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patriot [66]
2 years ago
6

What is a passing mechanic cues to remember when you are the receiver of a pass in hockey?

Physics
1 answer:
VARVARA [1.3K]2 years ago
6 0

Answer:

Positions in Hockey: 6 players for each team on the ice

1 Goalie – the player in the goal who tries to stop the puck from going in the net.

1 Center – plays in between the two wings and is usually the best passer on the team

2 Wings – offensive players who plays on both sides of the center. They are usually goal scorers

2 Defensemen – main job is to play defense and help defend the goal

Passing Cues

1. Stick blade faces target

2. Puck in center of blade

3. Transfer weight rear to front as you pass

4. Use wrist movement to drive the puck

5. Follow through at target

Receiving Cues:

1. athletic position

2. catch puck with middle of blade and control

3. slow the puck when it contacts the stick by giving with it

Explanation:

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A car travels due east 14 miles in 18 minutes. Then, the car turns around and returns to its starting point, taking an additiona
N76 [4]

Answer:

(a) The average speed is 0.85 milles/minute

(b) The average velocity is zero

Explanation:

In order to answer part (a) and (b) you have to apply the formulas for average speed and average velocity which are:

<em>-Average speed formula:</em>

v=\frac{d}{t}

where d is the total distance traveled and t is the tota time

Replacing the given values:

v=\frac{14+14}{18+15}\\v=\frac{28}{33} \\v=0.85 milles/minute

Notice that you have to replace the total distance, which is 14 milles for the go plus 14 milles for the return. The same for the total time.

<em>-Average velocity formula:</em>

V = Δx/Δt

Where V is the velocity vector, Δx is the displacement and Δt is the change in time

V= \frac{X2-X1}{t2-t1}

Where X2 is the final position and X1 is the initial position

In this case X1= 0 i and X2=0 i (i is the unit vector in the x direction). So, the displacement is zero.

Therefore, the average velocity is:

V= 0 i [milles/minute]

4 0
3 years ago
Show that the acceleration of any object down an incline where friction behaves simply (that is, where fk=μkN ) is a=g(sinθ−μkco
11Alexandr11 [23.1K]

Answer:

a=g(sinθ-μkcosθ)

Explanation:

In an inclined plane the forces that interact with the object can be seen in the figure. The normal force, the weight w and the decomposition of the force vector of weight can be observed.

wx=m*g*sinθ

wy=m*g*cosθ

As the objects moves down an incline, acceleration in y axis is 0.

Then, by second Newton's Law:

Fy = m*ay

FN - m*g cos θ = 0,

FN=m*g cos θ

In x axis the forces that interacs are the x component of weight and friction force:

Fx = m*ax

mg sen u-FN*μk=m*a

Being friction force, Fr=FN*μk, we replace with its value in below formula:

m*g *sinθ-(m*g*cosθ*μk)=m*a

Then, isolating a:

a=(m*g sinθ-(m*g*cosθ*μk))/m

Solving, we have next equation:

a=g sinθ-(g*cosθ*μk)

Applying distributive property we have:

a=g*(sinθ-μk*cosθ)

5 0
3 years ago
Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 4.0 m behind the apertur
devlian [24]

Answer:

d = 0.247 mm

Explanation:

given,

λ = 633 nm

distance from the hole to the screen = L = 4 m

width of the central maximum = 2.5 cm

                                             2 y = 0.025 m

                                               y = 0.0125 m

For circular aperture

  sin \theta = 1.22\dfrac{\lambda}{d}

using small angle approximation

  \theta = \dfrac{y}{D}

now,

   \dfrac{y}{D} = 1.22\dfrac{\lambda}{d}

   y = 1.22\dfrac{\lambda\ D}{d}

   d = 1.22\dfrac{\lambda\ D}{y}

   d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}

         d =0.247 x 10⁻³ m

         d = 0.247 mm

the diameter of the hole is equal to 0.247 mm

5 0
3 years ago
What type of fingerprints is invisible to the naked eye?
luda_lava [24]
Well latent fingerprints are made of oil and sweat and generally materials that you can't see very easily, so it should be that.

Hope this helps :D

3 0
4 years ago
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite
sladkih [1.3K]

Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

       The gravitational force is  F_g  = 6600 \ N

The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

     v  = \sqrt{\frac{G  M}{r^2} }

=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

7 0
3 years ago
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