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Kisachek [45]
3 years ago
13

According to Newton's first law of motion, what happens to a marble traveling in a straight line on a flat hardwood floor that g

oes on forever?
Question Options:

The force applied by the floor to the marble will cause it to roll backward at the same speed it was moving forward.


A force applied by the hardwood floor will oppose the motion of the marble, causing the marble to slow to a stop.


Gravity will cause the marble to stop after 6ft.


The marble will continue to roll along the hardwood floor at the same speed in the same direction 20 more seconds.
Chemistry
2 answers:
quester [9]3 years ago
8 0
According to Newton 's first law of motion, which states that an object in motion will keep on moving unless an external balance is applied to it. The solution is the last option, the marble will continue to roll along the hardwood floor at the same speed in the same direction for 20 more seconds.
Temka [501]3 years ago
6 0

A force applied by the hardwood floor will oppose the motion of the marble, causing the marble to slow to a stop.

This is the correct answer. gravity will cause it to slow down as well as the friction of the floor.

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Some amount of hydrogen peroxide (H2O2) breaks down to produce 3 molecules of oxygen (O2) and 6 molecules of water (H2O). How ma
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Atoms_H=12Atoms

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Hello,

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Atoms_H=6moleculesH_2O*\frac{1molH_2O}{6.022x10^{23}moleculesH_2O}\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}AtomsH}{1molH} \\Atoms_H=12Atoms

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Linda performed the following trials in an experiment. Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0
nexus9112 [7]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

To calculate the amount of heat absorbed or released, we use the following equation:

q=mc\Delta T    .....(1)

where, q = amount of heat absorbed or released.

m = mass of the substance

c = heat capacity of  water = 4.186 J/g ° C      

\Delta T = Change in temperature

  • <u>For Trial 1:</u>

We are given:

m=30g\\\Delta T=[40-0]^oC=40^oC\\q=?J

Putting values in equation 1, we get:

q=30g\times 4.186J/g^oC\times 40^oC

q = 5023.2 J

  • <u>For trial 2:</u>

We are given:

m=40g\\\Delta T=[40-30]^oC=10^oC\\q=?J

Putting values in equation 1, we get:

q=40g\times 4.186J/g^oC\times 10^oC

q = 1674.4 J

Heat gained by Trial 1 than trial 2 = (5023.2-1674.4)J=3347J

Hence, the amount of heat gained in Trial 1 about 3347 J more than the heat released in Trial 2.

Thus, the correct answer is Option b.

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