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aleksklad [387]

3 years ago

11

If the distance between two masses is tripled, the gravitational force between changes by a factor of

1 answer:

Kruka [31]3 years ago

6 0

Changes by a factor of 7n.

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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, $h=1.8\ m$

<u>Using the equation of motion:</u>

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

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 :

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

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

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

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

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, $f=105\ N$

<u>Now we find the force along the slide due to the body weight:</u>

$F=m_J.g.sin\theta$

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

$F=176.4\ N$

<em><u>Hence the net force along the slide:</u></em>

$F_R=71.4\ N$

<em>Now the acceleration of John:</em>

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

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

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

<u>Now the new velocity:</u>

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

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

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

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

A ball rolls off a desk at a speed of 3 m/s and lands .40 seconds later. How far from the base of the desk does the ball land?

Salsk061 [2.6K]

Is the velocity constant? Is there any friction?

3 meters per second

then after 40 seconds it must 3*40 = 120 meters

120 meters or 0.12 km if you will

7 0

3 years ago

Hydrogen is the second most abundant gas in the atmosphere? True or false?

Alenkinab [10]

False as oxygen is the second most abundant and nitrogen is the most abundant at 78%.

7 0

3 years ago

f your risk-aversion coefficient is A = 4 and you believe that the entire 1926–2015 period is representative of future expected

siniylev [52]

Answer:

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

Explanation:

As the complete question is not given here ,the table of data is missing which is as attached herewith.

From the maximized equation of the utility function it is evident that

$Weight=\frac{E_M-r_f}{A\sigma_M^2}$

For the equity, here as

$Weight$ is percentage of the equity which is to be calculated
${E_M-r_f}$ is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
$A$ is the risk aversion factor which is given as 4.
$\sigma_M$ is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59

By substituting values.

$Weight=\frac{E_M-r_f}{A\sigma_M^2}\\Weight=\frac{8.30\%}{4(20.59\%)^2}\\Weight=0.4894 =48.94\%$

So the weight of equity is 48.94%.

Now the weight of T bills is given as

$Weight_{T-Bills}=1-Weight_{equity}\\Weight_{T-Bills}=1-0.4894\\Weight_{T-Bills}=0.5105=51.05\%\\$

So the weight of T-bills is 51.05%.

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

7 0

2 years ago

Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball

Nimfa-mama [501]

Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

b) in this system the equations of motion are

y = v₀ t + ½ g t²

where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth

e) y = 0 + ½ g t²

t = √ (2y / g)

t = √(2 125 / 9.8)

t = 5.05 s

6 0

2 years ago