Answer:
0.71 m/s
Explanation:
We find the time it takes the stone to hit the water.
Using y = ut - 1/2gt² where y = height of bridge, u = initial speed of stone = 0 m/s, g = acceleration due to gravity = -9.8 m/s² (negative since it is directed downwards)and t = time it takes the stone to hit the water surface.
So, substituting the values of the variables into the equation, we have
y = ut - 1/2gt²
82.2 m = (0m/s)t - 1/2( -9.8 m/s²)t²
82.2 m = 0 + (4.9 m/s²)t²
82.2 m = (4.9 m/s²)t²
t² = 82.2 m/4.9 m/s²
t² = 16.78 s²
t = √16.78 s²
t = 4.1 s
This is also the time it takes the raft to move from 5.04 m before the bridge to 2.13 m before the bridge. So, the distance moved by the raft in time t = 4.1 s is 5.04 m - 2.13 m = 2.91 m.
Since speed = distance/time, the raft's speed v = 2.91 m/4.1 s = 0.71 m/s
Hey there!
Seems like you're looking for the size and direction to the final velocity of the two cars. To find it, you must solve it like this.
0.4 kg(3 m/s) + 0.8kg(–2 m/s) = 1.2 kg m/s -1.6 kg m/s = –0.4 kg m/s
–0.4 kg m/s = 1.2 kg(v) = (–0.4 kg m/s)/(1.2 kg) = v = –0.33 m/s
So, the cars are traveling at -0.33 m/s in the direction of the second car.
Hope this helps
<em>Tobey</em>
"Balanced" means that if there's something pulling one way, then there's also
something else pulling the other way.
-- If there's a kid sitting on one end of a see-saw, and another one with the
same weight sitting on the other end, then the see-saw is balanced, and
neither end goes up or down. It's just as if there's nobody sitting on it.
-- If there's a tug-of-war going on, and there are 300 freshmen pulling on one
end of a rope, and another 300 freshmen pulling in the opposite direction on
the other end of the rope, then the hanky hanging from the middle of the rope
doesn't move. The pulls on the rope are balanced, and it's just as if nobody
is pulling on it at all.
-- If a lady in the supermarket is pushing her shopping cart up the aisle, and her
two little kids are in front of the cart pushing it in the other direction, backwards,
toward her. If the kids are strong enough, then the forces on the cart can be
balanced. Then the cart doesn't move at all, and it's just as if nobody is pushing
on it at all.
From these examples, you can see a few things:
-- There's no such thing as "a balanced force" or "an unbalanced force".
It's a <em><u>group</u> of forces</em> that is either balanced or unbalanced.
-- The group of forces is balanced if their strengths and directions are
just right so that each force is canceled out by one or more of the others.
-- When the group of forces on an object is balanced, then the effect on the
object is just as if there were no force on it at all.
Answer:
The balloon hit the ground with velocity -15.34 m/s
Explanation:
<em>Lets explain how to solve the problem</em>
You found that the best height to pitch a water balloon in order for it to
burst when it hits the ground is 12 meters.
We consider that the 12 meters is the maximum height, so the velocity
at this height is zero.
To find the velocity when the balloon hits the ground lets use the rule
<em>v² = u² + 2gh</em>, where v is the final velocity, u is the initial velocity, g is
the acceleration of gravity and h is the height.
u = 0 , h = 12 m , g = 9.8 m/s²
<em>Substitute these values in the equation above</em>
v² = 0 + 2(9.8)(12)
v² = 235.2
<em>Take square root for both sides</em>
v = ± 
The velocity is downward, then it's a negative value
Then v = -15.34 m/s
<em>The balloon hit the ground with velocity -15.34 m/s</em>
To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have
Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s
