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3 years ago

CHAPTER 6: KINETICS OF A PARTICLE

Physics

1 answer:

vekshin13 years ago

8 0

Explanation:

Work done by winch = kinetic energy of car

∫ T ds = ½ mv²

∫ 225s ds = ½ mv²

225/2 s² = ½ mv²

225 s² = mv²

v = 15s / √m

Given s = 10 m and m = 2500 kg:

v = 15 (10) / √2500

v = 3 m/s

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Work is done on a locked door that remains closed while you try to pull it open. True False.

dsp73

Answer:False

Explanation:

Work is being done on a body when it causes displacement of body on the application of force

$Work\ done=Force\times displacement$

When we pull the door by a force it causes zero displacements of the door. So we can say that work done on it is zero.

Thus the above-given statement is false

8 0

3 years ago

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

3 years ago

Calculate the total displacement of a mouse walking along a ruler, if it begins at the x=5cm, and then does the following: It wa

Lana71 [14]

To begin, the mouse walks from 5 to 12 cm, for a displacement of 7 cm. Next, it walks 8 cm in the opposite direction, for a total displacement of (7 + [-8]) or (-1) cm. This leaves the mouse on 4 cm, and then it walks from there to the 7cm location, for a displacement of 7-4 or +3 cm. Adding 3cm to -1cm gives a final displacement of +2cm.

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3 years ago

A life preserver is thrown from an helicopter straight down to a person in distress. The initial velocity of the life preserver

Leno4ka [110]

Answer:29.627 m

Explanation:

Given

Initial velocity of life preserver(u) is 1.6 m/s

it takes 2.3 s to reach the water

using equation of motion

v=u+at

$v=1.6+9.81\times 2.3$

v=24.163 m/s

Let s be the height of life preserver

$v^2-u^2=2gs$

$24.163^2-1.6^2=2\times 9.81\times s$

$s=\frac{581.29}{2\times 9.81}$

s=29.627 m

6 0

3 years ago

Suppose a current flows through a copper wire. Which two things occur?

Bingel [31]

Answer:

Explanation:

as the copper wire is very dangerous so now if these two thing happens then it would easily help the current flows through it so it might be a little bit easy for the current to flow through it

4 0

2 years ago