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lana66690 [7]
3 years ago
12

Please tell me if this is Newton's first law, second, or third law of motion. There can be more than one of the same answer very

desperate to pass science!

Physics
1 answer:
AfilCa [17]3 years ago
5 0

Answer: Hope This Helps!

Explanation:

1: Newton’s first law of motion can explain how a magician pulls a tablecloth from underneath the dishes. A negligible horizontal force is applied during the process. As per Newton’s first law of motion, the dishes and glasses remain in their state of motion (rest); as a result, they remain undisturbed.

2: Newton's First Law of Motion is defined as "An object at rest and an object in motion will stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force."In soccer, when the soccer ball is in the soccer field and it is not moving, that means that it is at rest and there is no force acting upon it. When there is a person that is ready to play soccer and wants to kick the ball and play, then the unbalanced force would be the power from the person's foot.

3: Newtons third law can explain, as the cannonball is pushed forwards by the expanding high-pressure gases created by the exploding gunpowder, it pushes back on these gases. The gases push back on the cannon itself, causing it to roll backwards. Alternative answer: the cannon pushes forward on the cannonball. the reaction force is the cannonball pushing backwards on the cannon.

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A 13 cm long tendon was found to stretch 3.7 mm by a force of 12.1 N . The tendon was approximately round with an average diamet
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Answer:

young's modulus=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2

Explanation:

We have given length of tendon L= 13 cm =0.13 m

Change in length \Delta L=3.7mm=3.7\times 10^{-3}m

Force = 12.1 N

Average diameter d =8.8 mm

So r=\frac{d}{2}=\frac{8.8}{2}=4.4mm=4.4\times 10^{-3}m

Area A=\pi\times  (4.4\times 10^{-3})^2=60.79\times 10^{-6}m^2

Now stress =\frac{force}{area}=\frac{12.1}{60.79\times 10^{-6}}=0.1990\times 10^6N/m^2

Strain =\frac{change\ in\ lenght}{length}=\frac{3.7\times 10^{-3}}{0.13}=28.46\times 10^{-3}

Now young's modulus=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2

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D on Edge

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