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zysi [14]
3 years ago
15

Best Answer will receive BRAINLIEST One consequence of Newton's third law of motion is that __________. A. every object that has

mass has inertia B. a force acting upon an object increases that objects acceleration C. all actions have equal and opposite reactions D. none of the above
Physics
2 answers:
In-s [12.5K]3 years ago
7 0

One consequence of Newton's third law of motion is that all actions have equal and opposite reactions. <em>(C)</em>

In fact, that's pretty much what the law itself says in so many words.

Monica [59]3 years ago
7 0

Answer:

Its C All actions have equal and opposite reactions

Explanation:

Ive done this before

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A weight lifter is trying to do a bicep curl with a weight of 300 N. At the "sticking point", the moment arm of this weight is 3
lesantik [10]

Answer:

The weight lifter would not get past this sticking point.

Explanation:

Generally torque applied on the weight is mathematically represented as

             T =  F z

To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm = \frac{2}{100} = 0.02m for z the perpendicular distance

So Elbow Torque is   T_e= 4000 * 0.02

                                   = 80Nm

 To obtain the torque required we substitute 300 N for F and 30cm =\frac{30}{100} = 0.3 m

  So the Required Torque is T_R = 300 *0.3

                                                     =90Nm

Now since   T_e < T_R it mean that the weight lifter would not get past this sticking point

                                   

7 0
3 years ago
g A thin-walled hollow cylinder and a solid cylinder, both have same mass 2.0 kg and radius 20 cm, start rolling down from rest
ArbitrLikvidat [17]

Answer:

a. i. 3.43 m/s ii. 2.8 m/s

b. The thin-walled cylinder

Explanation:

a. Find translational speed of each cylinder upon reaching the bottom

The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy

So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.

i. translational speed of thin-walled cylinder upon reaching the bottom

So, For the thin-walled cylinder, I = mr², we find its translational velocity, v

So, mgh = 1/2mv² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²)(v/r)²  

mgh = 1/2mv² + 1/2mv²

mgh = mv²

v² = gh

v = √gh

v = √(9.8 m/s² × 1.2 m)

v = √(11.76 m²/s²)

v = 3.43 m/s

ii. translational speed of solid cylinder upon reaching the bottom

So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'

So, mgh = 1/2mv'² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²  

mgh = 1/2mv'² + mv'²

mgh = 3mv'²/2

v'² = 2gh/3

v' = √(2gh/3)

v' = √(2 × 9.8 m/s² × 1.2 m/3)

v' = √(23.52 m²/s²/3)

v' = √(7.84 m²/s²)

v' = 2.8 m/s

b. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Since v = 3.43 m/s > v'= 2.8 m/s,

the thin-walled cylinder has the greatest translational speed upon reaching the bottom.

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2 years ago
Real world System of equations
valina [46]
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