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3 years ago

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In physics, it is established that the acceleration due to gravity, g (in meters divided by sec squared) in meters/sec², at a

height h meters above sea level is given by the formula below. $g(h) = \frac{3.99 x 10^{14}}{(6.374 x 10^6 + h)^2}$ where 6.374 x 10⁶⁶ is the radius of Earth in meters.
A certain building is 458 meters tall. What is the acceleration due to gravity at the top at the top of the building? g(h) = _______ m/sec².

1 answer:

kompoz [17]3 years ago

5 0

Answer:

$g(h)=9.81943\ m/s^2$

Explanation:

The given function is

$g(h)=\dfrac{3.99\times 10^{14}}{(6.374\times 10^6+h)^2}$

Now h = the height from the surface of the Earth

Here the building is 458 m tall

$g(458)=\dfrac{3.99\times 10^{14}}{(6.374\times 10^6+458)^2}$

$\Rightarrow g(458)=9.81943\ m/s^2$

So,

$g(h)=9.81943\ m/s^2$

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A car is travelling at a constant speed of 26.5 m/s. Its tires have a radius of 72 cm. If the car slows down at a constant rate

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Answer:

Magnitude of angular acceleration = -3.95 rad/s²

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$a=\frac{11.7-26.5}{5.2}=-2.85m/s^2$

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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

$\Delta v =(v_J-v_J_n)$

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

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3 years ago