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3 years ago

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A typical adult can deliver about 12.5 N·m of torque when attempting to open a twist-off cap on a bottle. Assume that bottle cap

is about 2.0 cm in diameter.1) What is the maximum force that the average person can exert with his fingers?

1 answer:

Nikitich [7]3 years ago

3 0

Uhhhhhhhhh just tryna get a point so I can ask a question so eh I’m using ur question heheheheheh

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Hipparcos estimated its distance at roughly 96 parsecs from Earth, or 310 ± 20 light years away.

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Meher is riding a bicycle on a slope. explain the different motions taking place during this time

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Explanation:

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3 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

2) The <em>potential energy</em> at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The <em>kinetic energy</em> of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

Read more here:

brainly.com/question/21288807?referrer=searchResults

I hope it helps you!

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2 years ago

A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth

BaLLatris [955]

The answer would be 2.8m height on earth takes
2.8=1/2*9.8*t^2 => <span>s = ut +1/2at^2 </span>

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3 years ago