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andrezito [222]
3 years ago
14

If a metal ball suspended by a rod is at rest, which force is responsible for balancing the force due to gravity?

Physics
2 answers:
sineoko [7]3 years ago
4 0
The normal force applied by the rod on tbe metal ball is the force balancing the force due to gravity.

The normal force is the force that keeps any object from going through another object and that is why tables and floors can keep things up.
cestrela7 [59]3 years ago
3 0
Any force coming from the surface and acting at a right angle to the surface is called the Normal Force<span>.
normal force </span><span>is responsible for balancing the force due to gravity.</span>
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The student used two different types of thermometer to measure the temperature
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5 0
8 months ago
A 0.11 kg bullet traveling at speed hits a 18.3 kg block of wood and stays in the wood. The block with the bullet imbedded in it
arlik [135]

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5 0
2 years ago
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
liq [111]

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

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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

8 0
3 years ago
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