>s</em> = final position - initial position
, where <em>u</em> is the initial velocity, <em>a</em> is the
constant acceleration and <em>t</em> is the time
So we can find the final velocity by using the rule:
final position - initial position = 
initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s
Substitute these values in the rule
final position - 3.2 = 
final position - 3.2 = -12.6 + 1.2375
final position - 3.2 = -11.3625
add 3.2 for both sides
final position = -8.1625
<em>That means the car is at 8.1625 meters in opposite direction</em>
<em>The position of the car at t = 1.5 s is at -8.1625 meters </em>
Answer:
at t=46/22, x=24 699/1210 ≈ 24.56m
Explanation:
The general equation for location is:
x(t) = x₀ + v₀·t + 1/2 a·t²
Where:
x(t) is the location at time t. Let's say this is the height above the base of the cliff.
x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0
v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.
a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².
Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.
Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²
Stone: x(t) = 0 + 22·t - 1/2*9.8 t²
Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:
46 = 22·t
so t = 46/22 ≈ 2.09
Put this t back into either original (i.e., with the quadratic term) equation and get:
x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m
