Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
The first law states that “objects at rest and objects in motion remain in motion in a straight line unless acted upon by an unbalanced force”. Keeping the ice smooth will make sure there is not friction, friction would slow the puck down
Answer:
the heavier skater has less momentum
hope it is helpful to you
Answer:
c. V = 2 m/s
Explanation:
Using the conservation of energy:
so:
Mgh = 
where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.
Also we know that:
V = WR
Where R is the radius of the disk, so:
W = V/R
Also, the moment of inertia of the disk is equal to:
I = 
I = 
I = 10 kg*m^2
so, we can write the initial equation as:
Mgh = 
Replacing the data:
(5kg)(9.8)(0.3m) = 
solving for V:
(5kg)(9.8)(0.3m) = 
V = 2 m/s
Answer:
The height of the tree is three (3) deep
Explanation:
It's 3 deep
Under 129, comes 125 and 685;
Under 125, comes 52 : Under 685, comes 511;
Under 52, comes 46 : Under 511, is 601.
This is illustrated below.
129
∧
125,685
|,|
52,511
|,|
46,601
