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barxatty [35]
3 years ago
15

Why do youths fall in love so quick?

Physics
1 answer:
Nuetrik [128]3 years ago
3 0

Kids fall in love so quick because they have no idea what it is. Many of them get "broken hearted" But in actuality they have no idea what that is either. Love is an intense feeling of deep affection. Youths don't understand the freedom they have without someone. They only focus on what life would be with someone. Even though they don't understand everything. What I am trying to get at is, adults have more experience in the open world and how people grow up to be. Kids are just in school with the same people everyday thinking "hey I'm gonna be stuck with these people for the next 4 years might as well become more than friends"

Get what I mean? Hope this helps!

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Which of the following would probably not work with circuits on a daily basis
Anni [7]
You need to attach the answer choices
8 0
2 years ago
An airplane is flying at a speed of 200 m/s in level flight at an altitude of 800 m. A package is to be dropped from the airplan
MArishka [77]

Answer:

2560m or 2.56km (rounded to 3 significant figures)

Explanation:

First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):

Vertical \ velocity \ (\frac{m}{s} ) =  u_{v} = 0 \\\\ Horizontal \ velocity \ (\frac{m}{s} ) =  u_{h} = 200\\\\ Vertical \ acceleration \ (\frac{m}{s^{2} } ) =  a_{v} =  9.8 \\\\ Horizontal \ acceleration \ (\frac{m}{s^{2} } ) =  a_{h} =  0 \\\\ Vertical \ displacement \ (m) = s_{v} = 800 \\\\ Horizontal \ displacement \ (m) = s_{h}

The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;

First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;

Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);

What we have is the horizontal velocity but we don't have the time taken;

One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;

What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:

800 = (0)t + ¹/₂·(9.8)t²

800 = 4.9t²

t² = 163.26...

t = 12.77...

We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:

u_{h} = \frac{s_{h} }{t} \\\\ 200 = \frac{s_{h} }{12.77...} \\\\ s_{h} = 200(12.77...) \\\\ s_{h} = 2555.5...

7 0
2 years ago
A Heavy crate applied a force of 1500 N on a 25m2 piston. What force needs to be applied on a 0.8m2 piston to lift the crate?
Aleks04 [339]
25/1500 is equal to 0.8/x
0.8*1500 is equal to 1200
1200/25 is equal to 48 N
7 0
3 years ago
Read 2 more answers
A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.
LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

W = \frac{1}{2}(50)(5.61^2) = 786.8025 J

Now, we can define work:

\large\boxed{W = Fdcos\theta}}

Now, plug in the values:

786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta

Solve for theta:

cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}

4 0
2 years ago
What is the momentum of a 72 kg baseball player who slides into home plate
Alina [70]

Answer:

B) 350 kg m/s

Explanation:

momentum or p is given by the equation p= mxv

We have the mass and velocity so we can use the equation directly

p= 72kg x 4.9 m/s

p= 352.8 kg m/s

4 0
2 years ago
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