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3 years ago

Please someone help, I’m very confused and it’s due soon, thanks

Physics

1 answer:

Anit [1.1K]3 years ago

5 0

Answer:

1 s
19.6 m
2 s
0.8 m/s^2
28 m/s
79 m/s
0.37 s
26 m/s
242 m/s
19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

PE = Mgh

At velocity v, the kinetic energy of the object is ...

KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

PE = KE

Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

2gh = v^2

This can be solved for height:

h = v^2/(2g) . . . . [eq1]

or for velocity:

v = √(2gh) . . . . [eq2]

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

t = 2(v^2/(2g))/v = v/g . . . . [eq3]

t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

g = 2h/t^2 . . . . [eq6]

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

2(9/49 s) ≈ 0.37 s

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

<em>Additional comment</em>

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

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dmitriy555 [2]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2> $\huge\boxed{\text{V = 9.5 m/s}}$ </h2><h2>_____________________________________</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

${\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)$

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

$W\ =\ \frac{1}{2}\:6\:x\:30$

W = 90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

${\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2$

$\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}$