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Monica [59]
3 years ago
11

A ball is dropped from 215m. How long will it take to reach the ground? Use 9.8

Physics
1 answer:
valkas [14]3 years ago
7 0

Answer:

The time is  t = 6.62 \  s

Explanation:

From the question we are told that

   The height of the platform where the ball was dropped from is s  =  215 m

   The acceleration due to gravity is  g  =  9.8 \  m/s^2

Generally from kinematic equations

     s =  ut + \frac{1}{2} gt^2

Here u is the initial velocity of the ball and the value is  u = 0 m/s given that the ball was at rest before it was dropped

So

     21 5 =  0 * t + \frac{1}{2} *  9.8t^2

=>  21 5 =  4.9t^2

=>   t^2  = \frac{215}{4.9}

=>  t = 6.62 \  s

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Explanation:

v = u + a t

12 = 0 + 2a

a= 12/2

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v^2 - u ^2 = 2 as

12 ^2 - 0 ^2 = 2 * 6 * s

144 = 12 s

s = 144/12

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8 0
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For the element 35 17 C1, give the number of... 1. protons 2. neutrons 3. electrons​
faltersainse [42]

Answer:

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Explanation:

3 0
3 years ago
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2 years ago
Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

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