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Nuetrik [128]
3 years ago
8

A hockey player swings her hockey stick and strikes a puck. According to Newtons 3rd law of motion which of the following is a r

eaction to the stick pushing on the puck
Physics
1 answer:
Korvikt [17]3 years ago
3 0

<u>Complete Question:</u>

A hockey player swings her hockey stick and strikes a puck. According to Newton’s third law of motion, which of the following is a reaction to the stick pushing on the puck?

A. the puck pushing on the stick .

B. the stick pushing on the player .

C. the player pushing on the stick .

D. the puck pushing on the player.

<u>Correct Option:</u>

According to Newton’s third law of motion the puck pushing on the stick is a reaction to the stick pushing on the puck.

<u>Option: A</u>

<u>Explanation:</u>

As when the hockey exert force on the puck (which is a flat ball basically used in ice hockey) then this action by hockey will receive equal and opposite reaction by puck. Thus when the stick is pushing on the this flat ball, then puck also push the stick. This is understood by newton's third law pf motion, where action and reaction forces are subject of discussion, displaying their is pair of forces applied among the interacting objects. This form is observed more practically in life and very frequent.

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5 0
3 years ago
A copper cylinder is initially at 21.1 ∘C . At what temperature will its volume be 0.163 % larger than it is at 21.1 ∘C?
fomenos

To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as V_1, the relation with the final volume as

V_2 = V_1 +0.163\% V_1

V_2 = V_1 +0.00163V_1

V_2 = 1.00163V_1

Initial temperature = 21.1\°C

Let T be the temperature after expanding by the formula of volume expansion

we have,

V_2 = V_1 (1+\gamma \Delta t)

Where \gamma is the volume coefficient of copper 5.1*10^{-5}/C

1.00163V_1 = V_1(1+\gamma(T-21.1\°))

1.00163 = 1+5.1*10^{-5}(T-21.1\°)

0.00163 = 0.000051T-0.0010761

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7 0
3 years ago
the chef was careless and put the aluminum cookie sheet directly on the hot stove, which melted 0.05 kg of the aluminum. how muc
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Heat required to melt 0.05 kg of aluminum is 28.7 kJ.

<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>

The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.

The formula to be used is given below:

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Assuming the aluminum sheet was at room temperature initially.;

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Melting point of aluminum = 660.3 °C

Temperature difference = (660.3 - 25) = 635.3 903

Heat capacity of aluminum = 903 J/kg/903

Heat required = 0.05 * 903 * 635.3

Heat required = 28.7 kJ

In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.

Learn more about heat capacity at: brainly.com/question/21406849

#SPJ1

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