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GalinKa [24]
3 years ago
6

A small bug is lying resting when nine ants simultaneously ambush it and begin pulling it in different directions. They are each

applying force to the meatloaf of a different magnitude and direction. The resultant of all these forces is zero, and the bug amazingly does not rip apart under the strain. (a) Will the bug move or remain motionless? Why?
Physics
1 answer:
xxMikexx [17]3 years ago
4 0

Answer: The bug will remain motionless

Explanation:

According to Newton's first Law of Motion (sometimes called Law of Inertia):

<em>An object at rest or describing a uniform straight line motion (moving at constant velocity), will remain at rest or moving unless an external force is applied to it and changes its state of rest or motion. </em>

In other words:

An object or body will keep its state of motion until an external force changes its state

This means that objects tend to remain in its state of motion, and is the definition of the inertia, as well.

In addition, according to his law, an object in rest can be in equilibrium (net force equals to zero), and a moving object can also be in equilibrium, as long as it keeps a constant velocity.

<h2>This is why the bug, which is at rest will remain at rest, although the ants are simultaneously pulling it in different directions, since the resultant of all these forces is zero.</h2>
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A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The
lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

E_{i} = E_{f}

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}                              

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s      

 

3) When the object is at the equilibrium position, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s

I hope it helps you!                                                                                        

8 0
3 years ago
Newton's laws of motion work well for ordinary situations on earth. However, these laws of motion do not work for all cases. Und
Alexandra [31]

Answer:

C: for objects at extremely fast speeds.

Explanation:

Newton's second law does not hold for extremely fast speeds, because then relativistic effects come into play, where Einstein's theory of special relativity is a more correct description.

The reason why F=ma does not hold for fast speeds is that, as an object moves faster and faster, the proportional relationship between force and acceleration does not hold. As an object moves faster and faster, it becomes harder and harder (requires more force) to accelerate it. because it gains mass as a virtue of its velocity (what's called relativistic mass).

For relativistic speeds, the correct modification of Newtons second law is:

F=\frac{dp}{dt}

where p is the relativistic momentum:

p=\frac{mv}{\sqrt{1-\frac{v^2}{c^3} } }

3 0
3 years ago
1)How does scientist use radioactive dating to determine how old a fossil is?
Lyrx [107]

Answer:

the answer is B

Explanation:

i got mine right when i did mine

8 0
3 years ago
Which characteristic does Earth share with the other
Verizon [17]

Answer: B) They all orbit the sun

Explanation:

3 0
2 years ago
Suppose a 60-kg boy and a 41-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface. If the acceleration
Digiron [165]

Answer:

Approximately 2.05\; {\rm m\cdot s^{-2}}.

Explanation:

The net force on the girl would be:

\begin{aligned}m(\text{girl}) \, a(\text{girl}) &= 41\; {\rm kg} \times 3.0\; {\rm m\cdot s^{-2}} \\ &= 123.0\; {\rm N} \end{aligned}.

Under the assumptions, the net force on this girl would be equal to the tension force in the rope. All other forces on the girl would be balanced.

In other words, the tension force that the rope exerted on the girl would be 123.0\; {\rm N}. The girl would exert a reaction force on the rope at the same magnitude (123.0\; {\rm N}\!) in the opposite direction. This force would translate to a 123.0\; {\rm N}\!\! force on the boy towards the girl.

Under similar assumptions, the net force on the boy would also be 123.0\; {\rm N}. Since the mass of the boy is m(\text{boy}) = 60\; {\rm kg}, the acceleration of the boy would be:

\begin{aligned}a(\text{boy}) &= \frac{(\text{net force})}{m(\text{boy})} \\ &= \frac{123.0\; {\rm N}}{60\; {\rm kg}} \\ &= 2.05\; {\rm m\cdot s^{-2}}\end{aligned}.

3 0
2 years ago
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