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3 years ago

A player uses a hockey stick to push a puck at a constant velocity across the ice. The weight of the puck is 170 N. The

Physics

2 answers:

soldi70 [24.7K]3 years ago

8 0

Answer: The minimum force by which player should hit the puck to move it is 0.102 N.

Explanation:

Weight of the puck = 1.70 N

This weight of the puck is the force acting normal to the surface that is:

N = 1.70 N

The coefficient of friction =

Frictional force =

The minimum force by which player should hit the puck to move it is 0.102 N.

Read more on Brainly.com - brainly.com/question/9177221#readmore

Bingel [31]3 years ago

6 0

Answer:

F=10.2N : Force with which the player must push the disc

Explanation:

Conceptual analysis

We apply Newton's first law because the speed is constant in horizontal direction (x):

∑Fx = 0 Formula (1)

We apply Newton's first law because the speed is zero in vertical direction (y):

∑Fy= 0 Formula (2)

∑F: algebraic sum of forces in Newton (N)

We calculate the friction force with the following formula:

Ff=f*N Formula (3)

Ff= friction force in Newtons (N)

f = coefficient of friction

N= Normal force in Newtons (N)

Known data:

Wp = 170 N : puck weight

f = 0.0600 :coefficient of friction

Problem development

We replace the data in formula (1) and (2), considering that the force is positive (+) if it goes in the direction of the movement and negative (-) if it opposes the movement:

∑Fy= 0

N-Wp=0

N= Wp=170 N

∑Fx =0

F - Ff=0 F :force with which the player must push the disc

F - 0.0600*170 =0

F=10.2N

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4 years ago

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Doc Brown holds on to the end of the minute hand of the clock atop city hall. The tangential velocity of the minute hand is 0.41

alexdok [17]

The Professor's centripetal acceleration is 0.044 m/s²

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It is given by:

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3 years ago

Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 mm away horizontal

Harrizon [31]

Answer:

a. 8.96 m/s b. 1.81 m

Explanation:

Here is the complete question.

a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.

What is her "takeoff" speed v 0 ?

b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.

If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?

a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.

So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.

b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45

R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.

So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m

8 0

3 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

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Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$