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3 years ago

What statement best describes the kind of change that could have taken place

1 answer:

5 0

In order to answer the question, we need to know what was there before
AND what was there after. Is there a reason why you're concealing that
information ?

While you're at it, you might pass along the answer choices too.

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What would happen if all the toilets were flushed at once?

Sliva [168]

Answer:

nothing would happen

Explanation:

5 0

3 years ago

Read 2 more answers

A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed

Tresset [83]

For this case we first think that the skateboard and the child are one body.
We have then:
1 = jug
2 = skateboard + boy
By conservation of the linear amount of movement:
M1V1i + M2V2i = M1V1f + M2V2f
Initial rest:
v1i = v2i = 0
0 = M1V1f + M2V2f
Substituting values
0 = (7.8) (3.2) + (M2) (- 0.65)
0 = 24.96 + M2 (-0.65)
-24.96 = (-0.65) M2
M2 = (-24.96) / (- 0.65) = 38.4 kg
Then, the child's mass is:
M2 = Mskateboard + Mb
Clearing:
Mb = M2-Mskateboard
Mb = 38.4 - 1.9
Mb = 36.5 Kg
answer:
the boy's mass is 36.5 Kg

4 0

3 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

6 0

3 years ago

The heat flux that is applied to the left face of a plane wall is q" = 20 w/m2. the wall is of thickness l = 10 mm and of therma

inessss [21]

Of course steady state condition occurs in almost any system but time it will occurs varies among system. for this kind of system, conduction, steady state conduction occurs when the temperature change from one point to the point is already constant. steady state is not achieved immediately because the heat travels and material will not be heated at the same way at the starting point.

5 0

3 years ago

Read 2 more answers

Why are household wall sockets alternating current (AC) instead of direct current (DC)? Select all that apply.

Bond [772]

Answer:

It is easier to scale the voltage of AC from high to low and low to high than with DC

Explanation:

typically power is used far away from the place where it's generated so to ensure that transmission losses( copper losses) are minimized voltage has to be stepped up during transmission..but due to the fact that most house hold equipment requires low voltage levels it has to be stepped down once it reaches a household/ domestic load...it's easier to do this for Ac than for DC.

8 0

2 years ago