Answer:
I'm not sure..but please refer to your teacher later.
Answer: Based on Newton's First law of motion (where inertia is involved), smooth ice increases the forceused to accelerate the hockey puck.
Explanation;
- smooth ice reduces the resistances between the surface of the figure skates and the ice itself.
- based on inertia theory ; the heavier the weight, the larger the inertia.. which explains it takes alot of force to move a heavier object than the lighter ones.. it also hard to *stop* the motion of heavier objects than the lighter ones.
- now let's look at the design of the player shoe itself, they have a sharp blade at the bottom of the figure stakes.. which takes us to the law of the force.. the smaller the surface area, the more forces acting on it. So, players force (weight, F= mg) acts on the tip of the blade and on the ice
- high inertia (run fast) and high force (attack opponent and pass puck) enables them to perform well in playing hockey
- Thus if there's no resistance and the inertia of the player is high then they could run and pass the puck quickly
We Know, F = m*a
F = 2200 * 3.4
F = 7480 Kg m/s²
So, your final answer is 7480
Answer:
is the distance from the obstacle of reflection.
wavelength 
Explanation:
Given that:
- frequency of sound,

- time taken for the echo to be heard,

- speed of sound,

We know,

<em>During an echo the sound travels the same distance back and forth.</em>


is the distance from the obstacle of reflection.
<u>Now the wavelength of sound waves:</u>



Answer:
21.42m/S
Explanation:
Hello!
To solve this problem we must perform the following steps.
1. Find the distance traveled from until the driver reacts, this is achieved using the equation for constant speed movement.
X1=VT
where
x= distance traveled
v=initial speed
T=time=0.75s
X1=0.75Vo
we must take into account that the total distance is the sum of the distance at which the pilot reacts (x1) and when it starts to decelerate (x2)
39=0.75Vo+X2
X2=39-0.75Vo
2. Now we use the equation that defines a movement with constant acceleration.

where
Vf = final speed=0m/s
Vo = Initial speed
A = acceleration
=-10m/s2
X2 = displacement
now we use the ecuation of step 1

solving

Now we solve the quadratic equation and find the value of Vo
the solutions are 21.42m/S, -36.41m/S
as the speed must be positive we conclude that the answer is 21.42m/S