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Shalnov [3]
3 years ago
6

The reckless risk taking found among many adolescents is most likely due to the effects of _______.

Physics
2 answers:
ivanzaharov [21]3 years ago
8 0
The reckless risk taking found among many adolescents is most likely due to the effects of        <span>Peer pressure</span>
alexdok [17]3 years ago
3 0
Oh that's nothing to worry about.  It's normal.  It's due to the effects of youth,
and the instinctive impression of immortality that comes with it at no extra
charge just pay shipping and handling. 

Adolescent reasoning goes something like this:  "Whaddaya mean
this is dangerous ?  Whaddaya mean I'm being reckless ? ! ?
Who says ?  I've been doing it for almost a week now and I haven't
died yet and I don't know anybody who died from it, so it can't be that bad,
and you don't know anything and you just don't want me to have any fun,
and I'm just gonna keep doing it at least until I need glasses."
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7 0
3 years ago
7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
artcher [175]

Answer:

20.4 m/s^{2}

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 m/s^{2}

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3 years ago
In mechanics in what do we apply v=u+at​
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Answer:

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