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V125BC [204]
3 years ago
5

Which situation is an example of an external conflict?

Physics
2 answers:
kodGreya [7K]3 years ago
8 0

Answer:

(C) A young boy on a fishing trip with his father must get the boat back to shore after his father becomes ill during the trip.

Explanation:

KiRa [710]3 years ago
3 0

Hi!


The correct option would be C.


An external conflict is a predicament where the person has to battle with an outside factor, which may be a force of nature, an animal, another person, the weather etc. External conflict may more often than not involve a physical exertion. This is opposed to internal conflicts where an individual has to struggle against a belief, a notion, or an impulse etc - <em>a conflict with oneself</em>. Such as choosing between two pair of dresses would be an internal conflict, and a brawl, or an argument would be an external conflict.

Option C is hence correct as it involves a physical struggle that requires a young boy to guide the boat back to shore, where the conflict is with a force of nature, i.e a water body.

Option A, B and D all deal with the individual in concern's internal conflicts.


Hope this helps!

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A man seeking to set a world record wants to tow a 101,000-kg airplane along a runway by pulling horizontally on a cable attache
NemiM [27]

Answer:

6.19 x 10^{-3} m/s^{2}

Explanation:

For this exercise we need to sum the forces on the y-axis and x-axis as follows:

∑F_{y} = N - mg = m.a_{y} = 0

From the exercise, we deduce there is no motion in y-axis, so:

N = mg

Then for x-axis we have:

∑F_{x} = H - f^{s} = m.a_{x} = 0

Now, from the exercise we deduce that we are looking for the greatest static friction which means to have the maximun static friction to start moving, so at this point the acceleration is zero, so we can find horizontal force (H), which then will act in the airplane to move it. Therefore we have:

H = f^{s} = f^{sma_{x} } = u_{s}N = u_{s}mg

H = (0.76)(84Kg)(9.8m/s^{2})

H = 625.63 N

Now we apply this force to the weight of the plane to find the greatest acceleration the mann can give to start moving the plane.

a = \frac{F}{m} = \frac{H}{m}

a = \frac{6325.63N}{101000Kg}

a = 6.19 x 10^{-3} m/s^{2}

7 0
3 years ago
You are riding n a bus moving slowly through heavy traffic at 2.0 m/s. You hurry to the front of the bus at 4.0 m/s relative to
kati45 [8]

Answer:

6 m/s

Explanation:

if the bus and you are moving in the same direction, then you add your speeds together

7 0
3 years ago
Iron's ability to rust is not a physical property because
gtnhenbr [62]
The answer should be c
6 0
3 years ago
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When a brick is dropped from the roof of a house, it lands on the ground with a speed v. What will the speed at ground level be
Nina [5.8K]
Twice as much by the first brick
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3 years ago
A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is
Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

\frac{\Delta l}{l_{o}}=\frac{F}{AY} (1)

T=2 \pi \sqrt{\frac{l_{o}}{g}} (2)

Where:

\Delta l=0.05 mm=5(10)^{-5} m is the length the steel wire streches (taking into account 1mm=0.001 m)

l_{o} is the length of the steel wire before being streched

F=mg=(2 kg)(9.8 m/s^{2})=19.6 N is the force due gravity (the weight) acting on the pendulum with mass m=2 kg

A is the transversal area of the wire

Y=2(10)^{11} Pa is the Young modulus for steel

T is the period of the pendulum

g=9.8 m/s^{2} is the acceleration due gravity

Knowing this, let's begin by finding A:

A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4} (3)

Where d=1.1 mm=0.0011 m is the diameter of the wire

A=\pi \frac{(0.0011 m)^{2}}{4} (4)

A=9.5(10)^{-7}m^{2} (5)

Knowing this area we can isolate l_{o} from (1):

l_{o}=\frac{\Delta l AY}{F} (6)

And substitute l_{o} in (2):

T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}} (7)

T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}} (8)

Finally:

T=1.39 s

3 0
3 years ago
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