How many additional valence electrons does fluorine need to have a full

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.

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

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

$\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}$

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

8 0

3 years ago

At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction betwe

Salsk061 [2.6K]

Here’s my work to your question. I used Newton’s Second Law and a kinematics equation to arrive at the answer.

5 0

2 years ago

What is gravity?what is one solar day?<br>

mafiozo [28]

Answer:

<h2>Gravity :</h2><h3>the force that attracts a body towards the centre of the earth, or towards any other physical body having mass.</h3>

<h2>Solar day</h2><h3>A solar day is the time it takes for the Earth to rotate about its axis so that the Sun appears in the same position in the sky.</h3><h2> or</h2><h3>It is the time between successive meridian transits of the sun at a particular place.</h3>

6 0

3 years ago

-

Sergeeva-Olga [200]

Answer:

(a) 32.5 kgm/s

(b) 32.5 Ns

(c) 10.8 N

Explanation:

The change in momentum can be calculated from the definition of linear momentum:

$\Delta p=\Delta (mv)= m \Delta v\\\\\Delta p= (5.00kg)(9.75m/s-3.25m/s)\\\\\Delta p=32.5kgm/s$

Then, the change in momentum of the body is of 32.5 kgm/s (a).

Now, from the impulse-momentum theorem, we know that the change in momentum of a body $\Delta p$ is equal to the impulse $I$ exerted to it. So, the impulse produced by the force equals 32.5 kgm/s (or 32.5 Ns) (b).