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ivanzaharov [21]

3 years ago

11

On a sunny day, a student poured a cup of water on the sidewalk to make a puddle. When he returned later, the puddle was gone. T

he student concluded that the puddle was gone because the water soaked into the sidewalk. 1. What would you say to this student about this conclusion? 2. What could you do to demonstrate your idea about where the water went? Text EntryMedia Copy and paste or type your submission right here.

1 answer:

ivanzaharov [21]3 years ago

8 0

Answer:

The students conclusion is wrong.

Explanation:

The students conclusion is wrong because liquids can not soak into concrete. The water evaporated because it was a hot sunny day. The water turned into a gas called water vapor.

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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

6 0

3 years ago

When dogs see police dogs do they think "oh no its the cops"?

ira [324]

No they say "Watch out it's the fuzz"

4 0

2 years ago

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If the sprinter accelerates at that rate for a distance of 15 m, and then maintains the velocity he has at that point for the re

ahrayia [7]

Answer:

The time for the entire race is 11.39 sec.

Explanation:

Given that,

Distance = 15 m

Remainder distance = 100 m

Suppose A sprinter begins a race with an acceleration of 3.4 m/s².

We need to calculate the time

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value in the equation

$15=0+\dfrac{1}{2}\times3.4\times t^2$

$t^2=\dfrac{30}{3.4}$

$t=2.97\ sec$

We need to calculate the final velocity of sprinter

Using equation of motion again

$v=u+at$

Put the value into the formula

$v=0+3.4\times2.97$

$v=10.09\ m/s$

We need to calculate the distance covers by sprinter

$d=100-15=85\ m$

The sprinter need to covers only 85 m.

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t'=\dfrac{85}{10.09}$

$t'=8.42\ sec$

We need to calculate the time for the entire race

$t''=t+t'$

Put the value into the formula

$t''=2.97+8.42$

$t''=11.39\ sec$

Hence, The time for the entire race is 11.39 sec.

4 0

3 years ago

A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the

liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0

2 years ago

When forces acting on an object

Ksju [112]

Answer:

it will not change

Explanation:

if the forces are of equal force then there would be no movement because one force was not stronger to move it

6 0

2 years ago

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